Chapter 27: Properties of Triangles & Solutions of Triangles (PDF)
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This chapter covers fundamental triangle properties, delving into relations between sides and angles through the sine rule, cosine rule, and projection formulas. It also discusses the area of triangles with varied given parameters. Applications like medians and Apollonius' theorem are highlighted.
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3.1 Relation between Sides and Angles. 60 76 Properties of Triangles and Solutions of Triangles A triangle has six components, three sides and three angles. The three angles of a ABC are denoted by letters A, B, C and the sides opposite to these angles by letters a, b and c respectively. Following...
3.1 Relation between Sides and Angles. 60 76 Properties of Triangles and Solutions of Triangles A triangle has six components, three sides and three angles. The three angles of a ABC are denoted by letters A, B, C and the sides opposite to these angles by letters a, b and c respectively. Following are some well known relations for a triangle (say ABC ) A a a b c, b c a, c a b | a b | c,| b c | a,| c a | b r r b r c r E3 A B C 180 o (or ) B r C ID U Generally, the relations involving the sides and angles of a triangle are cyclic in nature, e.g. to obtain the second similar relation to a b c , we simply replace a by b, b by c and c by a. So, to write all the relations, follow the cycles given. (1) The law of sines or sine rule : The sides of a triangle are proportional to the sines of the angles opposite to D YG a b c k (say) sin A sin B sin C them i.e. , A A More generally, if R be the radius of the circumcircle of the triangle ABC, a b c 2R sin A sin B sin C Note : The above rule may also be expressed as F B c E O D C B b a C sin A sin B sin C a b c U The sine rule is very useful tool to express sides of a triangle in terms of sines of angle and vice-versa in the a b c following manner K (Let) a K sin A, b K sin B, c K sin C. sin A sin B sin C sin A sin B sin C (Let) sin A a, sin B b, sin C c. a b c ST Similarly, Example: 1 If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side to the perimeter is (a) Solution: (a) 3 : (2 3 ) (b) 1 : 6 (c) 1 : (2 3) (d) 2:3 4 x x x 180 6 x 180 x 30 o sin 120 o sin 30 o sin 30 o a b c a : (a b c) (sin 120 o ) : (sin 120 o sin 30 o sin 30 o ) = 3 3 2 : 3 : 3 2. 2 2 [IIT Screening 2003] Properties of Triangles and Solutions of Triangles 77} In a triangle ABC, B Example: 2 3 and C 4 and D divides BC internally in the ratio 1 : 3. Then sin BAD sin CAD is equal to [UPSEAT 2003, 2001; IIT 1995] 1 (b) (c) 1 2 3 (d) 6 3 Let BAD , CAD A x In ADB , applying sine formulae, we get sin In ADC , applying sine formulae, we get, AD sin 3 3x sin ........(i) AD sin 4..........(ii) /3 B Dividing (i) by (ii), we get, sin BAD sin 1 . sin CAD sin 6 x D /4 C 3x 2 sin 2 3 6 3 sin 3 U ID 1 sin sin x AD sin 4 2 3 sin AD sin 3x 3 sin 2 3 E3 Solution: (c) 1 3 60 (a) In a ABC , A : B : C 3 : 5 : 4. Then [a b c 2 ] is equal to Example: 3 Solution: (c) (b) 2c D YG (a) 2b A : B : C 3 : 5 : 4 A B C 12 x 180 o x 15 (c) 3b [DCE 2001] (d) 3a o A 45 o , B 75 o , C 60 o a b c k (say) o o sin 45 sin 75 sin 60 o i.e., a 1 K, b 2 K, c In any triangle ABC if 2 cos B U Example: 4 3 1 2 2 (a) Right angled 3 K. Hence [a b c 2 ] 3 b. 2 a , then the triangle is c (b) Equilateral a k sin A sin A 2 cos B sin C sin A c k sin C sin C ST Solution: (c) 2 cos B (c) Isosceles (d) None of these sin(B C) sin(B C) sin A sin(180 o A) sin(B C) = sinA sin A sin(B C) sin A sin(B C ) 0 B C 0 B C Triangle is isosceles. 3.2 The Law of Cosines or Cosine Rule. In any triangle ABC, the square of any side is equal to the sum of the squares of the other two sides diminished by twice the product of these sides and the cosine of their included angle, that is for a triangle ABC, (1) a 2 b 2 c 2 2bc cos A cos A b 2 c 2 a2 2bc 78 Properties of Triangles and Solutions of Triangles (2) b 2 c 2 a 2 2ca cos B cos B c 2 a2 b 2 2ca (3) c 2 a 2 b 2 2ab cos C cos C a2 b 2 c 2 2ab a b c , sin B , sin C 2R 2R 2R abc abc abc , tan B We have by division, tan A , tan C 2 2 2 2 2 2 2 R(b c a ) R(c a b ) R(a b 2 c 2 ) 60 Combining with sin A E3 Where R, be the radius of the circum-circle of the triangle ABC. The smallest angle of the ABC , when a 7, b 4 3 and c 13 , is o o cos C (a) (d) None of these bc ca ab , then cos C 11 12 13 7 5 (b) bc ca ab = (Let) 11 12 13 5 7 (c) D YG Solution: (b) 45 [MP PET 2003] b 2 a 2 c 2 49 48 13 3 3 C 30 o. 2ab 2 27 4 3 2 3 In a ABC , if Example: 6 (c) o ID Solution: (a) (a) 30 (b) 15 Smallest angle is opposite to smaller side U Example: 5 b c 11 (d) 16 17......(i).......(ii) and a b 13 c a 12 17 36 [Karnataka CET 2003].......(iii) From (i) + (ii) + (iii), 2(a b c) 36 a b c 18 .......(iv) Now subtract (i), (ii) and (iii) from (iv), a 7, b 6, c 5. Now cos C (7 )2 (6 )2 (5 )2 49 2 36 2 25 2 60 2 5 a2 b 2 c 2 = = . 2 2ab 2. 7 .6 84 2 7 84 ABC In a ABC , 2 ac sin is equal to 2 U Example: 7 (a) 2ac sin (b) c 2 a 2 b 2 Example: 8 Solution: (b) (c) b 2 c 2 a2 (d) c 2 a 2 b 2 ABC 2B c2 a2 b 2 2ac sin 2ac cos B = 2ac c2 a2 b 2. 2 2 2ca ST Solution: (b) a2 b 2 c 2 [IIT Screening 2000] In ambiguous case if a, b and A are given and if there are two possible values of third side, are c1 and c 2 , then [UPSEAT 1999] (a) c1 c2 2 (a 2 b 2 sin 2 A) (b) c1 c 2 2 (a 2 b 2 sin 2 A) (c) c1 c2 4 (a 2 b 2 sin 2 A) (d) c1 c 2 3 (a 2 b 2 sin 2 A) cos A b 2 c 2 a2 or c 2 (2b cos A)c (b 2 a2 ) 0 2bc Which is quadratic equation in c. Let there be two roots, c1 and c 2 of above quadratic equation then c1 c2 2b cos A and c1c 2 b 2 a 2 c1 c2 [(c1 c2 )2 4 c1c2 ] = [(2b cos A)2 4 (b 2 a 2 ) = [4 a 2 4 b 2 (1 cos 2 A)] = 2 (a 2 b 2 sin 2 A). Properties of Triangles and Solutions of Triangles 79} sin A sin( A B) , then a 2 , b 2 , c 2 are in sin C sin(B C) In a ABC , if (a) A.P. (Using sine formula) ab cos C ac cos B ac cos B bc cos A ab cos C bc cos A 2ac cos B a2 b 2 c 2 b 2 c 2 a2 c 2 a2 b 2 c 2 a2 b 2 c 2 a2 b 2 b 2 a 2 , b 2 , c 2 are in A.P. 2 2 2 1 In a triangle ABC, AD is altitude from A. Given b > c, C 23 o and AD (a) Solution: (c) (d) None of these 67 o cos B (b) 44 o (c) a c b a (b c ) 2ac 2ac 2 2 2 2 abc Now, AD 2 ; b c2 2 2 cos B Also AD b sin 23 ; cos B a2 a c sin 23 o 2c abc , then B equal to b c2 2 113 o [IIT 1994] (d) None of these sin(B 23 o ) 1 o sin 23 o sin 23 cos B 2 U o By sine formulae abc AD 2ac 60 Example: 10 (c) H.P. a a cos B b cos A c b cos C c cos B E3 Solution: (a) (b) G.P. sin A sin A cos B cos A sin B sin C sin B cos C cos B sin C [Karnataka CET 1999] ID Example: 9 a sin(B 23 o ) ; c sin 23 o D YG sin(23 o B) 1 sin(90 o ) ; therefore 23 o B 90 o or B 113 o. 3.3 Projection Formulae. In any triangle ABC, b cos C c cos B k sin B cos C k sin C cos B (from sine rule) A = k[sin( B C)] k sin( A) k sin A a Similarly, we can deduct other projection formulae from sine rule. (i) a b cos c c cos B (ii) b c cos A a cos C c (iii) c a cos B b cos A i.e., any side of a triangle is equal to the sum of the projection of other two sides on it. B U In a ABC , cos C cos A cos B is equal to ca b ST Example: 11 (a) Solution: (b) 1 b (c) 1 c (d) ca b 1 ac (b cos C c cos B) (b cos A a cos B) = (Using projection formulae) =. b b(c a) b(c a) If k be the perimeter of the ABC , then b cos 2 (a) k Solution: (c) (b) b cos 2 (b) 2k a C [EAMCET 2001] cos C cos A cos B (b cos C b cos A) (c cos B a cos B) = ca b b(c a) = Example: 12 1 a b C B c cos 2 is equal to 2 2 (c) k 2 (d) None of these C B b c abc k b c 1 . c cos 2 (1 cos C ) (1 cos B) = (b cos C c cos B) = 2 2 2 2 2 2 2 2 2 80 Properties of Triangles and Solutions of Triangles Example: 13 (a b c)(cos A cos B cos C) (a) Solution: (d) a sin 2 A 2 (b) A 2 a cos 2 (c) 2 a sin 2 A 2 2 a cos 2 (d) A 2 (a b c)(cos A cos B cos C) 9 terms. = a cos A (b cos C c cos B) = a cos A b cos B c cos C (a b c) = a(1 cos A) 2 a cos 2 (3 ) 3.4 Theorem of the Medians: (Apollonius Theorem) A E3 In every triangle the sum of the squares of any two sides is equal to twice the square on half the third side together with twice the square on the median that bisects the third side. 60 (6 ) For any triangle ABC, b 2 c 2 2(h2 m 2 ) 2{m 2 (a / 2)2 } by use of cosine rule. b c m B h D C h ID If be right angled, the mid point of hypotenuse is equidistant from the three vertices so that DA DB DC A. 2 b 2 c 2 a2 which is pythagoras theorem. This theorem is very useful for solving problems of height and distance. If AD, BE and CF are the medians of a ABC then ( AD 2 BE 2 CF 2 ) : (BC 2 CA 2 AB 2 ) is equal to U Example: 14 (a) 4 : 2 (c) 3 : 4 We have, AB 2 AC 2 2( AD 2 BD 2 ) c b a AD 2 2 4 2 2 a2 c 2 b 2 BE 2 2 4 a2 b 2 c 2 a2 b 2 c 2 CF 2 2 4......(ii) and (d) 2 : 3 2 D YG Solution: (c) (b) 3 : 2......(i).........(iii) a2 b 2 c 2 AD 2 BE 2 CF 2 4 Adding (i), (ii) and (iii) we get, ( AD 2 BE 2 CF 2 ) : (a 2 b 2 c 2 ) 3 : 4. Example: 15 AD is a median of the ABC , if AE and AF are medians of the triangles ABD and ADC respectively and AD m1 , AE m2 , a2 is equal to 8 U AF m3 , then (a) In ABC , AD 2 m12 ST Solution: (a) m 22 m 32 2m12 (b) m12 m 22 2m 32 a AD 2 b 2 2 2 2 In ADC , AF m 3 2 4 m 22 m 32 2m12 m 22 m 32 m12 (d) None of these c 2 b 2 a2 2 4 a c 2 AD 2 2 2 2 In ABD , AE m 2 2 4 m 22 m 32 AD 2 (c) 2 A 2 b 2 c 2 a2 a2 a2 = m12 m12 2 8 4 8 a2 a2 m 22 m 32 2m12. 8 8 B E D F C Properties of Triangles and Solutions of Triangles 81} 3.5 Napier's Analogy (Law of Tangents) For any triangle ABC, A B C b c (2) tan cot 2 2 b c If tan BC A x cot , then x equal to 2 2 ca ca (a) (b) ab ab We know, tan Example: 17 If in a ABC a 6, b 3 and cos( A B) (b) 4 AB ab C 1 C cot cot 2 2 3 2 ab (c) ar(ABC ) 9 [Roorkee 1997] (d) None of these AB C 1 1 cot 2 2 9 2 AB C 1 1 tan 2 1 cot 2 2 9 2 5 tan 2 6 3 6 2 2 3 2 5. and b : c 2 : 3. If tan U In a ABC , A B 60 o tan 3 , 0 , then 5 2 (b) C 60 o (c) B 60 o (d) C 60 o CB 1 3 C B cb A cot 30 o tan cot tan 2 5 5 2 2 cb ST (a) 2 C 1 1 ar (ABC ) ab .6.3 9 1 C 2 2 2 2 Also, sin A Solution:(b,c) 2 5 1 tan 4 cos( A B) Example: 18 A sin 1 4 , then 5 U tan C B 2 and C B 180 o 60 o 120 o i.e., B 60 o , C 60 o . Example: 19 In a ABC , a 2b and | A B | (a) Solution: (b) 4 (b) 3. The measure of C is 3 Clearly, A B ( a b) Now tan 1 C AB ab C cot tan 30 o cot 2 2 ab 3 2 3 cot C C or C . 2 6 2 3 (c) 6 [MP PET 1992, 2002] (d) None of these D YG Solution:(b,c) C bc bc (c) BC bc bc A cot x. bc 2 2 bc Solution: (c) (a) E3 Example: 16 1 1 cos ( A B) sin ( A B) a b 2 2 , . 1 1 c sin C cos C 2 2 ID Note ab : Mollweide's formula: For any triangle, c B C A c a (3) tan cot 2 2 c a 60 C A B ab (1) tan cot 2 2 ab (d) None of these 82 Properties of Triangles and Solutions of Triangles 3.6 Area of Triangle. Let three angles of ABC are denoted by A, B, C and the sides opposite to these angles by letters a, b, c respectively. 60 (1) When two sides and the included angle be given: The area of triangle ABC is given by, 1 1 1 1 bc sin A ca sin B ab sin C i.e., (Product of two sides) × sine of included angle 2 2 2 2 (2) When three sides are given: Area of ABC s(s a)(s b)(s c) where semiperimeter of triangle s a b c 2 E3 A (3) When three sides and the circum-radius be given: Area of triangle where R be the circum-radius of the triangle. Example: 20 In a ABC if a 2 x , b 2 y and C 120 o , then the area of the triangle is (b) a C xy 3 (c) 3 xy [MP PET 1986, 2002] (d) 2 xy 1 1 ab sin C .2 x.2 y sin 120 o 3 xy. 2 2 Example: 21 In a ABC , D YG Solution: (b) cos A cos B cos C and a 2, then the area of a triangle is a b c (a) 1 (b) 2 (c) 3 2 [MP PET 2000; IIT 1993] (d) 3 By sine rule, tan A tan B tan C ; Triangle is equilateral. 1 1 3 3..a.a. sin 60 o . 2. 2. 2 2 2 U Hence, Example: 22 b 1 2 sin B sin C 1 2 sin A sin C 1 2 sin A sin B a b c 2 sin( B C) 2 sin( A C) 2 sin( A B) (a) xy Solution: (d) B U c ID (4) When two angles and included side be given : abc , 4R In a triangle ABC, a, b, A are given and c1, c2 are two values of third side c. The sum of the areas of triangles with sides a, b, ST c1 and a, b, c2 is (a) Solution: (b) 1 2 a sin 2 A 2 (b) 1 2 b sin 2 A 2 (c) b 2 sin 2 A (d) a 2 sin 2 A Let the triangles be 1 ABC 1 and 2 ABC 2 A, b, a are given and c has two values. Hence we apply cosine formulae cos A b 2 c 2 a2 or c 2 2bc cos A b 2 a 2 0. 2bc Above is quadratic in c If c1, c2 be the two values of c, then c1 c 2 2b cos A, c1c2 b 2 a 2 1 1 1 ab sin C1 , 2 ab sin C 2 2 2 1 2 1 ab(sin C1 sin C2 ) 1 abk (2b cos A) b 2ak cos A b 2 sin A cos A = 1 b 2 sin 2 A. 2 Example: 23 2 If stands for the area of a triangle ABC, then a 2 sin 2 B b 2 sin 2 A 2 [WB JEE 1988] Properties of Triangles and Solutions of Triangles 83} (a) 3 (b) 2 (c) 4 (d) 4 Use sine rule, ab sin C Solution: (c) L.H.S. = k 2 (sin 2 A.2 sin B cos B sin 2 B.2 sin A cos A) = k 2 [2 sin A. sin B. sin( A B)] = 2ab sin C 4 . (a) 9 3 2 cm 2. Then a is , b c 3 3 and ar (ABC ) 2 3 (b) 9cm. 6 3 cm. 60 In a ABC , A Example: 24 (c) 18 cm. (d) None of these 1 9 3 2 9 3 1 3 bc sin or..bc = bc =18 2 2 2 2 3 2 Also, cos 2 b 2 c 2 a 2 1 (b c)2 2bc a 2 or (b c)2 3bc a 2 0 or 27 54 a2 a 9. 3 2bc 2 2bc E3 Solution: (b) If p1 , p2 , p3 are altitudes of a triangle ABC from the vertices A, B, C and , the area of the triangle, then p12 p 22 p 32 is Example: 25 equal to We have (b) a2 b 2 c 2 4 2 (c) a2 b 2 c 2 2 ID Solution: (b) ab c (d) None of these 2 2 2 1 1 1 , p2 , p3 ap1 , bp 2 , cp 3 p1 a b c 2 2 2 1 1 1 a2 b 2 c 2. p12 p 22 p32 4 2 D YG 3.7 Half Angle Formulae. U (a) If 2s shows the perimeter of a triangle ABC then, i.e., 2 s a b c, then (1) Formulae for sin A B C : , sin , sin 2 2 2 B (s a)(s c) A (s b)(s c) (ii) sin 2 ca 2 bc A B C (2) Formulae for cos , cos , cos : 2 2 2 U (i) sin A 2 s(s a) bc ST (i) cos (3) Formulae for tan (i) tan sin A 2 sin s(s b) ca (iii) cos C 2 (ii) tan B (s c)(s a) 2 s(s b) (iii) tan C 2 s(s c) ab (s a)(s b ) s(s c) 2 A A (s b)(s c) s(s a) 2 = s(s a)(s b)(s c) cos 2. bc 2 2 bc bc bc 2 2 , sin C ca ab A (s b)(s c) B (s c)(s a) C (s a)(s b) , tan tan , tan 2 2 2 Similarly sin B B 2 C (s a)(s b) 2 ab A B C , tan , tan : 2 2 2 A (s b)(s c) 2 s(s a) Note : (ii) cos (iii) sin 84 Properties of Triangles and Solutions of Triangles Important Tips 1/ 2 tan (s b)(s c) (s c)(s a) A B = tan. 2 2 s(s b) s(s a) tan A B tan = 2 2 1/ 2 s c s b s a c s(s c) c C = s (s a)(s b) s cot 2 s ( s a )( s b ) s c s b s a = s s a s b 60 Another form: sc A B s cot cot 2 2 sc s c.(s c) c (s c) 1/ 2 {s(s a)(s b)(s c)} s tan A B tan tan A B 2 2 c cot C. cot cot A B 2 2 sc 2 tan tan 2 2 (a) Solution: (b) cot A B C , cot , cot are in A.P., then 2 2 2 A B cot 4 2 2 (b) cot (b) 2 In a triangle ABC, tan (c) tan Example: 29 Solution: (b) tan (c) 3 (d) 2 B C 2a. cot 2. 2 2 a [EAMCET 1994] (b) cos A, cos B, cos C are in A.P. sin A, sin B, sin C are in A.P. (d) Both (a) and (c) A C (s b). tan 2 2 s 5 2 sb. 2 s 3b a b c 3b a c 2b. 6 5 s If the sides of triangle a, b, c be in A.P. then tan (a) B C tan 0 2 2 [Roorkee 1986; MP PET 1990, 97, 98] a, b, c are in A.P. ST Solution: (d) (d) cot A 5 C 2 and tan , then 2 6 2 5 U (a) B C cot 1 2 2 B C s(s b) s(s c) s. cot . 2 2 (s a)(s c) (s a)(s b) s a Given 3a b c a b c 4 a cot Example: 28 cot B C cot is 2 2 D YG cot (c) [MP PET 2003] A B C , cot and cot are in A.P. with common difference zero. Now option (b) 2 2 2 In a ABC , if 3a b c ,then the value of cot (a) 1 Solution: (b) A C cot 3 2 2 Trick: Take A B C 60 o , then cot satisfies. Example: 27 ID If in any ABC ; cot U Example: 26 E3 A B ab tan (s c) 2 2 s 2 A cot 3 2 (b) 2 B cot 3 2 A C (s b)(s c) tan 2 2 s(s a) Hence, tan 2b B cot 3b 2 A C 2 B tan cot. 2 2 3 2 (c) 2 C cot 3 2 (s a)(s b) b B 2b B cot = cot = s 2 2s 2 s(s c) a, b, c in A.P. a c 2b 2s 3b = A C tan equal to 2 2 [Roorkee 1993] (d) None of these Properties of Triangles and Solutions of Triangles 85} A B B A In ABC , cot cot a sin 2 b sin 2 equal to 2 2 2 2 Example: 30 (a) (b) c cot C cot C [Roorkee 1988] (c) cot C 2 C 2 (d) c cot s(s c) a ab (s a)(s c) ac b (s b)(s c) bc (s b)(s c) bc (s a)(s c) ac s(s c) sb s a ab ab ab s b s a E3 Solution: (d) 60 C B A cos sin sin A B C 2 B 2 A 2 B 2 A 2 2 2 b sin b sin b cot cot a sin a sin cos a A B A B 2 2 2 2 2 2 2 sin sin sin sin 2 2 2 2 s(s c) C sasb 2s a b s(s c) c cot. s(s c) c (s a)(s b) 2 (s a)(s b) (s a)(s b) ID Trick : Such type of unconditional problems can be checked by putting the particular values for a 1 , b 3 , c 2 and A 30 o , B 60 o , C 90 o , Here expression is equal to 2 which is given by (d). U 3.8 Circle Connected with Triangle. D YG (1) Circumcircle of a triangle and its radius (i) Circumcircle : The circle which passes through the angular points of a triangle is called its circumcircle. The centre of this circle is the point of intersection of perpendicular bisectors of the sides and is called the circumcentre. Its radius is always denoted by R. The circumcentre may lie within, outside or upon one of the sides of the triangle. (ii) Circum-radius : The circum-radius of a ABC is given by A a b abc c (b) R R [ area of ABC ] F 2 sin A 2 sin B 2 sin C 4 E O (2) Inscribed circle or incircle of a triangle and its radius B C D (i) In-circle or inscribed circle : The circle which can be inscribed within a triangle so as to touch each of its sides is called its inscribed circle or in circle. The centre of this circle is the point of intersection of the bisectors of the angles of the triangle. The radius of this circle is always denoted by r and is equal to the length of the perpendicular from its centre to any one of the sides of triangle. A (ii) In-radius : The radius r of the inscribed circle of a triangle ABC is given by U (a) s ST (a) r (b) r 4 R sin B A C , r (s b) tan , r (s c) tan 2 2 2 B C A C B A a sin sin b sin sin c sin sin 2 2 ,r 2 2 ,r 2 2 (d) r A B C cos cos cos 2 2 2 A/2 A B C sin sin 2 2 2 E F (c) r (s a) tan (e) cos A cos B cos C 1 r R (3) Escribed circles of a triangle and their radii r r I r C/2 B B 2 D C 86 Properties of Triangles and Solutions of Triangles (i) Escribed circle : The circle which touches the side BC and two sides AB and AC produced of a triangle ABC is called the escribed circle opposite to the angle A. Its radius is denoted by r1. Similarly, r2 and r3 denote the radii of the escribed circles opposite to the angles B and C respectively. The centres of the escribed circles are called the ex-centres. The centre of the escribed circle opposite to the angle A is the point of intersection of the external bisectors of angles B and C. The internal bisectors of angle A also passes through the same point. The centre is generally denoted by I1 A 60 (ii) Radii of ex-circles : In any ABC , we have A B C (b) r1 s tan , r2 s tan , r3 s tan , r2 , r3 sa 2 s b 2 sc 2 B C C A A B a cos cos b cos cos c cos cos 2 2 ,r 2 2 ,r 2 2 (c) r1 2 3 A B C cos cos cos 2 2 2 1 1 1 1 (d) r1 r2 r3 r 4 R (e) r1 r2 r3 r A 2 1 1 1 1 a2 b 2 c 2 r 2 r12 r22 r32 2 (h) r1r2 r2r3 r3r1 s 2 A 2 D B C F r1 E r1 I1 1 1 1 1 bc ca ab 2 Rr (i) 2 R 2 sin A. sin B. sin C 4 Rr cos A B C. cos. cos 2 2 2 A B C A B C A B C cos cos ; r2 cos. sin. cos ; r3 4 R cos cos sin 2 2 2 2 2 2 2 2 2 D YG (j) r1 4 R sin (g) U (f) ID E3 (a) r1 U (4) Centroid (G) : Common point of intersection of medians of a triangle. Divides every median in the ratio 2:1. Always lies inside the triangle. A B G D C ST (5) Orthocentre of a triangle : The point of intersection of perpendicular drawn from the vertices on the opposite sides of a triangle is called its orthocentre. 3.9 Pedal Triangle. Let the perpendiculars AD, BE and CF from the vertices A, B and C on the opposite sides BC, CA and AB of ABC respectively, meet at O. Then O is the orthocentre of the ABC. The triangle DEF is called the pedal triangle of the ABC. A F Othocentre of the triangle is the incentre of the pedal triangle. If O is the orthocentre and DEF the pedal triangle of the ABC , where AD, BE, CF are the perpendiculars drawn from A, B, C on the opposite sides BC, CA, AB respectively, then (i) OA 2 R cos A, OB 2 R cos B and OC 2 R cos C (ii) OD 2 R cos B cos C, OE 2 R cos C cos A and OF 2 R cos A cos B E O O B D C Properties of Triangles and Solutions of Triangles 87} (1) Sides and angles of a pedal triangle: The angles of pedal triangle DEF are: 180 2 A, 180 2 B, 180 2C and sides of pedal triangle are: A EF a cos A or R sin 2 A ; FD b cos B or R sin 2 B ; DE c cos C or R sin 2 C b cos B F If given ABC is obtuse, then angles are been represented by 2 A, 2 B, 2C 180 o and a cos B O O 1 2 R. sin 2 A. sin 2 B. sin 2C 2 Circum-radius of pedal triangle = 180 – 2A D C E3 B 1 (Product of the sides) × (sine of included angle) 2 Area of pedal triangle = 60 the sides are a cos A, b cos B, c cos C. (2) Area and circum-radius and in-radius of pedal triangle c cos B E EF R sin 2 A R . o 2 sin FDE 2 sin(180 2 A) 2 ID 1 2 R sin 2 A. sin 2 B. sin 2C area of DEF In-radius of pedal triangle = = 2 = 2 R cos A. cos B. cos C. semi - perimeter of DEF 2 R sin A. sin B. sin C Circum-centre, Centroid and Orthocentre are collinear. In any right angled triangle, the orthocentre coincides with the vertex containing the right angled. The mid-point of the hypotenuse of a right angled triangle is equidistant from the three vertices of the triangle. The mid-point of the hypotenuse of a right angled triangle is the circumcentre of the triangle. The length of the medians AD, BE, CF of ABC are given by D YG U Important Tips 1 1 1 2 1 2c 2 2a 2 b 2 2b 2 2c 2 a 2 c a 2 2ca. cos B b 2 c 2 2bc cos A , BE 2 2 2 2 1 1 CF 2 a 2 2b 2 c 2 a 2 b 2 2ab. cos c 2 2 AD A E F G The distance between the circumcentre O and centroid G of ABC is given by B U 1 1 OG OH R 1 8 cos A. cos B. cos C , Where H is the orthocentre of ABC. 3 3 ST The distance between the orthocentre H and centroid G of ABC is given by HG C 2 R 1 8 cos A. cos B. cos C. 3 The distance between the circumcentre O and the incentre I of ABC given by OI R 1 8 sin If I 1 is the centre of the escribed circle opposite to the angle B, then OI1 R 1 8 sin Similarly, OI2 R 1 8 cos D A B C. sin. sin 2 2 2 A B C. cos. cos 2 2 2 A B C A B C. sin. cos , OI3 R 1 8 cos. cos. sin 2 2 2 2 2 2 Circle circumscribing the pedal triangle of a given triangle bisects the sides of the given triangle and also the lines joining the vertices of the given triangle to the orthocentre of the given triangle. This circle is known as "Nine point circle". Circumcentre of the pedal triangle of a given triangle bisects the line joining the circum-centre of the triangle to the orthocentre. 3.10 Ex-central Triangle. 88 Properties of Triangles and Solutions of Triangles Let ABC be a triangle and I be the centre of incircle. Let I1 , I2 and I3 be the centres of the escribed circles which are opposite to A, B, C respectively then I1 I2 I3 is called the Ex-central triangle of ABC. 90o – C/2 I1 I2 I3 is a triangle, thus the triangle ABC is the pedal triangle of its B Area and circum-radius of the ex-central triangle 1 (Product of two sides) × (sine of included angles) 2 8 R 2 cos I2 I3 2 sin I 2 I1 I 3 4 R cos A 2 A 2 sin 90 o 2 2R. In a ABC r1 r2 r3 , then (a) Solution: (a) A B C. cos. cos 2 2 2 D YG Example: 31 ID B C 1 o A 4 R cos . 4 R cos sin 90 2 2 2 2 Circum-radius = 90o – A/2 U I1 E3 C B A ; I1 I2 4 R cos ; I2 I3 4 R cos 2 2 2 C 60 A B C 90 , 90 o ,90 o 2 2 2 o and sides are I1 I3 4 R cos 90o – B/2 I ex-central triangle I1 I2 I3. The angles of ex-central triangle I1 I2 I3 are Area of triangle = I2 I3 (b) a b c ab c [EAMCET 2003] (c) bac (d) a c b In a ABC , r1 r2 r3 1 1 1 sa sb sc (s a) (s b) (s c) a b c a b c. r1 r2 r3 U Which of the following pieces of data does NOT uniquely determine an acute-angled triangle ABC (R being the radius of the circum-circle) [IIT Screening 2002] ST Example: 32 (a) Solution: (d) (b) a, b, c a, sin A, sin B (d) a, sin B, R (d) a, sin A, R a R and b 2 R sin B. So two sides and two angles are known. So C is known. Therefore, two sides and included sin A angle is known. So, is uniquely known in case (a). If a, b, c are known the is uniquely known in case (b). b 2 R sin B, sin A a. So, sides a, b and angle A, B are known. So 2R C is known. Therefore two sides and included angle is known. So, is uniquely known in case (c). a R. So, only a side and an angle is known. So, is not uniquely known in case (d). sin A Example: 33 In a triangle ABC, let C 2. If r is the in radius and R is the circum-radius of the triangle, then 2(r R ) is equal to Properties of Triangles and Solutions of Triangles 89} [IIT Screening 2000] ab Solution: (a) (b) b c c 2R , sin C (c) ca (d) a b c c 2 R sin 90 o 2 R Also r (s c) tan C (s c) 2 [ tan 45 o 1] 2r 2s 2c a b c a b 2R , 2(r R) a b. If R is the radius of the circumcircle of the ABC and is its area, then Solution: (c) (b) Area of the triangle ABC () R Example: 35 abc abc 4 (c) (b) 2.5 unit Given, Sides are 3, 4, 5. (d) R abc [UPSEAT 2000] (c) 3.0 unit (d) 3.5 unit U 5 2.5 unit. 2 If x, y, z are perpendicular drawn from the vertices of triangle having sides a, b, and c, then the value of a2 b 2 c 2 2R (b) a2 b 2 c2 R2 D YG (a) Solution: (a) abc 4 If the length of the sides of a triangle are 3, 4 and 5 units, then R (the circum-radius) is Since 3 2 4 2 5 2. So, triangle is a right angle triangle. R Example: 36 R 1 a a abc bc sin A. From the sine formula, a 2 R sin A or sin A bc. or 2R 2 2R 4 R 2 abc. 4 (a) 2.0 unit Solution: (b) R E3 R (a) [Karnataka CET 2000] ID Example: 34 60 (a) (c) Let area of triangle be , then according to question, a2 b 2 c2 4R (d) bx cy az will be c a b 2(a 2 b 2 c 2 ) R 1 1 1 ax by cz 2 2 2 2 2 2 2 2 2 a2 b 2 c 2 bx cy az b 2 c 2 a 2 = 2(b c a ) 2(a b c ). abc =. c Example: 37 a b c a a b b c abc abc 2R 4R If r1 , r2 , r3 in a triangle be in H.P., then the sides are in (a) A.P. (c) H.P. (d) None of these 1 1 1 1 sb sa scsb ab b c sb sa sc sb U 1 1 1 1 r2 r1 r3 r2 (b) G.P. [EAMCET 1993] Solution: (a) b a cb 2b a c. ST Example: 38 If the sides of a triangle are 13,14 15, then the radius of its incircle is (a) Solution: (c) r 67 8 (b) 65 4 (c) 4 (d) 24 a b c 13 14 15 and s 21 s 2 2 21 8 7 6 84 r Example: 39 [EAMCET 1987] 84 4. 21 In an equilateral triangle the inradius and the circumradius are connected by (a) r 4R (b) r R 2 (c) r R 3 [EAMCET 1983] (d) None of these 90 Properties of Triangles and Solutions of Triangles r 4 R sin Solution: (b) A B C. sin. sin 2 2 2 For an equilateral triangle, A B C 60 o R 1 1 1 . 2 2 2 2 r 4 R sin 30 o. sin 30 o. sin 30 o 4 R... 60 3.11 Cyclic Quadrilateral. A quadrilateral PQRS is said to be cyclic quadrilateral if there exists a circle passing through all its four vertices P, Q, R and S. E3 Let a cyclic quadrilateral be such that PQ a, QR b, RS c and SP d. Then Q S 180 o and A C 180 o. Let 2 s a b c d and = Area of cyclic quadrilateral PQRS S 1 1 ab sin Q cd sin S 2 2 d ID = Area of PQR + Area of PRS = a b Q From cosine rule, PR 2 PQ 2 QR 2 2 PQ.QR cos Q = a2 b 2 2ab cos Q D YG R P U 1 1 1 = ab sin Q cd sin( Q) (ab cd) sin Q 2 2 2 1 = (ab cd) sin Q......(i) 2 In PQR and PRS , c......(ii) and PR 2 PS 2 RS 2 2 PS.RS cos S PR 2 d 2 c 2 2cd cos( Q) PR 2 d 2 c 2 2cd cos Q From (ii) and (iii) we have, = (s a)(s b)(s c)(s d ) Therefore, (1) Area of cyclic quadrilateral = U (2) Area of cyclic quadrilateral =........(iv) 1 (ab cd) sin Q 2 (s a)(s b)(s c)(s d ) , where 2 s a b c d a2 b 2 c 2 d 2 2(ab cd) ST (3) cos Q ........(iii) (4) Circumradius of cyclic quadrilateral : Circum circle of quadrilateral PQRS is also the circumcircle of PR PR(ab cd) PQR. Hence circumradius of cyclic quadrilateral PQRS = R = circumradius of PQR = 2 sin B 4 But PR (ac bd )(ad bc) (ab cd) Hence R 1 1 (ac bd )(ad bc)(ab cd) 4 4 (ac bd )(ad bc)(ab cd) (s a)(s b)(s c)(s d ) (5) Ptolemy's theorem : In a cyclic quadrilateral PQRS, the product of diagonals is equal to the sum of the products of the length of the opposite sides i.e., According to Ptolemy's theorem, for a cyclic quadrilateral PQRS A PR.QS PQ.RS RQ.PS. B C Example: 40 A cyclic quadrilateral ABCD of area 3 3 is inscribed in a unit circle. If one of its sides AB = 1 and the diagonal BD = 4 the lengths of the other sides are (a) 2, 1, 1 (b) 2, 1, 2 By sine formula in ABC , Now, AB x 1 3 x 2 y2 3 1 1 y2 3 y y2 2 2 xy 2 2y y y 2 0 (y 2)(y 1) 0 y 2 [ y 1 ] 2 ID (d) None of these 3 3 3 2R 2 sin A A sin A sin A 2 3 By cosine formula in ABD cos AD 2 Since A 60 o C 120 o In BDC , 3 p 2 q 2 2 pq cos 120 o 3 p 2 q 2 pq.........(i) 3 pq 3 3 3 4 3 3 2 3 2 4 3 D B U 3 4 q p C pq 1 4 (i) gives, 3 p 2 q 2 1 p 2 q 2 2 , y 3 3 1 1 3 3 ABD BCD .1.2 sin 60 o p.q. sin 120 o pq 4 2 2 2 4 D YG A x 3 3 Also area of quadrilateral ABCD 4 3 then [Roorkee 1995] (c) 3, 1, 2 E3 Solution: (a) 60 Properties of Triangles and Solutions of Triangles 91} [p, q 0] 1 p 2 2 2 p 4 2 p 2 1 0 ( p 2 1)2 0 p 2 1, p 1, q 1 p AB 1, AD 2, BC CD 1. The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60 o. If the third side is 3, the remaining fourth side is [UPSEAT 1994] (a) 2 (b) 3 (c) 4 (d) 5 U Example: 41 Solution: (a) Since ABCD is cyclic quadrilateral and ABC 60 o D ADC 180 60 120 Let AB 2, BC 5 and CD 3 o o ST o 120o A In ABC , AB 2 AC 2 2 AB.BC cos 60 o AC 2 or 4 25 2 2 5 1 AC 2 2 C 2 AC 2 19 ; In ABC AD 2 CD 2 2 AD.CD cos 60 o AC 2 5 60o 1 19 or AD 2 3 AD 10 0 or AD 2 5 AD 2 AD 10 0 2 or AD( AD 5) 2( AD 5) 0 or ( AD 2)( AD 5) 0. Therefore, fourth side is AD 2. or AD 2 9 2 AD. 3. Example: 42 B Two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60 o. If the area of the quadrilateral is 4 3 , then the remaining two sides are (a) 2, 3 Solution: (a) 3 (b) 3, 4 [Roorkee 1991] (c) 4, 5 (d) 5, 6 Let a PQ 2 , b QR 5 , RS c , SP d S 120o c d R P a 60o b 92 Properties of Triangles and Solutions of Triangles Given, Area of quadrilateral PQRS = 4 3 Area of ( PQR PRS ) = 4 3 5 3 3 cd cd 6 2 4........(i) 60 1 [2 5 sin 60 o c d sin 120 o ] 2 Now by cosine formula RS 2 SP 2 2 RS.SP. cos 120 o PR 2 PQ 2 QR 2 2 PQ.QR cos 60 o 1 1 c 2 d 2 2cd 4 25 2(2)(5 ) c 2 d 2 cd 19 c 2 d 2 19 6 13 ……..(ii) 2 2 E3 Solving (i) and (ii) we get c 2 and d 3. 3.12 Regular Polygon. A regular polygon is a polygon which has all its sides equal and all its angles equal. a π .cosec 2 n U R ID 2n 4 π 2n 4 (1) Each interior angle of a regular polygon of n sides is radians. right angles = n 2 n (2) The circle passing through all the vertices of a regular polygon is called its circumscribed circle. If a is the length of each side of a regular polygon of n sides, then the radius R of the circumscribed circle, is given by D YG (3) The circle which can be inscribed within the regular polygon so as to touch all its sides is called its inscribed circle. Again if a is the length of each side of a regular polygon of n sides, then the radius r of the inscribed circle is given by r a π .cot 2 n O R r A C /n F R B (4) The area of a regular polygon is given by n × area of triangle OAB 1 π n a 2 cot 4 n ST U π nr 2.tan n Example: 43 n 2 2π .R sin 2 n (in terms of side) (in terms of in-radius) (in terms of circum-radius) The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a, is (a) a cot n a (b) cot 2 2n (c) a cot 2n [AIEEE 2003] a (d) cot 4 2n Solution: (c) a a a a tan and sin r R cot cosec cot . 2 n n 2 n 2 r n 2 R 2n Example: 44 The area of the circle and the area of a regular polygon of n sides and its perimeter equal to that of the circle are in the ratio of (a) Solution: (a) tan : n n (b) cos : n n Let r be the radius of the circle and A1 be its area A1 r 2 (c) sin : n n (d) cot : n n Properties of Triangles and Solutions of Triangles 93} Since the perimeter of the circle is the same as the perimeter of a regular polygon of n sides 2r na , where 'a' is the length 2r n of one side of the regular polygon, a Let A2 be the area of the polygon, then n n A1 : A2 r 2 : r 2.. cot n cot n tan n co sec (b) cosec 9 We know that radius of the circumcircle is given by R . 9 9 cot (d) 9 tan [IIT 1994] 9 a cosec ; Here, a 2, n 9 2 n ID 2 If the number of sides of two regular polygons having the same perimeter be n and 2n respectively, their areas are in the ratio (b) 2 cos n 1 cos n (c) U 2 cos n cos 2n (a) cos n sin n (d) None of these Let s be the perimeter of both the polygons. Then the length of each side of the first polygon is D YG Solution: (b) n (c) 3 Required radius = 2 cosec cosec . Example: 46 : A regular polygon of nine sides, each of length 2 is inscribed in a circle. The radius of the circle is (a) Solution: (a) E3 Example: 45 = 1: 60 1 2 1 4 2r 2 = n. na. cot cot r 2.. cot 2 4 n 4 n n n n A2 s and that of second polygon is n s. 2n If A1 , A 2 denote their areas, then A1 2 1 s s n cot n and A2 4.(2n) 2n . cot 2n 2 2 cot 2 cos sin 2 cos sin 2 cos A n n n 2 n 2 n n. 1 A2 2 sin cos cos cot sin cos 1 cos 2 2 n n 2 n n 2 2 n n n U A1 A2 n 4 ST 3.13 Solutions of Triangles. Different formulae will be used in different cases and sometimes the same problem may be solved in different ways by different formulae. We should, therefore, look for that formula which will suit the problem best. (1) Solution of a right angled triangle (2) Solution of a triangle in general (1) Solution of a right angled triangle (i) When two sides are given: Let the triangle be right angled at C. Then we can determine the remaining elements as given in the following table Given Required 94 Properties of Triangles and Solutions of Triangles a, b tan A a a , B 90 o A , c sin A b a, c sin A a , b c cos A, B 90 o A c 60 (ii) When a side and an acute angle are given : In this case, we can determine the remaining elements as given in the following table Given Required a, A B 90 o A, b a cot A, c c, A B 90 o A, a c sin A, b c cos A ID E3 a sin A (2) Solution of a triangle in general (i) When three sides a, b and c are given in this case, the remaining elements are determined by using the following 2 , bc sin C B , 2 s(s b ) tan sin B 2 , ac D YG sin A U formulae, s(s a)(s b)(s c) , where 2s a b c = perimeter of triangle tan A , 2 s(s a) tan 2 ab C 2 s(s c) (ii) When two sides a, b and the included angle C are given : In this case, we use the following formulae A B ab C 1 a sin C AB C ; and c cot 90 o ab sin C ; tan sin A 2 2 2 2 ab 2 (iii) When one sides a and two angles A and B are given : In this case, we use the following formulae to determine U the remaining elements A B C 180 o C 180 o A B a sin B a sin C 1 ca sin B and c sin A sin A 2 ST b (iv) When two sides a, b and the angle A opposite to one side is given : In this case, we use the following formulae b..........(i) sin B sin A a C 180 o ( A B), c a sin C sin A Special Cases C Case I : When a is an acute angle (a) If a b sin A, there is no triangle. When a b sin A, from (i), sin B 1, which is impossible. a b B b sin A A A N X Properties of Triangles and Solutions of Triangles 95} From the following figure, If AC b, CAX A , then perpendicular CN b sin A. Now taking C as centre, If we draw an arc of radius a then it will never intersect the line AX and hence no triangle ABC can be constructed in this case. (b) If a b sin A , then only one triangle is possible which is right angled at C B. When a b sin A, then from sine rule. sin B 1, B 90 o Thus, in this case, only one triangle is possible which is right angled at B. b (c) If a b sin A, then three possibilities C will arise: a A A (i) a b b sinA N B a = b sinA A 90o A E3 b 60 from fig. It is clear that CB a b sin A In this case, from sine rule B X sin B sin A ; B A or B 180 o A X In this case, we get A B. ID But B 180 o A A B 180 o , which is not possible in a triangle. Hence, if b a b sin A then only one isosceles triangle ABC is possible in which A B. (ii) a b In the following figure, Let AC b, CAX A, and a b, also a b sin A. D YG U Now taking C as centre, if we draw an arc of radius a, it will intersect AX at one point B and hence only one ABC is constructed. Also this arc will intersect XA produced at B and A B C is also C formed but this is inadmissible (because CA B is an obtuse angle in this a a triangle) b Hence, if a b and a b sin A, then only one triangle is possible. A B A b sinA B N X (iii) b a (i.e., b a b sin A) In fig. let AC b, CAX A. Now taking C as centre, if we draw an arc of C2 radius a, then it will intersect AX at two points B1 and B 2. Hence if b a sin A, b C1 a a A then there are two triangles. U C A B2 b sinA N B1 X ST Case II : When A is an obtuse angle: In this case, there is only one triangle, if a b Case III: b c and B 90 o C b sinA a b A N A B X Again the circle with A as centre and b as radius will cut the line only in one point. So, only one triangle is possible. C B b c A 96 Properties of Triangles and Solutions of Triangles Case IV: b c and B 90 o The circle with A as centre and b as radius will not cut the line in any point. So, no triangle is possible. Alternative method: By applying cosine rule, we have cos B a2 c 2 b 2 2 ac a 2 (2c cos B)a (c 2 b 2 ) 0 a c cos B (c cos B)2 (c 2 b 2 ) 60 This is, sometimes called an ambiguous case. A c E3 B b a c cos B b 2 (c sin B)2 This equation leads to following cases: ID Case I : If b c sin B ,no such triangle is possible. Case II : Let b c sin B, there are further following case. (a) B is an obtuse angle cos B is negative. There exists no such triangle. U (b) B is an acute angle cos B is positive. There exists only one such triangle. Case III : Let b c sin B. There are further following cases : D YG (a) B is an acute angle cosB is positive. In this case two values of a will exists if and only if c cos B b 2 (c sin B)2 or c b Two such triangle is possible. If c b, only one such triangle is possible. (b) B is an obtuse angle cosB is negative. In this case triangle will exist if and only if b 2 (c sin B)2 | c cos B | b c. So, in this case only one such triangle is possible. If b c there exists no such triangle. If one side a and two angles B and C are given, then A 180 o (B C) and b U Note : a sin B a sin C ,c sin A sin A ST If the three angles A, B, C are given. We can only find the ratios of the sides a, b, c by using sine rule (since there are infinite similar triangle possible). Important Tips A triangle which does not contain a right angle is called an oblique triangle. If triangle is an acute angled triangle means every angle is less than 90o. If triangle is an obtuse angled triangle means one of its angle is greater than 90o. If A and B are complementry angles then A + B = 90o. If A and B are supplementary angle, then A + B = 180o. ( p q ) ( p q ) (q r) (r p ) 0 , p(q r) p(q r) q(r p ) r( p q ) 0. ( p a)(q r) p(q r) a(q r) 0. Properties of Triangles and Solutions of Triangles 97} Example: 47 If in a right angled triangle the hypotenuse is four times as long as the perpendicular drawn to it from opposite vertex, then one of its acute angle is [MP PET 1998] (a) 15 o Solution: (a) (b) 30 o (c) 45 o (d) None of these If x is length of perpendicular drawn to it from opposite vertex of a right angled triangle, so, length of diagonal AB y1 y2........(i) 60 B From OCB, y2 x cot and from OCA, y1 x tan Put the values in equation (i), then AB x (tan cot ) (90o – ).......(ii) c sin 2 cos 2 x (tan cot ) 4 x 4 sin . cos sin 2 ID 1 2 sin 2 sin 2 = sin 30 o 15 o. sin 2 2 The number of triangles ABC that can be formed with a 3, b 8 and sin A (b) 1 D YG (a) 0 Solution: (a) C A Length of hypotenuse 2 Length of perpend icular drawn from opposite vertex to hypotenuse sin 2 4 Example: 48 U Trick : 1 2 30 o 15 o. 2 x E3 Since, length of hypotenuse = 4 (Length of perpendicular) Given, a 3, b 8 and sin A 5 is 3 [Roorkee 1998] (c) 2 (d) 3 5. 13 5 40 b sin A 8 a( 3) 13 13 Thus in this case no triangle is possible. In any triangle AB 2, BC 4, CA 3 and D is mid point of BC, then 11 6 U Example: 49 (a) From ABC , cos B ST Solution: (d) cos B From ABD , cos B (b) cos B 7 8 (c) [Roorkee 1995] AD 2.4 (d) 2 2 4 2 3 2 11 224 16 A 2 2 2 2 AD 2 11 222 16 11 2 2 2 2 AD 2 16 222 If b 3, c 4 and B B (a) Infinite 3 3 2 AD 2 2. 5. Example: 50 AD 2 2. 5 2 , then the number of triangles that can be constructed is (b) Two (c) One D 2 C [Roorkee 1992] (d) Nil 98 Properties of Triangles and Solutions of Triangles Solution: (d) Here, c sin B 4 sin 2 3 b( 3) 3 Thus, we have b c sin B Hence no triangle is possible, i.e., the number of triangles that can be constructed is nil. If two sides of a triangle are 2 3 and 2 2 the angle opposite the shorter side is 45 o. The maximum value of the third side is (b) 2 6 2 3 2 2 c a b c o sin A sin B sin C sin A sin 45 sin C.......(i) 3 A 60 o. 2 sin A C 180 o A B 75 o From (i) c 4 sin C 4 sin(45 o 30 o ) = 2 6. In an ambiguous case, If the remaining angles of the triangles formed with a, b sin C1 sin C2 sin B1 sin B2 2 cos A (b) cos A In ' s ACB 1 and ACB 2 (c) and A be B1, C1 and B2 , C2 then C2 D YG sin C1 sin C 2 c AB 1 c 2 1 and = AC b sin B2 b sin B1 C C1 b sin C1 sin C 2 c1 c 2 sin B1 sin B 2 b b 2 c 2 a2 or c 2 (2b cos A)c (b 2 a 2 ) cos A 2 bc c1 c 2 2b cos A (d) sinA 2 sin A U (a) Solution: (a) (d) None of these Let a 2 3 , b 2 2 B 45 o Example: 52 6 2 (c) E3 Solution: (b) 2 6 60 (a) ID Example: 51 A A B2 a b sinA N B1 X 0 ST U sin C1 sin C2 2b cos A 2 cos A. = b sin B1 sin B2 99 Properties of Triangles and Solutions of Triangles 99} Relation between Sides and Angles Basic Level The triangle PQR of which the angles P, Q, R satisfy cos P (a) Equilateral 2. sin Q is 2 sin R (b) Right angled [Orissa JEE 2002] (c) Any triangle If angles of a triangle are in the ratio of 2 : 3 : 7, then the sides are in the ratio of 60 1. (d) Isosceles [MP PET 1996, 1993; BIT Ranchi 1992] 3 (b) (b) D YG 3abc 2 If in a triangle, a cos 2 (b) ab c (c) (d) None of these [MP PET 1985] (d) 2(a b c) [ISM Dhanbad 1973] a2 b 2 c 2 3 a 2 bc (c) (d) 2(a 2 b 2 c 2 ) (b) G.P. [MP PET 1994] (d) 3abc C A 3b c cos 2 , then its sides will be in 2 2 2 (s b)(s c) s(s a) U (s c)(s a) s(s b) (b) 3 ab 2 c [MP PET 1982; AIEEE 2003] (c) H.P. (d) None of these (s a)(s b) s(s c) (c) [MP PET 1994] (s a)s (s b)(s c) (d) A B C (c a) cot (a b) cot is equal to 2 2 2 ST In a ABC (b c) cot (b) 1 [WB JEE 1989] (c) 1 (d) 2 In a triangle ABC , a 2cm, b 3cm, c 4 cm then angle A is 1 cos 1 24 11 (b) cos 1 16 [MP PET 2002] (c) 7 cos 1 8 1 (d) cos 1 4 In a ABC , if (sin A sin B sin C)(sin A sin B sin C) 3 sin A sin B, then the angle C is equal to (a) 13. 3 A If a, b, c are the sides and A, B, C are the angles of a triangle ABC, then tan is equal to 2 (a) 12. (c) (b) a b c (a) 0 11. (c) In any triangle ABC, the value of a(b 2 c 2 ) cos A b(c 2 a 2 ) cos B c(a 2 b 2 ) cos C is (a) 10. (d) None of these [MP PET 1983, 1989, 1990] 4 (b) 1 (a) A.P. 9. 6 In a ABC , a sin(B C) b sin(C A) c sin( A B) (a) 8. 2 : ( 3 1) : 2 In a triangle ABC , (b c) cos A (c a) cos B (a b) cos C (a) 0 7. (c) ID 6 (a) 0 6. 4 (d) [EAMCET 1985] In a ABC , if c 2 a 2 b 2 ac, then B (a) 5. 2 : ( 3 1) : 2 (c) The smallest angle of the triangle whose sides are 6 12 , 48 , 24 is (a) 4. 2 : 2 : ( 3 1) U 3. (b) E3 2 : 2 : ( 3 1) (a) 2 (b) 3 (c) 4 (d) In a triangle ABC, right angled at C, the value of tan A tan B is (a) ab a bc 6 [Pb. CET 1990; Karnataka CET 1999; MP PET 2001] 2 2 (b) [AMU 1999] (c) b ac (d) c2 ab 100 Properties of Triangles and Solutions of Triangles In a ABC , if b 20, c 21 and sin A (a) 12 [EAMCET 2003] (b) 13 (c) 14 [MP PET 1988] 1 3 (b) (c) 4 4 The angles of a triangle are in the ratio 1 : 3 : 5 then the greatest angle is 5 2 (a) (b) (c) 9 9 (a) 17. ab (c) 30 In ABC , (b) 60 o (c) (d) 3 4 [IIT 1994] (c) 2 3 [Roorkee 1999] (d) 5 6 (b) a c [MP PET 1986, 1995] (c) 1 (d) c ab x 5 (b) x 6 [MP PET 1984] (c) x 4 (d) x 7 3 1 cm 2 3 3 1 cm 2 2 (b) a2 b 2 c 2 cm 2 (d) 3 1 cm 2 3 2 2 2 b c , then 2c ST A 2 3 1 (c) [MP PET 1997] [MP PET 1990] (b) b 2 c 2 a 2 (c) c 2 a2 b 2 (d) b c c a b ab [MP PET 1989] (b) b c (c) c b (d) None of these If B 60 o , then (a) 28. c2 a b2 2 sin B In ABC , = sin( A B) (a) 27. 2 [MP PET 1986] (d) 3 2 2 , ( 0). Its largest angle is D YG c a In ABC , if cos (a) 26. (d) 75 o If in a triangle ABC side a ( 3 1) cms and B 30 o , C 45 o , then the area of the triangle is (a) 25. 15 If a 9, b 8 and c x satisfies 3 cos C 2, then (a) 24. (b) 2a bc In ABC , cosec A(sin B cos C cos B sin C) (a) 23. 3 4 11 9 [IIT 1981; Haryana CEE 1998] o U The lengths of the sides of a triangle are , and (a) 22. (d) a2 b 2 c2 a2 b 2 (b) (c) 2 2 2 a b2 c c If the lengths of the sides of a triangle are 3, 5, 7 then the largest angle of the triangle is 5 2 (a) (b) (c) 2 6 3 U 21. ca sin( A B) sin( A B) (a) 20. (d) [MP PET 1996] (b) b c o 1 3 [Kerala (Engg.) 2002] 7 9 If in a triangle the angles are in A.P. and b : c 3 : 2 , then A is equal to (a) 19. (d) If in a triangle ABC, 2 cos A sin B cosec C, then (a) 18. 1 2 ID 16. (d) 15 B In ABC , if a 16, b 24 and c 20 , then cos equal to 2 60 15. 3 , then a 5 E3 14. (a b) 2 ab c 2 [Karnataka CET 1992] (b) (b c)2 bc a 2 (c) (c a) 2 ca b 2 (d) a 2 b 2 c 2 2b 2 ac If b 2 c 2 3 a 2 , then cot B cot C cot A = (a) 1 (b) ab 4 [MP PET 1991] (c) 0 (d) ac 4 Properties of Triangles and Solutions of Triangles 101} In triangle ABC, A = 30o, b = 8, a = 6, then B sin 1 x , where x = In ABC , 1 tan (a) 2c abc (b) (c) Isosceles 1 1 ac bc 2 ab c (c) (b) 3/20 30 o (b) 3 abc (c) 24/25 45 o (c) 4a abc (d) None of these 75 o (b) Equilateral [MP PET 1983] (d) 1/50 [MP PET 1983] (d) 15 o [IIT 1984] (c) Right angled (d) None of these If a 2 , b 2 , c 2 are in A.P., then which of the following are also in A.P. (a) sin A, sin B, sin C cot AB AB. tan 2 2 (b) ab ab (b) tan A, tan B, tan C ab ab [ISM Dhanbad 1989] (c) cot A, cot B, cot C (d) None of these [Roorkee 1975] (c) a ab (d) None of these ST If the sides of a triangle are in A.P., then the cotangent of its half the angles will be in (b) G.P. (c) A.P. [MP PET 1993] (d) No particular order In ABC , if cot A, cot B, cot C be in A.P., then a 2 , b 2 , c 2 are in (a) H.P. (b) G.P. [IIT 1975] (d) None of these 3 , then the triangle is 2 D YG If in a triangle ABC, cos A cos B cos C (a) H.P. 39. (d) If in a triangle ABC , b 3 , c 1 and B C 90 o , then A is (a) 38. 2 abc [MP PET 1991] In ABC , if a 3, b 4, c 5, then sin 2 B U 37. (c) (b) Right angled 1 abc (a) Isosceles 36. a abc If in a triangle ABC, C 60 o , then (a) 35. (d) 1 [Roorkee 1973] (b) (a) 4/5 34. 2 3 If cos 2 A cos 2 C sin 2 B, then ABC is (a) 33. (c) A B tan 2 2 (a) Equilateral 32. 1 3 E3 31. (b) ID 30. 1 2 60 (a) [Karnataka CET 1990] U 29. [MP PET 1997] (c) A.P. (d) None of these Advance Level 40. If the perpendicular AD divides the base of the triangle ABC such that BD, CD and AD are in the ratio 2, 3 and 6, then angle A is equal to (c) 3 4 2 cos A cos B 2 cos C a b , then the value of angle A is In a triangle ABC , a b c bc ca (a) 41. 2 (b) (d) 6 [IIT 1993] 102 Properties of Triangles and Solutions of Triangles (a) 45 o (b) 30 o (c) o 43. o 1 1 If the two angles on the base of a triangle are 22 and 112 , then the ratio of the height of the triangle to the length of the base is 2 2 (a) 1 : 2 (b) 2 : 1 (c) 2 : 3 (d) 1 : 1 If a 2 (b c)2 , where is the area of triangle ABC, then tan A is equal to (a) 15 16 (b) 8 15 (c) 8 17 [Karnataka CET 1990] (d) 1 2 60 42. (d) 60 o 90 o 44. The perimeter of a ABC is 6 times the arithmetic mean of the sines of its angles. If the side a is 1, then the angle A is (a) (b) (c) (d) 6 3 2 45. If A1 A2 A3..... An be a regular polygon of n sides and 49. E3 sin( x y ) sin(y z ) sin x sin z (a) 1 (c) 4 AC In a ABC , 2 cos 2 (b) 2 ac a 2 c 2 ac , then n7 ID 48. (b) The altitudes are in H.P. (d) The medians are in A.P. BD=DE=EC. If BAD x, DAE y, EAC z, then the values of (c) (d) None of these [IIT 1998] (d) None of these U 47. [IIT 1994] (a) n 5 (b) n 6 If an triangle PQR, sin P, sin Q, sin R are in A.P., then (a) The altitudes are in A.P. (c) The medians are in G.P. Points D, E are taken on the side BC of a triangle ABC such that (b) B C (c) A, B, C are in A.P. (d) Both (a) and (c) 3 If P is the product of the sines of angles of a triangle, and q the product of their cosines, then the tangents of the angle are roots of the equation (a) B (a) qx 3 px 2 (1 q)x p 0 (b) (c) (1 q)x 3 px 2 qx p 0 (d) None of these D YG 46. 1 1 1 , then A1 A 2 A1 A 3 A1 A 4 [IIT 1992] px 3 qx 2 (1 p)x q 0 Circle Connected with Triangle 50. U Basic Level Which is true in the following (a) (b) a cos A b cos B c cos C 2R sin A sin B sin C a cos A b cos B c cos C 4 R sin A sin B sin C (d) a cos A b cos B c cos C 8 R sin A sin B sin C ST (c) [UPSEAT 1999] a cos A b cos B c cos C R sin A sin B sin C 51. In ABC , a cos A b cos B c cos C 52. (a) 4 R sin A sin B sin C (b) 3 R sin A sin B sin C The in-radius of the triangle whose sides are 3, 5, 6 is (a) 53. 8 7 55. 8 (c) sin A sin B sin C (d) 4 R cos A cos B cos C [EAMCET 1982] (c) 7 (d) 7 8 In an equilateral triangle of side 2 3 cm, the circum-radius is (a) 1 cm 54. (b) [WB JEE 1971] (b) 3 cm [EAMCET 1978] (c) 2 cm If the lengths of the sides of a triangle are 3, 4 and 5 units, then R the circum-radius is (a) 2.0 (b) 2.5 (c) 3.0 In a triangle ABC, a : b : c = 4 : 5 : 6. The ratio of the radius of the circumcircle to that of the incircle is (d) 2 3 cm [MNR 1990] (d) 3.5 [IIT 1996] Properties of Triangles and Solutions of Triangles 103} (a) R ab c (b) R ab c 4 (c) (d) 7 16 [Karnataka CET 2000] R abc 4 (d) 2 2 a 12 (b) a 6 (c) The diameter of the circum-circle whose sides are 61, 60, 11 (a) 60 (b) 61 a 3 R abc (b) sin A B C : sin : sin 2 2 2 1:2:3 (b) cos 2a 2 [EAMCET 1988] (c) 62 (d) 63 [Kurukshetra CEE 1999] (c) 2 : 1: 3 In a ABC , I is the in-centre. The ratio IA : IB : IC is equal to (a) [IIT 1994] (d) In a equilateral triangle, circum-radius : in-radius : ex-radius i.e., R : r : r1 = (a) 1 : 1 : 1 60. 11 7 60 (a) 59. (c) A circle is inscribed in an equilateral triangle of side a. The area of any square inscribed in this circle is 2 58. 16 7 If R is the radius of the circumcircle of the ABC and is its area, then (a) 57. (b) E3 56. 16 9 A B C : cos : cos 2 2 2 (c) cosec (d) 3 : 2 : 4 A B C : cosec : cosec 2 2 2 (d) sec A B C : sec : sec 2 2 2 In a triangle ABC if r1 2r2 = 3r3 , then 62. (a) 2a b c (b) a c 2b (c) a b 2c 0 (d) None of these If in a triangle R and r are the circumradius and in-radius respectively, then the H.M. of the ex-radii of the triangle is (b) 2R 3r (c) R r (d) None of these U (a) ID 61. 63. If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ (= PR), then the angle P is (a) 64. D YG Advance Level 6 (b) 3 (c) 2 [IIT 1992] (d) 2 3 If p1 , p 2 , p 3 are respectively the perpendiculars from the vertices of a triangle to the opposite sides, then p1. p 2. p 3 2 (a) 2 2 a b c R2 2 (b) [DCE 1997; EAMCET 1994] 2 2 2 a b c 4R2 (c) 2 2 2 4a b c R2 (d) a b 2c 2 8R2 If r1 , r2 , r3 are the radii of the escribed circles of a triangle ABC and if r is the radius of its incircle, then r1r2 r3 r(r1r2 r2 r3 r3 r1 ) is equal 66. to (a) 0 (b) 1 (c) 2 In a triangle, the line joining the circum centre to the incentre is parallel to BC , then cos B cos C U 65. 3 2 ST (a) 67. (b) 1 (c) 3 4 (d) 3 [EAMCET 1994] (d) The sides of a triangle inscribed in a given circle subtend angles , , 1 2 at the centre. The minimum value of the A.M. of cos , cos and cos is equal to 2 2 2 (a) 3 2 (b) 3 2 [UPSEAT 1996; IIT 1994] (c) 2 (d) 2 3 Solution of Triangle Basic Level 104 Properties of Triangles and Solutions of Triangles 68. In a ABC , if C 30 o , a 47 cm and b 94 cm , then the triangle is (a) Right angled 69. (b) Right angled isosceles [MP PET 1986] (c) Isosceles (d) Obtuse angled In a triangle ABC, if a sin A b sin B, then the nature of the triangle [MP PET 1983] (b) a b (c) a b ab If the sides of a triangle be 6, 10 and 14 then the triangle is (a) Obtuse angled (b) Acute angled (c) Right angled If in a triangle ABC, a, b, c and angle A is given and c sin A a c, then (d) a b c (a) (a) 72. b1 b2 2c cos A In a triangle ABC if (b) b1 b2 c cos A (b) Obtuse angled (c) Equilateral (b) Equilateral (d) Isosceles (c) Right angled In any ABC if a cos B b cos A, then the triangle is [EAMCET 1994] (d) None of these [MP PET 1984] (b) Isosceles Triangle (d) Right angled Triangle ID (a) Equilateral Triangle (c) Scalene Triangle (d) b1 b2 4 c sin A [Karnataka CET 1991] A bc If cot , then the ABC is 2 2 (a) Isosceles 74. b1 b2 3c cos A cos A cos B cos C , then the triangle is a b c (a) Right angled 73. (c) [UPSEAT 1999] 60 71. [MP PET 1982] (d) Equilateral E3 70. If one side of a triangle is twice the other side and the angles opposite to these sides differ by 60 o , then the triangle is (a) Equilateral (b) Isosceles (c) Right angled (d) None of these 76. The sides of a triangle are 3 x 4 y, 4 x 3y and 5 x 5 y units where x 0, y 0. The triangle is (a) Right angled If A 60 , a 5, b 4 3 in ABC , then B (a) 78. (b) 60 o 30 o If one angle of a triangle is 30 (a) Right angled 80. If in ABC , a 5; b 4 ; A (b) Two solutions o (b) Isosceles 2 (d) None of these (c) No solution (d) None of these (c) Both (a) and (b) (d) None of these (b) Is equal to tan 1 1 9 (c) Is equal to tan 1 1 40 (d) Is equal to 2 tan 1 We are given b, c and sinB such that B is acute and b c sin B. Then (a) No triangle is possible (b) One triangle is possible (c) Two triangles are possible (d) A right angled triangle is possible 1 9 [Karnataka CET 1993] Advance Level In a right angled triangle the hypotenuse is 2 2 times the length of perpendicular drawn from the opposite vertex on the hypotenuse, then the other two angles are [UPSEAT 1999, 94; MP PET 1998] (a) 83. 90 o B , then the value of C ST 82. (d) None of these and the lengths of the sides adjacent to it are 40 and 40 3 , the triangle is U (a) Cannot be evaluated 81. (c) [AIEEE 2002] If A 30 , a 7, b 8 in ABC , then B has o (a) One solution 79. (c) Obtuse angled D YG 77. (b) Equilateral o U 75. , 3 6 (b) , 4 4 (c) 3 8 , 8 In a triangle ABC, a 4 , b 3, A 60 o. Then c is the root of the equation (a) c 2 3c 7 0 (b) c 2 3c 7 0 (c) c 2 3c 7 0 (d) 5 , 12 12 [Roorkee 1993] (d) c 2 3c 7 0 Properties of Triangles and Solutions of Triangles 105} In a triangle ABC, angle A is greater than angle B. If the measure of angles A and B satisfy the equation 3 sin x 4 sin 3 x k 0, 0 k 1, 84. then the measure of angle C is 3 (a) [IIT 1990; DCE 2001] (b) 2 (c) 2 3 (d) 5 6 In a ABC a, b , A are given and b1 ,b 2 are two values of the third side b such that b 2 2b1. Then sin A 9a 2 c 2 8a2 (a) 9a2 c 2 8c 2 (c) 9a 2 c 2 8a2 (d) None of these The exists a triangle ABC satisfying tan A tan B tan C 0 (c) (a b)2 c 2 ab and 2 (sin A cos A) 3 (b) sin A sin B sin C 2 3 7 (d) sin A sin B 3 1 3 , cos A cos B sin A sin B 4 2 2 a2 b 2 (a) 88. (b) a 2 b 2 sin 2 A The sides of a triangle are in A.P. and its area is (a) 1 : 2: 3 (c) 2 a 2 b 2 sin 2 A (d) [AMU 1998] a2 b 2 3 × (Area of an equilateral triangle of the same perimeter). Then the ratio of the sides is 5 (b) 3 : 5 : 7 (c) 1 : 3 : 5 (d) None of these There exists a triangle ABC satisfying the conditions (a) b sin A a, A (c) b sin A a, A or b sin A a, A ,b a D YG 89. ID In the ambiguous case, given a, b and A. Then the difference between the two values of c is U 87. (a) E3 86. (b) 60 85. 2 2 (b) b sin A a, A 2 (d) None of these ST U 106 2 [IIT 1996] Properties of Triangles and Solutions of Triangles 1 2 3 4 5 6 7 8 Assignment (Basic & Advance Level) 9 10 11 12 13 14 15 16 17 18 19 20 106 Properties of Triangles and Solutions of Triangles a c c c a c a b a c b d b a a c d a c 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 c c d b a b c c c a b c c a b c b c c c 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 c a b a c b c d a c a a c b b c b b c c 61 62 63 64 65 66 67 68 69 70 71 72 73 74 b a d d a b b d c a a c c b 81 82 83 84 85 86 87 88 89 a c a c b c c b a 75 76 77 78 79 80 c c d b b d E3 ID U D YG U ST 60 d
