Solutions of Triangle PDF

# SOLUTIONS OF TRIANGLE ## Nishant Jindal (MATHEMATICS) ### **KEY CONCEPTS** - **Sine Formula**: In any triangle ABC, sin A / a = sin B / b = sin C / c - **Cosine Formula**: - cos A = (b² + c² - a²) / 2bc or a² = b² + c² - 2bc cos A - cos B = (c² + a² - b²) / 2ca - cos C = (a² +...

# SOLUTIONS OF TRIANGLE ## Nishant Jindal (MATHEMATICS) ### **KEY CONCEPTS** - **Sine Formula**: In any triangle ABC, sin A / a = sin B / b = sin C / c - **Cosine Formula**: - cos A = (b² + c² - a²) / 2bc or a² = b² + c² - 2bc cos A - cos B = (c² + a² - b²) / 2ca - cos C = (a² + b² - c²) / 2ab - **Projection Formula**: - a = b cos C + c cos B - b = c cos A + a cos C - c = a cos B + b cos A - **NAPIER'S ANALOGY - TANGENT RULE**: - tan (C-A)/2 = (c-a) cot B/2 / (c+a) - tan (B-C)/2 = (b-c) cot A/2 / (b+c) - **Trigonometric Functions of Half Angles**: - sin A/2 = √((s-b)(s-c)/bc) ; sin B/2 = √((s-c)(s-a)/ca) ; sin C/2 = √((s-a)(s-b)/ab) - cos A/2 = √(s(s-a)/bc) ; cos B/2 = √(s(s-b)/ca) ; cos C/2 = √(s(s-c)/ab) - tan A/2 = √((s-b)(s-c)/s(s-a)) ; tan B/2 = √((s-c)(s-a)/s(s-b)) ; tan C/2 = √((s-a)(s-b)/s(s-c)) - where s = (a+b+c)/2 & ∆ = area of triangle. - **Area of triangle** = √s(s – a)(s – b)(s - c) - **M – N Rule**: In any triangle, - (m + n) cot θ = m cot α = n cot β = n cotB = m cotC - **Area of triangle ABC** = ab sinC/2 = bc sinA/2 = ca sinB/2 - **Circumradius**: R = abc/4Δ where R is the radius of circumcircle & A is the area of triangle - **Inradius**: - r = Δ/s = (s - a)tan A/2 = (s – b)tan B/2 = (s - c)tan C/2 - r = a sin B sin C / cos A/2 = b sin A sin C / cos B/2 = c sin A sin B / cos C/2 - r = 4RsinsinAsinC/2 - **Radius of the Ex-circles r₁, r2 & r3 are given by**: - r₁ = Δ/(s-a) ; r2 = Δ/(s-b) ; r3 = Δ/(s-c) - r₁ = s tan B/2 ; r2 = s tan A/2; r3 = s tan C/2 - r₁ = acosA / (cosB - cosC) cos C/2 ; r2 = acosB / (cosC - cosA) cos A/2 ; r3 = acosC / (cosA - cosB) cos B/2 - r₁ = 4RsinA/2 cosB/2 cosC/2 ; r2 = 4RsinB/2 cosC/2 cosA/2; r3 = 4RsinC/2 cosA/2 cosB/2 - **Length Of Angle Bisector & Medians**: - ma = √(2b² + 2c² - a²)/2 and βa = bc cos A/2 / (b+c). - Note that m²a + m²b + m²c = (a² + b² + c²)/4 - **Orthocentre and Pedal Triangle**: - The triangle KLM which is formed by joining the feet of the altitudes is called the pedal triangle. - The distances of the orthocentre from the angular points of the △ ABC are 2RcosA,2RcosB and 2RcosC. - The distances of P from sides are 2RcosBcosC, 2RcosCcosA and 2RcosAcosB. The sides of the pedal triangle are acos A(= Rsin2A), b cos B(= Rsin2B) and ccos C(= Rsin2C) and its angles are π - 2Α, π - 2B and π - 2C. - Circumradii of the triangles PBC, PCA, PAB and ABC are equal . - **Excentral Triangle**: - The triangle formed by joining the three excentres I1, I2 and I3 of △ ABC is called the excentral or excentric triangle. - Note that: - Incentre I of △ ABC is the orthocentre of the excentral AI1 I213. - △ ABC is the pedal triangle of the AI1 I2I3. - The sides of the excentral triangle are 4Rcos, 4Rcos and 4Rcos - **The Distances Between The Special Points**: - The distance between circumcentre and orthocentre is = R · √1 – 8cosAcosBcosC - The distance between circumcentre and incentre is = √R2 – 2Rr - The distance between incentre and orthocentre is √2r2 – 4R2cosAcosBcosC - **Perimeter (P) and area (A) of a regular polygon of n sides inscribed in a circle of radius r are given by P = 2nrsin and A = (1/2)nr²sin(2π/n).** - **Perimeter and area of a regular polygon of n sides circumscribed about a given circle of radius r is given by P = 2nrtan(π/n) and A = nr2tan(π/n).** ## EXERCISE-I With usual notations, prove that in a triangle ABC : 1. (b-c)/r1 + (c-a)/r2 + (a-b)/r3 = 0 2. a cotA + b cotB + c cotC = 2(R + r) 3. r1/(s-b)(s-c) + r2/(s-c)(s-a) + r3/(s-a)(s-b) = 3/r 4. r1-r/a + r2-r/b + r3-r/c = 1 5. abc cosA/2 cosB/2 cosC/2 = Δ 6. (r1 + r₂)tanC/2 = (r3 - r)cotC/2 = c 7. (r1 - r)(r2 - r)(r3 - r) = 4Rr² 8. (r + r₁)tan(B-C)/2 + (r + r₂)tan(C-A)/2 + (r + r3)tan(A-B)/2 = 0 9. 1/r1 + 1/r2 + 1/r3 = 1/r + (a² + b² + c²)/2Δ 10. (r3 + r₁)(r3 + r2)sinC = 2r3√(r2r3 + r3r1 + r1r2) 11. 1/bc + 1/ca + 1/ab = 1/2Rr 12. r1/bc - r2/ca = ab-r1r2/2Rr 13. bc-r2r3/r2 + ca-r3r1/r3 + ab-r1r2/r1 = 4R 14. (r1 + r2 + r3)/4 = r 15. Rr(sinA + sinB + sinC) = Δ 16. 2RcosA = 2R + r - r1 17. cotA + cotB + cotC = a² + b² + c² / 4Δ 18. cotA + cotB + cotC = s²/Δ 19. Given a triangle ABC with sides a = 7, b = 8 and c = 5. If the value of the expression (EsinA)(Scot) can be expressed in the form p/q, where p, q ∈ N and q is in its lowest form find the value of (p + q). 20. If r₁ = r + r2 + r3 then prove that the triangle is a right angled triangle. 21. If two times the square of the diameter of the circumcircle of a triangle is equal to the sum of the squares of its sides then prove that the triangle is right angled. 22. In acute angled triangle ABC, a semicircle with radius ra is constructed with its base on BC and tangent to the other two sides. r♭ and rĉ are defined similarly. If r is the radius of the incircle of triangle ABC then prove that, 1/r² + 1/ra² + 1/r♭² + 1/rĉ² = 1/r. 23. Given a right triangle with ∠A = 90°. Let M be the mid-point of BC. If the inradii of the triangle ABM and ACM are r₁ and r₂ then find the range of r1/r2. 24. If the length of the perpendiculars from the vertices of a triangle A, B, C on the opposite sides are P1, P2, P3 then prove that 1/P1 + 1/P2 + 1/P3 = 1/r + 1/r1 + 1/r2 + 1/r3. 25. Prove that in a triangle bc/r1 + ca/r2 + ab/r3 = 2R[(b+c)/a + (c+a)/b + (a+b)/c - 3]. ## EXERCISE-II With usual notation, if in a △ ABC, (b+c)/11 = (c+a)/12 = (a+b)/13, then prove that cos A/7 = cos B/19 = cos C/25. 2. For any triangle ABC, if B = 3C, show that cosC = (b-c)/2c * √3 & sinC = √3 / (b+c) * √3. 3. In a triangle ABC, BD is a median. If l(BD) = √3/4 * (b+c) / 2c, then prove that ∠ABC = π/2. 4. ABCD is a trapezium such that AB, DC are parallel & BC is perpendicular to them. If angle ADB = θ, BC = p & CD = q, show that AB = (p² + q²)sinθ / (pcosθ + qsinθ). 5. If sides a, b, c of the triangle ABC are in A.P., then prove that sin²A/2 cosec²A, sin² B/2 cosec²B, sin² C/2 cosec²C are in H.P. 6. Find the angles of a triangle in which the altitude and a median drawn from the same vertex divide the angle at that vertex into 3 equal parts. 7. In a triangle ABC, if tanA/2, tanB/2, tanC/2 are in AP. Show that cosA, cosB, cosC are in AP. 8. ABCD is a rhombus. The circumradii of △ ABD& △ ACD are 12.5&25 respectively. Find the area of rhombus. 9. In a triangle ABC if a² + b² = 101c² then find the value of cotC / (cotA+cotB). 10. The two adjacent sides of a cyclic quadrilateral are 2 & 5 and the angle between them is 60°. If the area of the quadrilateral is 4√3, find the remaining two sides. 11. If I be the in-centre of the triangle ABC and x, y, z be the circum radii of the triangles IBC, ICA & IAB, show that 4R3 – R(x² + y² + z²) xyz = 0. 12. Sides a, b, c of the triangle ABC are in H.P., then prove that cosecA(cosecA + cotA), cosecB(cosecB + cotB) & cosecC(cosecC + cotC) are in A.P. 13. A point 'O ' is situated on a circle of radius R and with centre O, another circle of radius 3R/2 is described. Inside the smaller crescent shaped area intercepted between these circles, a circle of radius R/8 is placed If the same circle moves in contact with the original circle of radius R, then find the length of the arc described by its centre in moving from one extreme position to the other. 14. ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, then prove that cosA cosC = 2(c2-a2) / 3ac 15. In a △ ABC, (i) cosA/a = cosB/b = cosC/c (ii) 2sinA cosB = sinC (iii) tan²A/2 + 2 tanA/2 tanB/2 - 1 = 0, prove that (i) ⇒ (ii) ⇒ (iii) ⇒ (i). 16. The sequence a1, a2, a3, is a geometric sequence. The sequence b1, b2, b3, ... ..... is a geometric sequence. b1 = 1; b2 = √7 – √28 + 1; a₁ = √28 and Σn=1 ∞1/an = Σn=1 ∞ bn. If the area of the triangle with sides lengths a1, a2 and a3 can be expressed in the form of p/q where p and q are relatively prime, find (p + q) 17. If P1, P2, P3 are the altitudes of a triangle from the vertices A, B, C&A denotes the area of the triangle, prove that 1/P1 + 1/P2 + 1/P3 = (a+b+c)/2ab * cos²C/2. 18. If a tanA + b tanB = (a + b)tan(A+B)/2 prove that triangle ABC is isosceles. 19. The triangle ABC (with side lengths a, b, c as usual) satisfies loga² = logb² + logc² – log (2bccosA). What can you say about this triangle? 20. With reference to a given circle, A₁ and B₁ are the areas of the inscribed and circumscribed regular polygons of n sides, A2 and B2 are corresponding quantities for regular polygons of 2n sides. Prove that - A2 is a geometric mean between A₁ and B₁. - B2 is a harmonic mean between A2 and B₁. 21. The sides of a triangle are consecutive integers n, n + 1 and n + 2 and the largest angle is twice the smallest angle. Find n. 22. The triangle ABC is a right angled triangle, right angle at A. The ratio of the radius of the circle circumscribed to the radius of the circle escribed to the hypotenuse is, √2/(√3 + √2). Find the acute angles B&C. Also find the ratio of the two sides of the triangle other than the hypotenuse. 23. ABC is a triangle. Circles with radii as shown are drawn inside the triangle each touching two sides and the incircle. Find the radius of the incircle of the △ ABC. 24. Line l is a tangent to a unit circle S at a point P. Point A and the circle S are on the same side of l, and the distance from A to l is 3 . Two tangents from point A intersect line l at the point B and C respectively. Find the value of (PB)(PC). 25. In a scalene triangle ABC the altitudes AD&CF are dropped from the vertices A&C to the sides BC&AB. The area of △ ABC is known to be equal to 18, the area of triangle BDF is equal to 2 and length of segment DF is equal to 2√2. Find the radius of the circle circumscribed. ## EXERCISE-III 1. In a △ ABC, if b² + c² = 3a², then cotB + cotC - cotA = - (A) 1 - (B) 2 - (C) 3 - (D) 0 2. In an acute triangle ABC if 2a²b² + 2b2c² = a² + b² + c², then angle B is equal to : - (A) π/4 - (B) π/3 - (C) π/6 - (D) π/2 3. Area of the triangle is 10√3sq. cm, ∠C = 60° and its perimeter is 20 cm, then side c will be : - (A) 8 - (B) 5 - (C) 6 - (D) 7 4. Point D, E are taken on the side BC of a triangle ABC such that BD = DE = EC. If ∠BAD = x, ∠DAE = y, ∠EAC = z, then the value of sin(x+y)sin(y+z) / sinx sinz is: - (A) 2 - (B) 3 - (C) 4 - (D) 6 5. In a △ ABC, a = 5, b = 4 and cos(A - B) = 31/32, then side c is equal to : - (A) 3 - (B) 6 - (C) 8 - (D) 9 6. If the sides of a triangle are in ratio 3: 7: 8, then R: r is equal to : - (A) 3: 1 - (B) 5:2 - (C) 7:2 - (D) 5:3 7. In triangle ABC, (b²sin²C + c²sin²B) / Δ is always equal to : - (A) 1 - (B) 2 - (C) 3 - (D) 4 8. In triangle ABC, 1/r1 + 1/r2 + 1/r3 = 3, then the value of tan(tanB/2 + tanC/2) is equal to: - (A) 2 - (B) 1/2 - (C) 1 - (D) None of these 9. In triangle ABC, (r1 + r2 + r3 - r) is equal to: - (A) 2a sinA - (B) 2a cosecA - (C) 2a sinB/2 - (D) 2a cosecA/2 10. In triangle ABC, (2cosA / a + 2cosB / b + 2cosC / c) = (a+b+c)/abc then ∠A is equal to : - (A) π/3 - (B) π/4 - (C) π/6 - (D) π/2 11. The expression (a+b+c)(b+c-a)(c+a-b)(a+b-c) / 4b²c² is equal to: - (A) cos²A - (B) sin²A - (C) cosAcosBcosC - (D) none of these 12. If the area of a triangle ABC is given by Δ = a² – (b − c)², then tanA is equal to: - (A) 1/4 - (B) 8/15 - (C) 4/15 - (D) 3/4 13. If in a triangle ABC, sinA sinB sinC = 4/5 * 5/6 * 6/7, the value of cosA + cosB + cosC is equal to: - (A) 4/15 - (B) 1/3 - (C) 2/5 - (D) 23/16 14. In triangle ABC, (acosA + bcosB + ccosC) / (a+b+c) is equal to: - (A) r + R - (B) R/r - (C) r/R - (D) rR 15. In triangle ABC, cos²A + cos²B – cos²C = 1, then the triangle is necessarily : - (A) right-angled - (B) obtuse-angled - (C) isosceles - (D) equilateral ## EXERCISE-IV 1. The sides of a triangle are 3x + 4y, 4x + 3y and 5x + 5y where x, y > 0 then the triangle is - (A) Right angled - (B) Obtuse angled - (C) Equilateral - (D) None of these [AIEEE-2002] 2. In a triangle with sides a, b, c, r₁ > r2 > r3 (which are the exradii) then - (A) a > b > c - (B) a < b < c - (C) a > b and b<c - (D)a<b and b > c [AIEEE-2002] 3. In a triangle ABC, medians AD and BE are drawn. If AD = 4, ∠DAB = π/6 and ∠ABE = π/3, then the area of the △ ABC is : - (A) 64/√3 - (B) 8/√3 - (C) 16/√3 - (D) 32/√3 [AIEEE-2003] 4. If in a △ ABC a cos²(A/2) + c cos²(A/2) = 3b/2, then the sides a, b and c - (A) satisfy a + b = c - (B) are in A.P. - (C) are in G.P. - (D) are in H.P. [AIEEE-2003] 5. The sides of a triangle are sina, cosa and √1 + sina cosa for some 0 < a < π/2. Then the greatest angle of the triangle is - (A) 150° - (B) 90° - (C) 120° - (D) 60° [AIEEE-2004] 6. In a triangle ABC, let ∠C = π/2. If r is the inradius and R is the circumradius of the triangle ABC, then 2(r + R) equals - (A) b + c - (B) a + b - (C) a + b + c - (D) c + a [AIEEE-2005] 7. If in a △ ABC, the altitudes from the vertices A, B, C on opposite sides are in H.P., then sinA, sinB, sinC are in - (A) G.P. - (B) А.Р. - (C) A.P. - G.P. - (D) H.P. [AIEEE-2005] 8. ABCD is a trapezium such that AB and CD are parallel and BC ⊥ CD. If ∠ADB = θ, BC = p and CD = q, then AB is equal to : - (A) (p²+q²)sine / pcose+qsine - (B) (p²+q²)cose / pcose+qsine - (C) p²+q2 / pcose+qsine - (D) None of these [JEE-Main 2013] 9. Let the orthocentre and centroid of a triangle be A(-3,5) and B(3,3) respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter is: - (A) 3√5 / 2 - (B) √10 - (C) 2√10 - (D) 3√3 / 2 [JEE-Main 2018] ## EXERCISE-V 1. If in a △ ABC, a = 6, b = 3 and cos(A - B) = 4/5 then find its area. [REE '97, 6] 2. If in a triangle PQR, sinP, sinQ, sinR are in A.P., then [JEE '98, 2] - (A) the altitudes are in A.P. - (B) the altitudes are in H.P. - (C) the medians are in G.P. - (D) the medians are in A.P. 3. Two sides of a triangle are of lengths √6 and 4 and the angle opposite to smaller side is 30°. How many such triangles are possible ? Find the length of their third side and area.[REE '98, 6] 4. The radii r₁, r2, r3 of escribed circles of a triangle ABC are in harmonic progression. If its area is 24 sq. cm and its perimeter is 24 cm, find the lengths of its sides. [REE '99, 6] 5. (a) In a triangle ABC, Let ∠C = π/2. If ' r' is the inradius and ' R' is the circumradius of the triangle, then 2(r + R) is equal to: - (A) a + b - (B) b + c - (C) c + a - (D) a + b + c (b) In a triangle ABC, 2ac sin (A – B + C) = [JEE '2000 (Screening) 1+1] - (A) a² + b² - c² - (B) c² + a² - b² - (C) b² — c² — a² - (D) c² — a² – b² 6. Let ABC be a triangle with incentre '1' and inradius ' r ' . Let D, E, F be the feet of the perpendiculars from I to the sides BC, CA & AB respectively . If r1, r2 & r3 are the radii of circles inscribed in the quadrilaterals AFIE, BDIF & CEID respectively, prove that r/(r-r₁) + r/(r-r₂) + r/(r-r₃) = (r - r₁)(r - r₂)(r - r₃)/r₁r₂r₃. [JEE '2000, 7] 7. If ∆ is the area of a triangle with side lengths a, b, c, then show that: A ≤ √( a+ b + c) abc 8. Also show that equality occurs in the above inequality if and only if a = b = c.[JEE' 2001] 9. Which of the following pieces of data does NOT uniquely determine an acute-angled triangle ABC (R being the radius of the circumcircle)? [JEE' 2002 (Scr), 3 ] - (A) a, sinA, sinB - (B) a, b, c - (C) a, sinB, R - (D) a, sinA, R 10. If In is the area of n sided regular polygon inscribed in a circle of unit radius and On be the area of the polygon circumscribing the given circle, prove that In / On = 2 / (1 + 1/n) * (1 - (2/n)^n) [JEE 2003, Mains, 4 out of 60] 11. The ratio of the sides of a triangle ABC is 1: √3: 2. The ratio A: B: C is [JEE 2004 (Screening)] - (A) 3:5:2 - (B) 1: √3:2 - (C) 3: 2:1 - (D) 1:2:3 12. (a) In △ ABC, a, b, c are the lengths of its sides and A, B, C are the angles of triangle ABC. The correct relation is [JEE 2005 (Screening)] - (A) (b - c)sin (B-C)/2 = acos(A/2) - (B) (b - c)cos(C/2) = asin (B-C)/2 - (C) (b + c)sin (B+C)/2 = acos(A/2) - (D) (b - c)cos(C/2) = 2asin(B+C)/2 (b) Circles with radii 3, 4 and 5 touch each other externally if P is the point of intersection of tangents to these circles at their points of contact. Find the distance of P from the points of contact. [JEE 2005 (Mains), 2] 13. (a) Given an isosceles triangle, whose one angle is 120° and radius of its incircle is √3. Then the area of triangle in sq. units is : - (A) 7 + 12√3 - (B) 12 - 7√3 - (C) 12 + 7√3 - (D) 4π [JEE 2006, 3] (b) Internal bisector of ∠A of a triangle ABC meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and the side AB at F. If a, b, c represent sides of △ ABC then [JEE 2006, 5] - (A) AE is HM of b and c - (B) AD = 2bc / (b+c) - (C) EF = sinA / 2 * 4bc / (b+c) - (D) the triangle AEF is isosceles 14. If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C respectively, then the value of the expression sin²C + sin²A is [JEE 2010] - (A) √3 / 2 - (B) a / 2 - (C) 1 - (D) √3 15. Let ABC be a triangle such that ∠ACB = π/6 and let a, b and c denote the lengths of the sides opposite to A, B and C respectively. The value(s) of x for which a = x² + x + 1, b = x² - 1 and c = 2x + 1 is (are) - (A) −(2 + √3) - (B) 1 + √3 - (C) 2 + √3 - (D) 4√3 [JEE 2010] 16. Consider a triangle ABC and let, a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively. Suppose a = 6, b = 10 and the area of the triangle is 15√3. If ∠ACB is obtuse and if r denotes the radius of the incircle of the triangle, then r² is equal to [JEE 2010] 17. Let PQR be a triangle of area A with a = 2, b = √5/2, where a, b and c are the lengths of the sides of the triangle opposite to the angles at P, Q and R respectively. Then (2sinP-sin2P) / (2sinP+sin2P) equals [JEE 2012] - (A) 3 / 4Δ - (B) 45 / 4Δ - (C) (√3 / 2) - (D) (√3 / 2)² 18. In a triangle PQR, P is the largest angle and cos P = 1/3. Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is/are [JEE (Adv.)2013] - (A) 16 - (B) 18 - (C) 24 - (D) 22 19. In a triangle XYZ, let x, y, z be the lengths of sides opposite to the angles X, Y, Z, respectively, and 2s = x + y + z. If (s-x)/4 = (s-y)/3 = (s-z)/2 and area of incircle of the triangle XYZ is π/6, then [JEE Advanced-2016] - (A) area of the triangle XYZ is 6√6 - (B) the radius of circumcircle of the triangle XYZ is √6 - (C) sinsinsinZ/2 = 3√5/35 - (D) sin²X/2 = 3/5 20. In a non-right-angled △ PQR, let p, q, r denote the lengths of the sides opposite to the angles at P, Q, R respectively. The median from R meets the side PQ at S, the perpendicular from P meets the side QR at E, and RS and PE intersect at O. If p = √3, q = 1 and the radius of the circumcircle of the △ PQR equals 1, then which of the following options is/are correct? [JEE Advanced-2019] - (A) Radius of incircle of △ PQR = (2 - √3) - (B) Area of A SOE = √3 / 12 - (C) Length of RS = √7 / 2 - (D) Length of OE = 1 / 6 21. Let x, y and z be positive real numbers, Suppose x, y and z are the length of the sides of aa triangle opposite to its angles X, Y and Z, respectively. If tanX/2 + tanZ/2 = 2y/(x+y+z), then which