Grade 10 Trigonometry PDF

CAMI Mathematics athematics: athematics: Grade 10 GRADE 10_CAPS Curriculum 10.9 Trigonometry 1.1 Define the trigonometric ratios sin θ , cos θ and tan θ using right-...

CAMI Mathematics athematics: athematics: Grade 10 GRADE 10_CAPS Curriculum 10.9 Trigonometry 1.1 Define the trigonometric ratios sin θ , cos θ and tan θ using right- right-angled triangle. triangle. (a) (b) cosA = … sinA = … sinC = … tanC = … tanA =... cosC = … (c) (d) sinZ = … sinY = … sinQ = … cosZ = … cosY = … tanR = … tanZ = … tanY = … cosQ = … 1.2 Extend the definitions of sin θ , cos θ and tan θ for 0° ≤ θ ≤ 360°. (a) cos 100° (b) tan 210° (c) sin 300° (d) tan 135° (e) sin 315° (f) cos 120° (g) sin 240° (h) cos 225° (i) tan 150° (j) sin 135° 1.3 Define the reciprocals of the trigonometric ratios cosec θ , sec θ and cot θ , using right- right-angled triangles. triangles.  CAMI Mathematics athematics: athematics: Grade 10 sinA = … cosec C = … cosA = sec C = … tanA = … cotC = … cosecA =… sin C = … secA = … cos C = … cot A = … tan C = … 1.4 Derive values of the trigonometric ratios for the special cases (without using a calculator). calculator). tan 225°. sin 135°. tan 300° (a) cos 315°. cos 225°. cos 150° tan 120° (b) tan 330° (c) sin60°.cos30°.tan60° (d) sin30°.tan45°.cos45° tan 120°. cos 210° (e) sin 240°. sin 240° cos 330° (f) cos 225°. cos 315°. tan 225° 1.5 Solve two- two-dimensional problems involving right- right-angled triangles. (a) The length of a mast is 8.5m, and the length of the shadow of the mast is 7.25m. Calculate the angle of elevation of the sun at the particular moment.  CAMI Mathematics athematics: athematics: Grade 10 (b) The angle of elevation of a glider according to a woman on the ground is 43°. If the glider is 2340m from the woman, calculate the altitude of the glider. (c) Two towers are 12m apart. From B the angle of elevation to DE is 29° and from D the angle of elevation to BC is 48°. Calculate the difference in the heights of the towers. (d) A building (DF) and a tower (CE) are 94m apart. From the roof of the building the angle of elevation to the top of the tower is 15° and the angle of depression to the bottom of the tower is 46°. Calculate the height of the tower. 1.6 Solve simple trigonometric equations for angles between 0° and 90°. (a) sin51° = cos β , β an acute angle. (b) cos33° = sin α  CAMI Mathematics athematics: athematics: Grade 10 (c) sin75° = cos3 θ (d) cos 4 α = sin 5 α (e) cos ( β – 43°) = sin 65° (f) sin( θ + 54°) = cos ( θ – 8°) 1.7 Use diagrams to determine the numerical values of ratios for angles from 0° and 360°. (a) If 17sinA = 15, 0°≤ A ≤ 90°, determine tan A. (b) If 9tan β = 40 and β is an acute angle, determine sin β. (c) If 6sin α – 5 = 0 and α ∈ [90°;180°], determine cos α. (d) If -5cos β – 4 = 0 and β ∈ [180°;270°], determine sin β. (e) If 5sin θ – 4 = 0 and 90° ≤ θ ≤ 180°, determine cos θ.  CAMI Mathematics athematics: athematics: Grade 10 MEMO 1.1 Define the trigonometric ratios sin θ , cos θ and tan θ using right- right-angled angled triangle. [7.2.1.1 [7.2.1.1] 7.2.1.1 (a) (b) c 6 cos A = sin A = b x c 8 sin C = tan C = b 6 a 6 tan A = cos C = c x (c) (d) 120 65 q sinZ = ; sinY = sinQ = x x p 65 120 r cosZ = ; cosY = tanR = x x q 120 65 r tanZ = ; tanY = cosQ = 65 120 p 1.2 Extend the definitions of sin θ , cos θ and tan θ for 0° ≤ θ ≤ 360°. [7.4.2.2; [7.4.2.2; 7.4.2.3] 7.4.2.3 (a) cos 100° = -cos 80° (b) tan 210° = tan 30°  CAMI Mathematics athematics: athematics: Grade 10 (c) sin 300° = -sin 60° (d) tan 135° = -tan 45° (e) sin 315° = -sin 45° (f) cos 120° = - cos 60° (g) sin 240° = -sin 60° (h) cos 225° = -cos 45° (i) tan 150° = -tan 30° (j) sin 135° = sin 45° 1.3 Define the reciprocals of the trigonometric ratios cosec θ , sec θ and cot θ , using right- right-angled triangles. [7.2.1.3; 7.2.1.4; 7.2.1.5; 7.2.1.2] 7.2.1.2] a b sin A = cosec C = b c c b cos A = sec C = b a a a tan A = cot C = c c b c cosec A = sin C = a b b a sec A = cos C = c b c c cot A = tan C = a a 1.4 Derive values of the trigonometric ratios for the special cases (without using a calculator). [7.3.2.1; 7.3.2.3; 7.3.1.5; 7.3.1.1] 7.3.1.1  CAMI Mathematics athematics: athematics: Grade 10 (a) tan 225°. sin 135°. tan 300° cos 315°. cos 225°. cos150° tan 45°. sin 45°.(− tan 60°) = cos 45°.(− cos 45°).( − cos 30°) 1 1..(− 3 ) = 2 1 1 3.(− ).(− ) 2 2 2 = −2 2 (b) tan 120° tan 330° − tan 60° = − tan 30° 3 = 1 3 =3 (c) sin60°.cos30°.tan60° 3 3 =.. 3 2 2 3 3 = 4 (d) sin30°.tan45°.cos45° 1 1 =.1. 2 2 1 = 2 2  CAMI Mathematics athematics: athematics: Grade 10 (e) tan 120°. cos 210° sin 240°. sin 240° (− tan 60°).( − cos 30°) = (− sin 60°).(− sin 60°) 3 3. = 2 3 2 ( ) 2 =2 (f) cos 330° cos 225°. cos 315°. tan 225° cos 30° = (− cos 45°). cos 45°. tan 45° 3 = 2 1 1 −..1 2 2 =− 3 1.5 Solve two- two-dimensional problems involving right- right-angled triangles. [7.7.1.1; 7.7.1.2; 7.7.1.3] 7.7.1.3 (a) The length of a mast is 8.5m, and the length of the shadow of the mast is 7.25m. Calculate the angle of elevation of the sun at the particular moment. AB = tan x BC 8.5 = tan x 7.25 8.5 x = tan −1 ( ) 7.25 x = 49.5°  CAMI Mathematics athematics: athematics: Grade 10 (b) The angle of elevation of a glider according to a woman on the ground is 43°. If the glider is 2340m from the woman, calculate the altitude of the glider. AC sin x BC x = sin 43° 2340 x = 2340 × sin 43° x = 1595.9m (c) Two towers are 12m apart. From B the angle of elevation to DE is 29° and from D the angle of elevation to BC is 48°. Calculate the difference in the heights of the towers. ∆BCD : ∆BDE : BC DE = tan 48° = tan 29° BD BD x y = tan 48° tan 29° 12 12 x = 12 × tan 48° y = 12 × tan 29° x = 13.33m y = 6.65m BC − DE = 13.33 − 6.65 = 6.68m  CAMI Mathematics athematics: athematics: Grade 10 (d) A building (DF) and a tower (CE) are 94m apart. From the roof of the building the angle of elevation to the top of the tower is 15° and the angle of depression to the bottom of the tower is 46°. Calculate the height of the tower. ∆BCD : ∆BDE : BC BE = tan 46° = tan 15° DB DB x y = tan 46° = tan 15° 94 94 x = 94 × tan 46° y = 94 × tan 15° x = 97.34m y = 25.19m EC = 97.34 + 25.19 = 122.53m 1.6 Solve simple trigonometric equations for angles between 0° and 90°. [7.6.2.1; 7.6.2.3; 7.6.2.5] 7.6.2.5] (a) sin51° = cos β , β an acute angle. sin 51° = cos β sin 51° = sin(90° − β ) ∴ 51° = 90° − β ∴ β = 90° − 51° ∴ β = 39° (b) cos33° = sin α  CAMI Mathematics athematics: athematics: Grade 10 cos 33° = sin α cos 33° = cos(90° − α ) ∴ 33° = 90° − α ∴ α = 90° − 33° ∴ α = 57° (c) sin75° = cos3 θ sin 75° = cos 3θ sin 75° = sin(90° − 3θ ) ∴ 75° = 90° − 3θ ∴ 3θ = 90° − 75° ∴ 3θ = 15° ∴ θ = 5° (d) cos 4 α = sin 5 α cos 4α = sin 5α cos 4α = cos(90° − 5α ) ∴ 4α = 90° − 5α ∴ 9α = 90° ∴ α = 10° (e) cos ( β – 43°) = sin 65° cos( β − 43°) = sin 65° cos( β − 43°) = cos(90° − 65°) ∴ β − 43° = 90° − 65° ∴ β = 68° (f) sin( θ + 54°) = cos ( θ – 8°) sin(θ + 54°) = cos(θ − 8°) sin(θ + 54°) = sin(90° − (θ − 8°) sin(θ + 54°) = sin(90° − θ + 8°) ∴θ + 54° = 90° − θ + 8° ∴ 2θ = 90° − 54° + 8° ∴ 2θ = 44° ∴θ = 22°  CAMI Mathematics athematics: athematics: Grade 10 1.7 Use diagrams to determine the numerical values of ratios for angles from 0° and 360°. [7.6.3.1; [7.6.3.1; 7.6.3.3; 7.6.3.5; 7.6.5.1] 7.6.5.1 (a) If 17sinA = 15, 0°≤ A ≤ 90°, determine tan A. x2 + y2 = r 2 x 2 + (15) 2 = (17) 2 x 2 = 64 x=8 15 ∴ tan A = 8 (b) If 9tan β = 40 and β is an acute angle, determine sin β. x2 + y2 = r 2 (9) 2 + (40) 2 = r 2 81 + 1600 = r 2 r 2 = 1681 r = 41 40 ∴ sin β = 41 (c) If 6sin α – 5 = 0 and α ∈ [90°;180°], determine cos α.  CAMI Mathematics athematics: athematics: Grade 10 x2 + y2 = r 2 x 2 + (5) 2 = (6) 2 x 2 = 36 − 25 x = − 11 − 11 ∴ cos α = 6 (d) If -5cos β – 4 = 0 and β ∈ [180°;270°], determine sin β. x2 + y2 = r 2 (−4) 2 + y 2 = (5) 2 16 + y 2 = 25 y = −3 −3 ∴ sin β = 5 (e) If 5sin θ – 4 = 0 and 90° ≤ θ ≤ 180°, determine cos θ.  CAMI Mathematics athematics: athematics: Grade 10 x2 + y2 = r 2 x 2 + (4) 2 = (5) 2 x 2 + 16 = 25 x = −3 −3 ∴ cos θ = 5