Question Bank Trigonometry 3 3 3 3 cos A sin A cos A sin A 1. Prove that cos A sina cos A sina 3 3 3 3 cos A sin A cos A sin A L.H.S. cos A sina cos A sina (cosa sina) (cos A sin A cosa sina) (cosa sina) cos A sin A cosa sina (cosa sina) (cosa sina) 3 3 3 3 a b ( a b) ( a b ab) and a b ( a b) ( a b ab) (cos A sin A cosasina) (cos A sin A cosa sina) (1 cosa sina) (1 cosa sina) cos A sin A 1 1 cosa sina 1 cosa sina R.H.S. Proved. cosa sina. Prove that cosa sina 1 tana 1 cota cosa sina L.H.S. 1 tana 1 cota cosa sina sina cosa 1 1 cosa sina cosa cosa cosa sina sina sina cosa sina sin A cosa sin A cos A cosa sina cos A sin A cosa sina (cosa sina) (cosa sina) (cosa sina) Math Class X 1 Question Bank
[ a b (a b) (a b)] cosa sina R.H.S. Proved. sina 3. Prove that cota coseca sina L.H.S. cota coseca sina cota coseca sina sina sina cosa 1 cosa 1 cosa 1 sina sina sin A 1 cos A sin 1 cos cosa 1 θ (1 cosa)(1 cosa) 1 cosa (cosa 1) sina R.H.S. cota coseca sina sin A cosa 1 cosa 1 sina sina 1 cos A sin 1 cos cosa 1 θ (1 cosa) (1 cosa) cosa 1 (1 cosa) (1 cosa) cosa 1 θ θ (1 cosa) 1 cos A L.H.S. Proved. Math Class X Question Bank
4. If sina cosa m and seca coseca n, prove that n (m 1) m. We have, m sina cosa m (sina cosa) sin A cos A sina cosa 1 sin A cosa m 1 1 sina cosa 1 sina cosa n (m 1) (seca coseca). sina cosa sina cosa seca sina cosa coseca sina cosa [ cosa seca 1 and sina cosec A 1] (sina cosa) m Hence, n (m 1) m. Proved. 5. Prove that 1 1 1 1 (sec A tana) cosa cosa (seca tana) 1 1 L.H.S. seca tana cosa 1 1 1 sina cosa cos A cosa cosa 1 cos A 1 sina 1 sina cosa cos A ( 1 sina) 1 sin A 1 sina sina(1 sina) cos A(1 sina) tana. cosa (1 sina) Math Class X 3 Question Bank
1 1 R.H.S. cosa seca tana 1 1 cosa 1 sina cosa cosa 1 cosa cosa 1 sina 1 sina cos A cosa (1 sina) 1 sina 1 sin A sina(1 sina) cos A(1 sina) cos A(1 sina) tana. Hence, LHS RHS. Proved. 6. If x sin 3 θ y cos 3 θ sinθ and x sinθ y 0, then prove that x y 1 We have x sin 3 θ y cos 3 θ sinθ... (i) x sinθ y 0... (ii) sin θ y x (iii) From (i) sin θ cos θ x. y. 1 sinθ sinθ x. sin θ y. cos θ 1 sinθ x. y x sin θ y.. 1 x y [From (iii)] y sinθ x 1 x y sinθ 1...(iv) Squaring (ii) and (iv) and adding, we get, (x sinθ y ) (x y sinθ) 0 1 x sin θ y cos θ xy sinθ x cos θ y i sin θ xy sinθ 1 Math Class X 4 Question Bank
x ( sin θ cos θ) y (cos θ sin θ) 1 x y 1. Proved. 7. Is an identity? If not solve for θ, cosec θ 1 cosec θ 1 where 0 < θ < 90. Here, LHS cosec θ 1 cosec θ 1 1 1 1 1 sin θ sinθ cos θ sinθ cos θ sinθ 1 sinθ 1 sinθ cos θ sin θ ( 1 sin θ 1 sin θ) 1 sin θ sin θ tanθ cos θ Thus, the given equality becomes tanθ If the equality holds true for all values of θ, then the equality is an identity. Let us take θ 30 So, tan θ tan30 3 tan θ for θ 30 Therefore the equality is not an identity. It is an equation. Now, tan θ tan θ 1 tan θ tan 45 θ 45. 8. If tan θ sec θ 3, where θ is acute, then prove that 5 sinθ 4. We have tanθ secθ 3 sinθ 1 3 1 sinθ 3 Math Class X 5 Question Bank
(1 sinθ) 9 cos θ [Squaring both sides] 1 sin θ sinθ 9 9 sin θ 10 sin θ sinθ 8 0 10 sin θ 10 sinθ 8 sinθ 8 0 10 sinθ (sinθ 1) 8 (sinθ 1) 0 (sinθ 1) (10 sinθ 8) 0 sinθ 1 or sinθ 8 4 10 5 sin θ 4 [Rejecting sinθ 1, since θ is acute] 5 5 sinθ 4. Proved. 9. Without using trigonometric tables, prove that : tan 10º tan 0º tan 1 30º tan 70º tan 80º 3 L.H.S. tan 10º tan 0º tan 30º tan 70º tan 80º (tan 10º tan 80º ), (tan 0º tan 70º) tan 30º tan (90º 80º) tan 80º. tan (90º 70º) tan 70º tan 30º cot 80º tan 80º. cot70º tan 70º tan 30º [ tan (90º θ ) cot θ ] 1 1. tan80º. tan 70º tan 30º tan 80º tan 70º 1.1. tan 30º 1 R.H.S. Proved. 3 10. Prove that sina cosa sec (90º A ) cosec (90º A) sin(90º A) cos(90º A) sina cosa L.H.S. sin(90º A) (cos(90º A) sina cosa cosa sina Math Class X 6 Question Bank
[ sin (90º A) cosa and cos (90º A) sina] sin A cos A 1 sin A cos A 1 sin A cos A sina cosa coseca seca R.H.S sec ( 90º A ) cosec (90º A) coseca seca [ sec (90º A) cosec A and cosec (90º A) seca] L.H.S. Hence, L.H.S. R.H.S. Proved. 11. Prove that L.H.S. sin 0º sin 70º sin (90º θ) sinθ cos(90º θ) cos 0º cos 70º tanθ cotθ sin 0º sin 70º sin (90º θ) sin θ cos(90º θ) cos 0º cos 70º tan θ cotθ sin (90º 70º ) sin 70º cos sin sin cos θ θ θ θ cos (90º 70º) cos 70º tan θ cotθ cos 70º sin 70º cos θ sinθ sin θ sin 70 cos 70º sinθ cos θ sin θ [ sin (90º θ) cos θ,cos (90º θ ) sinθ] 1 cos θ sin θ 1 1 1 R.H.S. Proved. 1. Using the tables, find the values of (i) sin 60º 3 (ii) cos 1º 56 (iii) tan 75º (iv) cot 40º 36 From trigonometric tables, we have (i) sin 60º 18 0.8689 Mean difference for 5 7 (To be added) sin 60º 3 0.8696 (ii) cos 51º54 0.978 Mean difference for (To be subtracted) Math Class X 7 Question Bank
1º 56 0.976 (iii) tan 75º 3.731 Mean difference for 93 (To be added) 75º 3.7414 (iv) cot 40º 36 cot (90º 49º 4 ) tan 49º 4 Now, tan 49º 4 1.1667 13. Find θ when (i) sin θ 0.0990 (ii) cos θ 0.5536 (iii) tan θ 5.010 (i) From the table, find the angle whose sine is just smaller than 0.0990 sin θ 0.0990 sin 5º 36 0.0976 Difference 0.0014 Mean difference 14 corresponds to 5 Required angle (5º 36 5 5º 41. (ii) From the table, find the angle whose cosine is just greater than 0.5536 cos θ 0.5536 sin 5º 36 0.5548 Difference 0.001 Mean difference 1 corresponds to 5 Required angle (56º 18 5 ) 56º 3. (iii) From the table, find the angle whose tangent is just smaller than 5.010 tan θ 5.010 tan 79º 6 5.193 Since mean differences are not given corresponding to 79, therefore required angle 79 6. Math Class X 8 Question Bank
14. A boy standing on a vertical cliff in a jungle observes two rest houses in line with him on opposite sides deep in the jungle below. If their angles of depression are 19 and 6 and the distance between them is m, find the height of the cliff. Let A be the top of the cliff and C and D be the two rest houses. Let AB h m and BC x m Then, BD ( x) m In ΔABC, tan 19 h x 0.3443 h x h x 0.3443..(i) h In ΔABD, tan 6 x 0.4877 h x h ( x) 0.4877..(ii) From (i) and (ii), we have ( x) 0.4877 x 0.3443 x (0.3443 0.4877) 0.4877 0.4877 x 0.83 From (i), we have, h 0.4877 0.3443 0.83 log h log log 0.4877 log 0.3443 log 0.83.3463 1.6881 1.5369 1.901.3463 ( 1 1 1) (0.6881 0.5369 0.901).3463 1 0.3049.3463 0.6951 1.651 h antilog 1.651 44.79 Hence, height of the cliff 44.80 m. Math Class X 9 Question Bank
13. An aeroplane is flying horizontally 4000 m above the ground and is going away from an observer on the level ground. At a certain instant the observer finds that the angle of elevation of the plane is 45. After 15 seconds, its elevation from the same point changes to 30. Find the speed of the aeroplane in km/h. Let A be the position of the observer, B be the point whose angle of elevation from A is 45. Let after 15 seconds the position of the plane be C, whose angle of elevation from A be 30. In ΔABD, tan 45 BD AD 1 BD AD BD AD..(i) CE 4000 In ΔACE, tan 30 AE AD DE [ AE AD DE] 13 4000 BD DE [From (i)] 13 4000 4000 DE 4000 3 4000 DE DE 4000( 3 1 ) DE 4000 0.73 98 Distance covered by the aeroplane in 15 seconds 98 m Speed of the aeroplane 98 98 18 m/s 15 15 5 km/h 70.7 km/h. Math Class X 10 Question Bank
14. At the foot of a mountain, the elevation of its summit is 45. After ascending 1000 m towards the mountain up a slope of 30 inclination, the elevation is found to be 60. Find the height of the mountain. Let AB be the mountain of height h m and C be its foot. CD 1000 m, ACB 45, DCB 30 and ADF 60. h In ΔACB, tan 45 CB h 1 CB h CB..(i) In ΔCDE, sin 30 DE 1000 1 DE 1000 DE 500..(ii) In ΔCDE, cos 30 CE 1000 3 CE 1000 CE 500 3..(iii) Now, BE BC EC h 500 3 [From (i) and (iii)] In Δ ADF, tan 60 AF DF h BE h 500 3 h 500 3 h 500 3 h 500 h 3 1500 h ( 3 1 ) 1000 h 1000 0.73 1369.86 [ DF BF and BF DE] Math Class X 11 Question Bank
Hence, height of the mountain is 1369.86 m. 15. A man is standing on the deck of a ship which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60 and the angle of depression of the base of the hill as 30. Calculate the distance of the hill from the ship and the height of the hill. In the figure. A is the deck of the ship and CD is the hill. Let BC x m and DE h m. In ΔABC, tan 30 8 x 1 3 8 x x 8 3m. In ΔADE, tan 60 DE AE h x 3 h x [AE BC x] h 3 x 3 8 3 4 cm. Distance of the hill from the ship 8 3m, and height of the hill (h 8) m (4 8) m 3 m. 16. A ladder rests against a house on one side of a street. The angle of elevation of the top of the ladder is 60. The ladder is turned over to rest against a house on the other side of the street and the elevation now becomes 4 50'. If the ladder is 40 m long, find the breadth of the street. In the figure, AB and CD are two houses. O is a point on the street, at which one end of the ladder rests. Let OB x m and OD y m. x In ΔAOB, cos 60 40 1 x x 0 m. 40 Math Class X 1 Question Bank
In ΔCOD, cos 4 50' 40 y 0.7333 40 y [From tables] y 0.7333 40 9.33 Hence, breadth of the street (x y) m (0 9.33) m 49.33 m. 17. A vertical tower stands on horizontal plane and is surmounted by a vertical flagstaff of height h m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is α and that of the top of the h tanα flagstaff is β. Prove that the height of the tower is tanβ tanα. Let AB be the tower, AC be the flagstaff of height h m and D be the point of observation. AB BD In ΔABD, tan α AB BD tan α..(i) In ΔCBD, tan β BC BD AB AC tan β BD AB AC BD tanβ BD tan α h BD tanβ h BD (tanβ tanα) h BD tanβ tan α Height of the tower AB BD tan α h tanα tanβ tanα [From (i)] Math Class X 13 Question Bank