Geometry · Textbook solutions

Trigonometry

Every solved example, exercise, and miscellaneous question — in the order the textbook teaches them. · 55 questions

Trigonometric identities

5 q

Solved Examples

Worked · 5
  1. Trigonometric identities SolvedEx.1
    If sin⁡θ=2029\sin\theta = \dfrac{20}{29} then find cos⁡θ\cos\theta.
  2. Trigonometric identities SolvedEx.3
    If 5sin⁡θ−12cos⁡θ=05\sin\theta - 12\cos\theta = 0, find the values of sec⁡θ\sec\theta and cosec⁡θ\operatorname{cosec}\theta.
  3. Trigonometric identities SolvedEx.4
    cos⁡θ=32\cos\theta = \dfrac{\sqrt{3}}{2} then find the value of 1−sec⁡θ1+cosec⁡θ\dfrac{1 - \sec\theta}{1 + \operatorname{cosec}\theta}.
  4. Trigonometric identities SolvedEx.5
    Show that sec⁡x+tan⁡x=1+sin⁡x1−sin⁡x\sec x + \tan x = \sqrt{\dfrac{1 + \sin x}{1 - \sin x}}.
  5. Trigonometric identities SolvedEx.6
    Eliminate θ\theta from the given equations. x=acot⁡θ−bcosec⁡θx = a\cot\theta - b\operatorname{cosec}\theta y=acot⁡θ+bcosec⁡θy = a\cot\theta + b\operatorname{cosec}\theta

Practice set 6.1

17 q
  1. Ex 6.1 Q.1
    If sin⁡θ=725\sin\theta = \dfrac{7}{25}, find the values of cos⁡θ\cos\theta and tan⁡θ\tan\theta.
  2. Ex 6.1 Q.2
    If tan⁡θ=34\tan\theta = \dfrac{3}{4}, find the values of sec⁡θ\sec\theta and cos⁡θ\cos\theta.
  3. Ex 6.1 Q.3
    If cot⁡θ=409\cot\theta = \dfrac{40}{9}, find the values of cosec⁡θ\operatorname{cosec}\theta and sin⁡θ\sin\theta.
  4. Ex 6.1 Q.4
    If 5sec⁡θ−12cosec⁡θ=05\sec\theta - 12\operatorname{cosec}\theta = 0, find the values of sec⁡θ\sec\theta, cos⁡θ\cos\theta and sin⁡θ\sin\theta.
  5. Ex 6.1 Q.5
    If tan⁡θ=1\tan\theta = 1 then, find the values of sin⁡θ+cos⁡θsec⁡θ+cosec⁡θ\dfrac{\sin\theta + \cos\theta}{\sec\theta + \operatorname{cosec}\theta}.
  6. Prove that:
    Ex 6.1 Q.6 (1)
    sin⁡2θcos⁡θ+cos⁡θ=sec⁡θ\dfrac{\sin^2\theta}{\cos\theta} + \cos\theta = \sec\theta
  7. Ex 6.1 Q.6 (2)
    cos⁡2θ(1+tan⁡2θ)=1\cos^2\theta(1 + \tan^2\theta) = 1
  8. Ex 6.1 Q.6 (3)
    1−sin⁡θ1+sin⁡θ=sec⁡θ−tan⁡θ\sqrt{\dfrac{1 - \sin\theta}{1 + \sin\theta}} = \sec\theta - \tan\theta
  9. Ex 6.1 Q.6 (4)
    (sec⁡θ−cos⁡θ)(cot⁡θ+tan⁡θ)=tan⁡θ sec⁡θ(\sec\theta - \cos\theta)(\cot\theta + \tan\theta) = \tan\theta\,\sec\theta
  10. Ex 6.1 Q.6 (5)
    cot⁡θ+tan⁡θ=cosec⁡θ sec⁡θ\cot\theta + \tan\theta = \operatorname{cosec}\theta\,\sec\theta
  11. Ex 6.1 Q.6 (6)
    1sec⁡θ−tan⁡θ=sec⁡θ+tan⁡θ\dfrac{1}{\sec\theta - \tan\theta} = \sec\theta + \tan\theta
  12. Ex 6.1 Q.6 (7)
    sin⁡4θ−cos⁡4θ=1−2cos⁡2θ\sin^4\theta - \cos^4\theta = 1 - 2\cos^2\theta
  13. Ex 6.1 Q.6 (8)
    sec⁡θ+tan⁡θ=cos⁡θ1−sin⁡θ\sec\theta + \tan\theta = \dfrac{\cos\theta}{1 - \sin\theta}
  14. Ex 6.1 Q.6 (9)
    If tan⁡θ+1tan⁡θ=2\tan\theta + \dfrac{1}{\tan\theta} = 2, then show that tan⁡2θ+1tan⁡2θ=2\tan^2\theta + \dfrac{1}{\tan^2\theta} = 2
  15. Ex 6.1 Q.6 (10)
    tan⁡A(1+tan⁡2A)2+cot⁡A(1+cot⁡2A)2=sin⁡Acos⁡A\dfrac{\tan A}{\left(1 + \tan^2 A\right)^2} + \dfrac{\cot A}{\left(1 + \cot^2 A\right)^2} = \sin A\cos A
  16. Ex 6.1 Q.6 (11)
    sec⁡4A (1−sin⁡4A)−2tan⁡2A=1\sec^4 A\,(1 - \sin^4 A) - 2\tan^2 A = 1
  17. Ex 6.1 Q.6 (12)
    tan⁡θsec⁡θ−1=tan⁡θ+sec⁡θ+1tan⁡θ+sec⁡θ−1\dfrac{\tan\theta}{\sec\theta - 1} = \dfrac{\tan\theta + \sec\theta + 1}{\tan\theta + \sec\theta - 1}

Application of trigonometry

5 q

Solved Examples

Worked · 5
  1. Application of trigonometry SolvedEx.1
    An observer at a distance of 10 m from a tree looks at the top of the tree, the angle of elevation is 60∘60^\circ. What is the height of the tree? (3=1.73)(\sqrt{3} = 1.73)
  2. Application of trigonometry SolvedEx.2
    From the top of a building, an observer is looking at a scooter parked at some distance away, makes an angle of depression of 30∘30^\circ. If the height of the building is 40 m, find how far the scooter is from the building. (3=1.73)(\sqrt{3} = 1.73)
  3. Application of trigonometry SolvedEx.3
    To find the width of the river, a man observes the top of a tower on the opposite bank making an angle of elevation of 61∘61^\circ. When he moves 50 m backword from bank and observes the same top of the tower, his line of vision makes an angle of elevation of 35∘35^\circ. Find the height of the tower and width of the river. (tan⁡61∘=1.8, tan⁡35∘=0.7)(\tan 61^\circ = 1.8,\ \tan 35^\circ = 0.7)
  4. Application of trigonometry SolvedEx.4
    Roshani saw an eagle on the top of a tree at an angle of elevation of 61∘61^\circ, while she was standing at the door of her house. She went on the terrace of the house so that she could see it clearly. The terrace was at a height of 4 m. While observing the eagle from there the angle of elevation was 52∘52^\circ. At what height from the ground was the eagle? (Find the answer correct upto nearest integer) (tan⁡61∘=1.80, tan⁡52∘=1.28, tan⁡29∘=0.55, tan⁡38∘=0.78)(\tan 61^\circ = 1.80,\ \tan 52^\circ = 1.28,\ \tan 29^\circ = 0.55,\ \tan 38^\circ = 0.78)
  5. Application of trigonometry SolvedEx.5
    A tree was broken due to storm. Its broken upper part was so inclined that its top touched the ground making an angle of 30∘30^\circ with the ground. The distance from the foot of the tree and the point where the top touched the ground was 10 metre. What was the height of the tree.

Practice set 6.2

6 q
  1. Ex 6.2 Q.1
    A person is standing at a distance of 80 m from a church looking at its top. The angle of elevation is of 45∘45^\circ. Find the height of the church.
  2. Ex 6.2 Q.2
    From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60∘60^\circ. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. (3=1.73)(\sqrt{3} = 1.73)
  3. Ex 6.2 Q.3
    Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60∘60^\circ. What is the height of the second building?
  4. Ex 6.2 Q.4
    Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops in 22 metre. Find the angle made by the wire with the horizontal.
  5. Ex 6.2 Q.5
    A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60∘60^\circ with the horizontal. Find the height of the tree.
  6. Ex 6.2 Q.6
    A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60∘60^\circ with the ground. Assuming that the string is straight, find the length of the string. (3=1.73)(\sqrt{3} = 1.73)

Problem set 6

22 q
  1. Choose the correct alternative answer for the following questions.
    PS6 Q.1 (1)
    sin⁡θ cosec⁡θ=\sin\theta\,\operatorname{cosec}\theta = ?
    1. A.
      11
    2. B.
      00
    3. C.
      12\dfrac{1}{2}
    4. D.
      2\sqrt{2}
  2. PS6 Q.1 (2)
    cosec⁡45∘=\operatorname{cosec} 45^\circ = ?
    1. A.
      12\dfrac{1}{\sqrt{2}}
    2. B.
      2\sqrt{2}
    3. C.
      32\dfrac{\sqrt{3}}{2}
    4. D.
      23\dfrac{2}{\sqrt{3}}
  3. PS6 Q.1 (3)
    1+tan⁡2θ=1 + \tan^2\theta = ?
    1. A.
      cot⁡2θ\cot^2\theta
    2. B.
      cosec⁡2θ\operatorname{cosec}^2\theta
    3. C.
      sec⁡2θ\sec^2\theta
    4. D.
      tan⁡2θ\tan^2\theta
  4. PS6 Q.1 (4)
    When we see at a higher level, from the horizontal line, angle formed is .........
    1. A.
      angle of elevation.
    2. B.
      angle of depression.
    3. C.
      00
    4. D.
      straight angle.
  5. PS6 Q.2
    If sin⁡θ=1161\sin\theta = \dfrac{11}{61}, find the values of cos⁡θ\cos\theta using trigonometric identity.
  6. PS6 Q.3
    If tan⁡θ=2\tan\theta = 2, find the values of other trigonometric ratios.
  7. PS6 Q.4
    If sec⁡θ=1312\sec\theta = \dfrac{13}{12}, find the values of other trigonometric ratios.
  8. Prove the following.
    PS6 Q.5 (1)
    sec⁡θ (1−sin⁡θ)(sec⁡θ+tan⁡θ)=1\sec\theta\,(1 - \sin\theta)(\sec\theta + \tan\theta) = 1
  9. PS6 Q.5 (2)
    (sec⁡θ+tan⁡θ)(1−sin⁡θ)=cos⁡θ(\sec\theta + \tan\theta)(1 - \sin\theta) = \cos\theta
  10. PS6 Q.5 (3)
    sec⁡2θ+cosec⁡2θ=sec⁡2θ×cosec⁡2θ\sec^2\theta + \operatorname{cosec}^2\theta = \sec^2\theta \times \operatorname{cosec}^2\theta
  11. PS6 Q.5 (4)
    cot⁡2θ−tan⁡2θ=cosec⁡2θ−sec⁡2θ\cot^2\theta - \tan^2\theta = \operatorname{cosec}^2\theta - \sec^2\theta
  12. PS6 Q.5 (5)
    tan⁡4θ+tan⁡2θ=sec⁡4θ−sec⁡2θ\tan^4\theta + \tan^2\theta = \sec^4\theta - \sec^2\theta
  13. PS6 Q.5 (6)
    11−sin⁡θ+11+sin⁡θ=2sec⁡2θ\dfrac{1}{1 - \sin\theta} + \dfrac{1}{1 + \sin\theta} = 2\sec^2\theta
  14. PS6 Q.5 (7)
    sec⁡6x−tan⁡6x=1+3sec⁡2x×tan⁡2x\sec^6 x - \tan^6 x = 1 + 3\sec^2 x \times \tan^2 x
  15. PS6 Q.5 (8)
    tan⁡θsec⁡θ+1=sec⁡θ−1tan⁡θ\dfrac{\tan\theta}{\sec\theta + 1} = \dfrac{\sec\theta - 1}{\tan\theta}
  16. PS6 Q.5 (9)
    tan⁡3θ−1tan⁡θ−1=sec⁡2θ+tan⁡θ\dfrac{\tan^3\theta - 1}{\tan\theta - 1} = \sec^2\theta + \tan\theta
  17. PS6 Q.5 (10)
    sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1=1sec⁡θ−tan⁡θ\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \dfrac{1}{\sec\theta - \tan\theta}
  18. PS6 Q.6
    A boy standing at a distance of 48 meters from a building observes the top of the building and makes an angle of elevation of 30∘30^\circ. Find the height of the building.
  19. PS6 Q.7
    From the top of the light house, an observer looks at a ship and finds the angle of depression to be 30∘30^\circ. If the height of the light-house is 100 meters, then find how far the ship is from the light-house.
  20. PS6 Q.8
    Two buildings are in front of each other on a road of width 15 meters. From the top of the first building, having a height of 12 meter, the angle of elevation of the top of the second building is 30∘30^\circ. What is the height of the second building?
  21. PS6 Q.9
    A ladder on the platform of a fire brigade van can be elevated at an angle of 70∘70^\circ to the maximum. The length of the ladder can be extended upto 20 m. If the platform is 2 m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin⁡70∘=0.94)(\sin 70^\circ = 0.94)
  22. PS6 Q.10
    While landing at an airport, a pilot made an angle of depression of 20∘20^\circ. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing. (sin⁡20∘=0.342)(\sin 20^\circ = 0.342) [Note: marked ⋆\star (challenging) in the textbook.]