The angle of elevation of the top of the tower from a point on the ground is sin−135. If the point of observation is 20 m away from the foot of the tower. What is the height of the tower?

Given that θ=sin−1(35)⇒sinθ=35PQ=52−32PQ=16=4Given that θ=sin−1(35)and PQ=20 mwe know that PQ:QR=4:3i.e., 20:QR=4:3

The angle of elevation of the top of the tower from a point on the ground is sin−135. If the point of observation is 20 m away from the foot of the tower. What is the height of the tower?

Given that θ=sin−1(35)⇒sinθ=35PQ=52−32PQ=16=4Given that θ=sin−1(35)and PQ=20 mwe know that PQ:QR=4:3i.e., 20:QR=4:3

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The angle of elevation of the top of the tower from a point on the ground is $\sin^{- 1} (\frac{3}{5})$ . If the point of observation is $20 m$ away from the foot of the tower. What is the height of the tower?

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$9 m$

$18 m$

$15 m$

$12 m$

answer is C.

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Detailed Solution

Given that $θ = \sin^{- 1} (\frac{3}{5})$

$\Rightarrow \sin θ = \frac{3}{5}$

$P Q = \sqrt{5^{2} - 3^{2}}$

$P Q = \sqrt{16} = 4$

Given that $θ = \sin^{- 1} (\frac{3}{5})$

and $P Q = 20 m$

we know that $P Q : Q R = 4 : 3$

i.e., $20 : Q R = 4 : 3$

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The angle of elevation of the top of the tower from a point on the ground is sin−135. If the point of observation is 20 m away from the foot of the tower. What is the height of the tower?

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The angle of elevation of the top of the tower from a point on the ground is $\sin^{- 1} (\frac{3}{5})$ . If the point of observation is $20 m$ away from the foot of the tower. What is the height of the tower?