The angle of elevation of a cloud at a height $h$ above the level of wager in a lake is $α$ and the angle of depression of its image in the lake is $β$ . Prove that the height of the cloud above the surface of the lake is
$\frac{h ( tan β + tan α )}{tan β - tan α}$ or $\frac{h sin ( α + β )}{sin ( β - α )}$
or $\frac{h ( cot α - cot β )}{cot α - cot β}$

Question

The angle of elevation of a cloud at a height

h

above the level of wager in a lake is

α

and the angle of depression of its image in the lake is

β

. Prove that the height of the cloud above the surface of the lake is

\frac{h ( tan β + tan α )}{tan β - tan α}

or

\frac{h sin ( α + β )}{sin ( β - α )}

or

\frac{h ( cot α - cot β )}{cot α - cot β}

Filo · Accepted Answer

M is the image of L in the water so that BL=BM.p is a point at height h above level of water at which the angle of elevation of L is α and the angle of depression of M is β. We have to find BL=x, say.Let AB=PR=yLR=ytanα, MR=ytanβ∴x−hx+h​=ytanα  ytanβ  ​=tanα  tanβ  ​.Apply comp. and divi.2h2x​=tanβ −tanα  tanβ +tanα  ​x=htanβ −tanα  tanβ +tanα  ​ ...(1)Another form. If we divide above and below of (1) by tanαtanβ, thenx=hcotβ −cotα  cotβ +cotα  ​If we change in terms of sin and cos, then from (1)x=hsin(β−α )  sin(β−α )  ​Note : You can do it by m−n theorem with θ=90o, m=MR= x+h,n=RL=x−h.

Question Text	The angle of elevation of a cloud at a height $h$ above the level of wager in a lake is $α$ and the angle of depression of its image in the lake is $β$ . Prove that the height of the cloud above the surface of the lake is $\frac{h ( tan β + tan α )}{tan β - tan α}$ or $\frac{h sin ( α + β )}{sin ( β - α )}$ or $\frac{h ( cot α - cot β )}{cot α - cot β}$
Answer Type	Text solution:1
Upvotes	151

Text solutionVerified

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