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Trigonometric Ratios of Compound Angles

A chimney lea...

Question

A chimney leans towards the north. At the points, equal distances due north and south of it in a horizontal plane, the elevations of the tops are $α$ and $β$ respectively. The inclination of the top of the chimney to the vertical is

A

{tan}^{- 1} [\frac{s i n (α - β)}{sin α sin β}]

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B

{tan}^{- 1} [\frac{2 s i n (α - β)}{sin α sin β}]

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C

{tan}^{- 1} [\frac{s i n (α - β)}{2 sin α sin β}]

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D

{tan}^{- 1} [\frac{s i n (α - β)}{2 cos α cos β}]

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Solution

The correct option is C ${tan}^{- 1} [\frac{s i n (α - β)}{2 sin α sin β}]$
OC-chimney and $O A = O B$
Them, according to the $(m - n)$ theorem,'if O be the point on the side AB of a $△ A B C$ which divided the side AB in the ratio $m : n$ , then
$(O A + O B) cot (\frac{π}{2} - θ) = O A cot β - O B cot α$
$(O A + O A) cot (\frac{π}{2} - θ) = O A cot β - O A cot α$
$2 cot (\frac{π}{2} - θ) = cot β - cot α$
$2 tan θ = cot β - cot α [∵ cot (\frac{π}{2} - θ) = tan θ]$
$2 tan θ = \frac{cos β}{sin β} - \frac{cos α}{sin α} = \frac{sin α cos β - cos α sin β}{sin α . sin β}$
$tan θ = \frac{sin (α - β)}{2 sin α sin β}$
$θ = {tan}^{- 1} [\frac{sin (α - β)}{2 sin α sin β}]$

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Similar questions

Q. A chimney leans towards North. At equal distances due North and South of it in a horizontal plane, the elevations of the top are

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{tan}^{- 1} [\frac{sin (α - β)}{2 sin α sin β}]

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Q. Question 8
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