Question

$A$ pole is slightly inclined towards the east. $A$t two points due west of it at distances a and $b$, the angles of elevation of the top of the pole are $\alpha$ and $\beta$ respectively. The inclination of the pole to the horizon is

• A. $ta{n}^{-1}[\frac{a+b}{bCot\alpha -aCot\beta }]$
• B. ${\tan }^{-1}[\frac{b-a}{bCot\alpha -aCot\beta }]$
• C. ${\cos }^{-1}\left[\frac{a-b}{bCos\alpha -aCos\beta }\right]$
• D. ${\sin }^{-1}\left[\frac{a-b}{bCot\alpha -aCot\beta }\right]$

Answer 1

The correct option is B ${\tan }^{-1}[\frac{b-a}{bCot\alpha -aCot\beta }]$

Let AB be the tree,

In $△ACB$,

$\tan \theta =\frac{h}{d}\to 1$

In $△APC$,

$\tan \alpha =\frac{h}{a+d}\Rightarrow h=(a+d)\tan \alpha \to 2$

In $△AQC,\tan \beta =\frac{h}{b+d}\Rightarrow h=(b+d)\tan \beta \to 3$

From $2$ and $3$, we get

$(a+d)\tan \alpha =(b+d)\tan \beta$

$a\tan \alpha +d\tan \alpha =b\tan \beta +d\tan \beta$

$d(\tan \alpha -\tan \beta )=b\tan \beta a\tan \alpha$

$d=\frac{b\tan \beta -a\tan \alpha }{\tan \alpha -\tan \beta }$

From $2$, $h=\frac{(b-a)\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }$

$\therefore \tan \theta =\frac{(b-a)\tan \alpha \tan \beta }{b\tan \beta -a\tan \alpha }$

$\theta ={\tan }^{-1}\left[\frac{(b-a)}{b\cot \alpha -a\cot \beta }\right]$