Question

$A$ tower standing at point $A$ leans towards west making an angle $\alpha$ with the vertical. The angular elevation of $B$, the top most point of the tower is $\beta$ as observed from a point $C$ due east of $A$ at a distance $d$ from $A$. lf the angular elevation of $B$ from a point due east of $C$ at a distance $2d$ from $C$ is $\gamma$, then $2\tan \alpha$ can be written as

• A. 3 $\cot \beta -2\cot \gamma$
• B. $\cot \gamma -\cot \beta$
• C. $3\cot \beta -\cot \gamma$
• D. $\cot \beta -3\cot \gamma$

Answer 1

The correct option is B $\cot \gamma -\cot \beta$

Let $AB,$ be the tower leaning towards west making an angle $\alpha$ with vertical at $C$, angle of elevation of $B$ is $\beta$ $abd$at $D$ us $\gamma$

$CA=AD=d$ in $△ABH$

$\tan \alpha =\frac{AH}{h}$
$\Rightarrow AH=h\tan \alpha .........1$

In $△BCH,\tan \beta =\frac{h}{CH}$
$\Rightarrow CH=h\cot \beta$

$d-AH=h\cot \beta ,$ put the value of AH from $1$

$d-h\tan \alpha =h\cot \beta \Rightarrow d=h(\cot \beta +\tan \alpha ).......2$

In $△BDH,\tan \gamma =\frac{BH}{HD}=\frac{h}{AH+d}=AH+d=h\cot \gamma$

$\Rightarrow h\tan \alpha +d=h\cot \gamma \Rightarrow d=h(\cot \gamma -\tan \alpha )\to 3$

From $2$ and $3$ we get

$h(\cot \beta +\tan \alpha )=h(\cot \gamma -\tan \alpha )$

$2\tan \alpha =\cot \gamma -\cot \beta$