Question

The angle of the elevation of the top of a vertical tower from two points at distances $a$ and $b(a>b)$ from the base and in the same line with it, are complimentary. If $\theta$ is the angle subtended at the top of the tower by the line joining these points, then $\sin \theta$ is equal to:

• A. $\frac{a+b}{a-b}$
• B. $\frac{a-b}{a+b}$
• C. $\frac{(a-b)b}{a+b}$
• D. $\frac{a-b}{(a+b)b}$

Answer 1

Answer: (B)

$\frac{a-b}{a+b}$

$\tan \alpha =\frac{h}{b}$
$\cot \alpha =\frac{h}{a}$
${\tan }^2\alpha =\frac{a}{b}$
$\theta +\frac{\pi }{2}-\alpha +{\pi }_{\alpha }=\pi$ (sum of angle of $△$)
$\theta =2\alpha -\frac{\pi }{2}$
$\sin 2\theta =\sin (2\alpha -\frac{\pi }{2})$
$2-\cos 2\alpha -\cos 2\alpha =-\left(\frac{ta{n}^2\alpha }{1+ta{n}^2\alpha }\right)$
$2-\left(\frac{1-\frac{a}{b}}{1+\frac{a}{b}}\right)=\frac{a-b}{a+b}$