Question

The angle of 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

The correct option is B $\frac{a-b}{a+b}$

$In△CAD,\tan \varphi =\frac{AC}{b}\_\_\_\left(1\right)$

$In△CAB,\tan (90-\varphi )=\frac{AC}{a}\Rightarrow \cot \varphi =\frac{AC}{a}\_\_\_\_\left(2\right)$

From (1) and (2),

$\Rightarrow \frac{AC}{b}=\frac{a}{AC}$

$\Rightarrow {AC}^2=ab$

$\Rightarrow AC=\sqrt{ab}$

$\sin \theta =-\sin (90-2\varphi )=-\cos 2\varphi$

$=-1+2{\sin }^2\varphi$

$=-1+\frac{2a}{a+b}$

As $\sin \varphi =\sqrt{\frac{ab}{ab+b^2}}$ and ${\sin }^2\varphi =\frac{a}{a+b}$ we get $\sin \theta =\frac{a-b}{a+b}$