Question

The angle of elevation of the top of a TV tower from three points A, B, C in a straight line, (in a horizontal plane) through the foot of the tower are $\alpha$,$2\alpha$,$3\alpha$ respectively. If AB=a, the height of the tower is

• A. a $tan\alpha$
• B. a $sin\alpha$
• C. a $sin2\alpha$
• D. a $sin3\alpha$

Answer 1

The correct option is C a $sin2\alpha$

Given $AB=a$

Let $D$ be foot of lower in the plane

Let $BD=y$

$\therefore AD=a+y$

$\tan \alpha =\frac{h}{y+a}$

$\tan 2\alpha =\frac{h}{y}$

$\Rightarrow (y+a)\tan \alpha =y\tan 2\alpha$

$(y+a)\tan \alpha =y\left(\frac{2\tan \alpha }{1-{\tan }^2\alpha }\right)$

$\frac{2y}{1-{\tan }^2\alpha }=y+a$

$2y=(y+a)-(y+a){\tan }^2\alpha$

$y+y{\tan }^2\alpha =a(1-{\tan }^2\alpha )$

$y=a\frac{(1-{\tan }^2\alpha )}{1+{\tan }^2\alpha }$

$\Rightarrow h=(\frac{a(1-{\tan }^2\alpha )}{1+{\tan }^2\alpha }+a)\tan \alpha$

$\Rightarrow h=\frac{2a\tan \alpha }{{\sec }^2\alpha }$

$=2a\sin \alpha \cos \alpha$

$\Rightarrow h=a\sin 2\alpha$

Hence, the answer is $a\sin 2\alpha .$