Height of the object

Question

An object is observed from A, B and C lying in a straight line in the horizontal plane just below the object. The angular elevation at B is twice that at A and at C three times that at A. If $AB=a, BC=b$, then show that height of the object is $\frac{a}{2b}\sqrt{(a+b)(3b-a)}$

My try: Let foot of object P is O. Let $OC=c$

$\tan3\theta=\frac{h}{c}$

$\tan 2\theta=\frac{h}{b+c}$

$\tan\theta=\frac{h}{a+b+c}$

I am not able to find height now.

Answer 1

$tan3θ=\frac{h}{c} $

$tan2θ=\frac{h}{b+c} $

$tanθ=\frac{h}{a+b+c} $

Now, $c=\frac{h}{tan3θ} $

Substituting the value of c in the second and third equation,

$a=\frac{h}{tan2θ}-\frac{h}{tan3θ} $

$b=\frac{h}{tanθ}-\frac{h}{tan2θ} $

On substituting the values of $a$ and $b$ in the equation: $\frac{a}{2b}\sqrt{(a+b)(3b-a)}$,we find it to be equal to $h$.