Question

An object is observed from three points A, B, C in the same horizontal line passing through the base of the object. The angle of elevation at B is twice and at C thrice that A. If AB = a, BC = b, prove that the height of the object is
$\frac{a}{2b}\sqrt{(a+b)(3b-a)}$

Answer 1

Let PD = h and $\theta$ be the angle subtended at A where $\theta$ is unknown. Clearly
$AB=PB=a$ from isosceles $△ABP$
$h=asin2\theta =2asin\theta cos\theta$ from $\Delta PBD$ ....(1)
(Also by sine rule on $\Delta PBC$)
$\frac{a}{sin(\pi -3\theta )}=\frac{b}{sin\theta }$
$\therefore \frac{a}{b}=\frac{3sin\theta -4si{n}^3\theta }{sin\theta }\therefore si{n}^2\theta =\frac{3b-a}{4b}$
$\therefore co{s}^2\theta =\frac{a+b}{4b}$ put in the above (1),
we get
$h=\frac{a}{2b}\sqrt{(a+b)(3b-a)}$