Question

An object is observed from three points A, B and 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 at A. If AB$=a$, BC$=b$, then the height of the object is.

• A. $\frac{a}{2b}\sqrt{(a+b)\left(\right(3b-a)}$
• B. $\frac{a}{2b}\sqrt{(a-b)(3b-a)}$
• C. $\frac{a}{2b}\sqrt{(a-b)(3b+a)}$
• D. $\frac{a}{2b}\sqrt{(a+b)(3b+a)}$

Answer 1

The correct option is A $\frac{a}{2b}\sqrt{(a+b)\left(\right(3b-a)}$

$h=a\sin \theta =2a\sin \theta \cos \theta ⟶1$
In $△PBC$, by sine rule
$\frac{a}{\sin (\pi -3\theta )}=\frac{b}{\sin \theta }\therefore \frac{a}{b}=\frac{3\sin \theta -4{\sin }^3\theta }{\sin \theta }$
$\therefore {\sin }^2\theta =\frac{3b-a}{4b}\&{\cos }^2\theta =\frac{a+b}{4b}$
Putting in equation 1
$h=\frac{a}{2b}\sqrt{(a+b)(3b-a)}$