Question

The angles of elevation of a flying kite at three points, $A,B,C$ on a horizontal line (lying in the same vertical plane with the kite) are in the ratio $1:2:3$ if $AB=a,BC=b$ then the height of the kite is

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

Answer 1

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

Let $PQ=h,$ $\angle PAQ=\alpha$ and $BP=AB=a$

In $△PBC$

$\Rightarrow$ $\frac{BC}{\sin \alpha }=\frac{PB}{\sin 3\alpha }$

$\Rightarrow$ $\frac{b}{\sin \alpha }=\frac{a}{3\sin \alpha -4{\sin }^3\alpha }$

$\Rightarrow$ $(3-4{\sin }^2\alpha )b=a$

$\Rightarrow$ $3-4{\sin }^2\alpha =\frac{a}{b}$

$\Rightarrow$ ${\sin }^2\alpha =(3-\frac{a}{b})\frac{1}{4}$

$\Rightarrow$ ${\sin }^2\alpha =\frac{3b-a}{4b}$

$\Rightarrow$ $\sin \alpha =\sqrt{\frac{3b-a}{4b}}$ ----- ( 1 )

Now,

${\sin }^2\alpha =\frac{3b-a}{4b}$

$\Rightarrow$ ${\cos }^2\alpha =1-\frac{3b-a}{4b}$

$\Rightarrow$ $\cos \alpha =\sqrt{\frac{b+a}{4b}}$ ---- ( 2 )

In $△QPB,$

$\Rightarrow$ $\frac{PQ}{BP}=2\sin \alpha \cos \alpha$

$\Rightarrow$ $h=2a\times \sqrt{\frac{3b-a}{4b}}\times \sqrt{\frac{b+a}{4b}}$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$ $h=\frac{2a\sqrt{(a+b)(3b-a)}}{4b}$

$\therefore$ $h=\frac{a}{2b}\sqrt{(a+b)(3b-a)}$