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

Let Q=kite, A=first point observation, B=second point observation, C= third point of observation.  Given AB =a, BC=b.  Let PQ = h. Since exterior angle = sum of the opposite interior angles,

$\angle PBQ=\angle BQA+\angle BAQ$    $and \angle PCQ=\angle CBQ+\angle CQB$

$\therefore AB=a=QB$

$Applying \sin e rule$$in \Delta BQC$, $we get \frac{BQ}{\sin \angle QCB}=\frac{BC}{\sin \angle CQB}$

$\Rightarrow \frac{a}{\sin \left({180}^0-3\theta \right)}=\frac{b}{\sin \theta }$

$$\begin{gathered} \Rightarrow 4{\sin }^2\theta =3-\frac{a}{b}=\frac{3b-a}{b} \\ \Rightarrow {\sin }^2\theta =\frac{3b-a}{4b} \end{gathered}$$

$\therefore {\cos }^2\theta =1-{\sin }^2\theta =1-\frac{3b-a}{4b}=\frac{4b-3b+a}{4b}=\frac{b+a}{4b}$

$From PBQ$, 

$$\begin{gathered} \sin 2\theta =\frac{h}{a}\Rightarrow 2\sin \theta \cos \theta =\frac{h}{a} \\ \Rightarrow 4{\sin }^2\theta {\cos }^2\theta =\frac{h^2}{a^2} \end{gathered}$$

$$\begin{gathered} \Rightarrow 4.\frac{3b-a}{4b}.\frac{a+b}{4b}=\frac{h^2}{a^2} \\ \Rightarrow h^2=\frac{a^2}{4b^2}\left(3b-a\right)\left(a+b\right) \end{gathered}$$

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