A boy standing on a horizontal plane is flying a kite with a string of length $\displaystyle 60$ m, at an angle of elevation of $\displaystyle 30$°. Another boy standing on the roof of a $\displaystyle 20$ m high building, finds the angle of elevation of same kite to be $\displaystyle 45$°. If both the boys are on opposite sides of the kite, find the distance of the first boy from the base of the building. Also, find the height of the kite from the ground. (Use $\displaystyle \sqrt{3}=1.73$ )

Let K be the position of kite. Let \(\displaystyle \mathrm{KL}=\mathrm{h}, \mathrm{CP}=x \& \mathrm{~PB}=y\) In rt. angled \(\displaystyle \triangle \mathrm{KPB}\) \[\frac{\mathrm{h}+20}{60}=\sin 30^{\circ}=\frac{1}{2} \] \(\displaystyle \Rightarrow \mathrm{h}=10\) Also, \(\displaystyle \frac{\mathrm{y}}{60}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\) \(\displaystyle \Rightarrow \mathrm{y}=30 \sqrt{3}=30 \times 1.73=51.9\) In rt. angled \(\displaystyle \Delta \mathrm{KLM}\) \[\frac{10}{x}=\tan 45^{\circ}=1 \] \(\displaystyle \Rightarrow x=10\) Height of kite from ground is \(\displaystyle 10+20=30 \mathrm{~m}\) Distance of first boy from the base of the building \(\displaystyle =51.9+10=61.9 \mathrm{~m}\)