Question

Two towers stand on a horizontal plane. P and Q where PQ - 30 m, are two points on the line joining their feet. As seen from P the angle of elevation of the tops of the towers are 30 and 60 but as seen from Q are 60 and 45. The distance between the towers is equal to

• A. 15 $(4+\sqrt{3})$m
• B. 15 $(4-\sqrt{3})$m
• C. 15 $(3+\sqrt{3})$m
• D. 15 $(2+\sqrt{3})$m

Answer 1

The correct option is A 15 $(4+\sqrt{3})$m

Also $PQ=30m$
If h and H be the heights of the towers
$PQ=30=AP-AQ=h\cot {30}^0-h\cot {60}^0$
30 = h[$\sqrt{3}$ -(1/$\sqrt{3}$)]
$h=15\sqrt{3}$
Also $PQ=30=BQ-BP$
=H $\cot {45}^0$ - H $\cot {60}^0$ = H$\frac{\sqrt{3}-1}{\sqrt{3}}$
H = $\frac{30\sqrt{3}}{\sqrt{3}-1}=\frac{30\sqrt{3}(\sqrt{3}+1)}{2}=15(3+\sqrt{3})$
Hence the distance between the towers is
AB = AQ + BQ = h cot ${60}^0$ + H cot ${45}^0$
=(h/$\sqrt{3}$) + H = $\frac{15\sqrt{3}}{\sqrt{3}}+15(3+\sqrt{3})$
$AB=15+45+15$$\sqrt{3}$ = $60+15$$\sqrt{3}$ = $15(4+\sqrt{3}$)