Question

Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be ${30}^{\circ }$ and ${38}^{\circ }$ respectively. Find the distance between them, if the height of the tower is 50 m. [$\tan {38}^{\circ }=0.7813$] [2 MARKS]

Answer 1

Two persons A and B are standing on the opposite side of the tower TR and height of tower TR = 50 m and angles of elevation with A and B are ${30}^{\circ }$ and ${38}^{\circ }$ respectively. Let AR = x and RB = y.

Now in right $\Delta TAR$

$tan\theta =\frac{TR}{AR}\Rightarrow tan{30}^{\circ }=\frac{50}{x}$

$\frac{1}{\sqrt{3}}=\frac{50}{x}$

$\therefore x=50\sqrt{3}m=86.60m$

Again in right $\Delta TRB$,

$tan{38}^{\circ }=\frac{50}{y}\Rightarrow ytan{38}^{\circ }=50$

$y=\frac{50}{tan{38}^{\circ }}=\frac{50}{0.7813}=63.99or64.00m$

$\therefore$ Distance between A and B

$=x+y=86.60+64.00=150.60m=150.6m$