Question

A tower is built on a river of width 80 m. One can see the top of the tower at an angle of elevation of ${55}^{\circ }$ and ${65}^{\circ }$ from either banks of the river. The height of tower measured from the water level is $[Tan{65}^{\circ }=2.14,Tan{55}^{\circ }=1.43]$

• A. 58.3 m
• B. 76.2 m
• C. 52.4 m
• D. 68.57 m

Answer 1

Let CD be the tower and A, B be the banks of the river. Let $AD=x,DB=(80-x)$

In $\Delta ACD$

$Tan{65}^{\circ }=\frac{CD}{AD}=\frac{CD}{x}$

$CD=x.Tan{65}^{\circ }=x\times 2.14=2.14x...\left(i\right)$

In $\Delta DBC$

$Tan{55}^{\circ }=\frac{CD}{BD}=\frac{CD}{(80-x)}$

$CD=(80-x).\left(1.43\right)=114.4-1.43x...\left(ii\right)$

From (i) $\&$ (ii)

$2.14x=114.4-1.43x$

$3.57x=114.4$

$x=32.04m$

From (i) $CD=2.14x$

$=2.14\times 32.04$

$CD=68.57$

So, height of tower from the sea level = 68.57 m