Question

the horizontal distance between two towers is 120m. the angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is30,20degree respectively. find the height of the 2 towers

Answer 1

Answer:

The height of two towers is 112.94 m and 43.66 m.

Step-by-step explanation:

Refer the attached figure

The horizontal distance between two towers is 120 m.i.e. EC = AB = 120 m

The angle of elevation of the top of the first tower as observed from the second tower is 30° i.e. ∠DEC = 30°

The angle of elevation of the bottom of the first tower as observed from the second tower is 20 degree i.e. ∠CEB = 20°

∠CEB =∠EBA = 20°(Alternate interior angles)

Now we are supposed to find the heights of the tower i.e. AE and DB

In ΔDEC

Using trigonometric ratio

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan30^{\circ} = \frac{DC}{EC}$

$\frac{1}{\sqrt{3}}= \frac{DC}{120}$

$\frac{1}{\sqrt{3}} \times 120=DC$

$69.28=DC$

Now In ΔAEB

Using trigonometric ratio

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan20^{\circ} = \frac{AE}{AB}$

$0.3639= \frac{AE}{120}$

$0.3639 \times 120= AE$

$43.66= AE$

AE = BC = 43.66

BD = BC+DC = 43.66+69.28=112.94 m

Thus the height of two towers is 112.94 m and 43.66 m.