the horizontal distance between two towers is 120 m the angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is 30 and 60 respectively find the height of two towers.
Answers
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}Tanθ=
Base
Perpendicular
Tan30^{\circ} = \frac{DC}{EC}Tan30
∘
=
EC
DC
\frac{1}{\sqrt{3}}= \frac{DC}{120}
3
1
=
120
DC
\frac{1}{\sqrt{3}} \times 120=DC
3
1
×120=DC
69.28=DC69.28=DC
Now In ΔAEB
Using trigonometric ratio
Tan\theta = \frac{Perpendicular}{Base}Tanθ=
Base
Perpendicular
Tan20^{\circ} = \frac{AE}{AB}Tan20
∘
=
AB
AE
0.3639= \frac{AE}{120}0.3639=
120
AE
0.3639 \times 120= AE0.3639×120=AE
43.66= AE43.66=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.