Question

from the top of the tower of 60 M high the angle of depression to the top and bottom of the pole is 30 degree and 60 degree respectively find the height of the tower​

Answer 1

height of the tower is 60- 20

=40m

Answer 2

$\huge\underline\mathbb {SOLUTION:-}$

CD is the height of the lamp and AB is the height of building.

From the figure we consider

$\mathsf {Tan\:30° = AE/EC}$

$\implies$EC = $\sqrt{3} \: \: \: EC$

$\mathsf \red {And,}$

$\mathsf {Tan\:60° = AB/BD}$

$\implies$AB = $\sqrt{3} \: BD \: = \sqrt{3} \: EC \:$ (EC = BD)

$\implies$ EC = $\frac{AB}{ \sqrt{3} } \: = \frac{60}{ \sqrt{3} } = 20 \sqrt{3}$$\implies$AE = 20 m

$\implies$EB=AB - AE = 60 - 20 = 40 m

• Therefore, Horizontal distance Between the building and the lamp Post is BD =$20 \sqrt{3} \: \: m$

$\underline \mathsf \blue {Height\:of\:lamp\: post\:=\:40\:m.}$