Question

The angle of elevation of the top of a

tower from the top and foot of a pole of
height 10 m are 30° and 60°. Then the
height of the tower is

Answer 1

15 m

Step-by-step explanation:

Let AB and CD be the pole and tower respectively. Let CD=h Then ∠DAC=60 ∘ and∠DBE=30 ∘ Now CA CD ​ =tan60 ∘ = 3 ​ ∴CD= 3 ​ CA⇒ 3 ​ h ​ =CA Again BE DE ​ =tan30 ∘ = 3 ​ h ​ ∴(h−10)= 3 ​ BE ​ = 3 ​ CA ​ ...[∵BE=CA] = 3 ​ h/ 3 ​ ​ = 3 h ​ ⇒3h−30=h⇒2h=30⇒h=15

Hence, height of the power =15 m