Question

Pls answer 5th question

Answer 1

Answer:

14.64 m

Step-by-step explanation:

From figure:

AB = 20m be the height of the building.

BC = h metres be the height of tower.

Given, ∠AOB = 45° and ∠AOC = 60°.

(i) From ΔAOB:

tan 45° = AB/OA

⇒ 1 = 20/OA

⇒ OA = 20 m.

(ii) From ΔAOC:

tan 60° = AC/OA

⇒ √3 = (h + 20)/20

⇒ 20√3 = h + 20

⇒ 20√3 - 20 = h

⇒ h = 20(√3 - 1).

⇒ h = 20(1.732 - 1)

⇒ h = 20(0.732)

⇒ h = 14.64 m

∴ Therefore,height of the rower = 14.64 m

Hope it helps!