Question

please answer with a figure aswell

Answer 1

HOPE IT HELPS YOU.......

Answer 2

Step-by-step explanation:

Let AB,CD be the two poles and BC be the width of the river.

∠ADE = 30° and ∠ACB = 60°, AB = 60 m.

(i) In ΔABC:

tan 60° = (AB)/BC

⇒ √3 = 60/BC

⇒ BC = (60/√3)

⇒ BC = 34.64 m

(ii) In ΔAED:

tan 30° = (AE)/AD

⇒ (1/√3) = (60 - DC)/60/√3

⇒ 60/3 = 60 - DC

⇒ -120/3 = -DC

⇒ DC = 40 m

Therefore, Width of the river = 34.64 m and height = 40 m.

Hope it helps!