Two straight roads O A and O B intersect at O . A tower is situated within the angle formed by them and subtends angles of 45 o and 30 o at the points A and B where the roads are nearest to it. If O A = 400   m e t e r s and O B = 300   m e t e r s , then the height of the tower is

Step-by-step solution

Let, P Q be the tower of height h . Let, P A , P B be the perpendiculars from P upon O A and O B respectively. Then, ∠ P A Q = 45 o and ∠ P B Q = 3 0 o O A = 400 ,   O B = 30 0 ⇒ P A h = cot ⁡ 45 o = 1 ∴   P A = h ⇒ P B h = cot ⁡ 30 o = 3 ∴   P B = 3 h Now, O P 2 = P A 2 + O A 2 = P B 2 + O B 2 ⇒ h 2 + 400 2 = 3 h 2 + 300 2 ⇒ 2 h 2 = 400 2 - 300 2 ⇒ h = 4 2 - 3 2 2 × 100 = 50 14 meters