Q) The angles of depression of the top and the bottom of a 50 m high building from the top of a tower are 45° and 60°, respectively. Find the height of the tower. (Use √3 = 1·73)

Ans: 

Let’s start with the diagram for this question:

Here we have tower AB is the tower and PQ is a building of 50 m height.

Angles of depression from A to P and Q are given as 45⁰ and 60⁰.

We need to find the height H of tower AB.

Let’s make a simplified diagram of the same for our better understanding:

Step 1: In Δ ACP, tan P =

Since AC = AB – CB = AB – PQ = H – 50;

PC = QB = D

and ∠ P = 45⁰

∴ tan 45 =

∴ 1 =

∴ D = H – 50 …………. (i)

Step 2: In Δ ABQ, tan Q = tan 60° =

∴ √ 3 =

∴ D = …………. (ii)

Step 3: By comparing equations (i) and (ii), we get:

H – 50 =

∴ √3 (H – 50) = H

∴ H √3 – 50 √3 = H

∴ H √3 – H = 50 √3

∴ H (√3 – 1) = 50 √3

∴ H (1.73 -1) = 50 x 1.73

∴ H (0.73) = 86.5

∴ H = = 118.5 m

Therefore, the height of the tower is 118.5 m.

Please press the “Heart” button if you like the solution.