Question

A tower subtends angles α, 2α, 3α respectively at points A, B and C all lying on a horizontal line through the foot of the tower. Then, \(\frac{AB}{BC}\) is equal to

(a) ) 1 + 2 cos 2α

(b) \(\frac{\text{sin}\,3\alpha}{\text{sin}\,2\alpha}\)

(c) 2 + cos 3α

(d) \(\frac{\text{sin}\,2\alpha}{\alpha}\)

Answer 1

(a) ) 1 + 2 cos 2α

Let the height of the tower PQ = h metres. 

Given, ∠PAQ = α, ∠PBQ = 2α, ∠PCQ = 3α.

∴ In rt. Δ PAQ, tan α = \(\frac{PQ}{AQ}\) = \(\frac{h}{AQ}\)

⇒ AQ = h cot α                 ...(i) 

In rt. Δ PBQ, tan 2α = \(\frac{PQ}{BQ}\) = \(\frac{h}{BQ}\)

⇒ BQ = h cot 2α                 ...(ii) 

In rt. Δ PCQ, tan 3α = \(\frac{PQ}{CQ}\) = \(\frac{h}{CQ}\)

⇒ CQ = h cot 3α                 ...(iii) 

∴ From (i) and (ii), AB = AQ – BQ = h cot α – h cot 2α        ...(iv) 

From (iv) and (iii), BC = BQ – CQ = h cot 2α – h cot 3α       ...(v) 

∴ From (iv) and (v), we have

= 3 – 2 (1 – cos 2α) = 1 + 2 cos 2α.