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The angle of elevation of the top of a tower observed from each of 3 points A, B, C on the ground are α, 2α and 3α respectively. If AB = a, BC = b, then the height of the tower is

(a) \(\frac{a}{2b}\)\(\sqrt{(3b-a)(a+b)}\)

(b) \(\frac{a}{b}\)\(\sqrt{(3b+a)(a+b)}\)

(c) \(\frac{2a}{b}\)\(\sqrt{(3b-a)(a-b)}\)

(d) None of these

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(a) \(\frac{a}{2b}\)\(\sqrt{(3b-a)(a+b)}\)

Let the height of the tower PQ = h metres 

Given, ∠QAP = α, ∠QBP = 2α, 

∠QCP = 3α, AB = a, BC = b, CP = x (say) 

Now, in rt. ΔQAP, tan α = \(\frac{QP}{AP}\) = \(\frac{h}{a+b+x}\)

⇒ a + b + \(x\) = h cot α                        ...(i)

In rt. Δ QBP,

tan α = \(\frac{QP}{BP}\) = \(\frac{h}{b+x}\)

b + \(x\) = h cot 2α                           ...(ii)

In rt. Δ QCP,

tan 3α = \(\frac{QP}{CP}\) = \(\frac{h}{x}\) ⇒ \(x\) = h cot 3α                    ...(iii)

Eq (i) – Eq (ii)

⇒ a = h (cot α – cot 2α) = h \(\bigg[\frac{\text{cos}\,\alpha}{\text{sin}\,\alpha}-\frac{\text{cos}\,2\alpha}{\text{sin}\,2\alpha}\bigg]\)

⇒ a = \(\frac{h[\text{cos}\,\alpha\,\text{sin}\,2\alpha-\text{sin}\,\alpha\,\text{cos}\,2\alpha}{\text{sin}\,\alpha\,\text{sin}\,2\alpha}\)

⇒ a = \(\frac{h\,\text{sin}\,(2\alpha-\alpha)}{\text{sin}\,\alpha\,\text{sin}\,2\alpha}\) = \(\frac{h\,\text{sin}\,\alpha}{\text{sin}\,\alpha\,\text{sin}\,2\alpha}\)

⇒ h = a sin 2α                 ...(iv) 

Eqn (ii) – Eqn (iii)

 ⇒ b = h (cot 2α – cot 3α) = h \(\bigg[\frac{\text{cos}\,2\alpha}{\text{sin}\,2\alpha}-\frac{\text{cos}\,3\alpha}{\text{sin}\,3\alpha}\bigg]\)

⇒ a = \(\frac{h\,.\,[\text{cos}\,2\alpha\,\text{sin}\,3\alpha-\text{cos}\,3\alpha\,\text{sin}\,2\alpha}{\text{sin}\,2\alpha\,\text{sin}\,3\alpha}\)

⇒ a = \(\frac{h\,\text{sin}\,(3\alpha-2\alpha)}{\text{sin}\,2\alpha\,\text{sin}\,3\alpha}\) ⇒ b = \(\frac{h\,\text{sin}\,\alpha}{\text{sin}\,2\alpha\,\text{sin}\,3\alpha}\)

⇒ h = \(\frac{b\,\text{sin}\,2\alpha\,\text{sin}\,3\alpha}{\text{sin}\,\alpha}\)                            ....(v)

∴ From eqn (iv) and (v), we have

a sin 2α = \(\frac{b\,\text{sin}\,2\alpha\,\text{sin}\,3\alpha}{\text{sin}\,\alpha}\) ⇒ sin α = \(\frac{b}{a}\) sin 3α

⇒ sin α = \(\frac{b}{a}\) (3 sin α - 4 sin3 α) ⇒ 1 = \(\frac{3b}{a}\) - \(\frac{4b}{a}\) sin2 α

⇒ a = 3b – 4b sin2α ⇒ 4b sin2α = 3b – a

⇒ sin α = \(\sqrt{\frac{3b-a}{4b}}\) ⇒ cos α = \(\sqrt{1-\text{sin}^2\,\alpha}\)

\(\sqrt{1-\frac{3b-a}{4b}}\) = \(\sqrt{\frac{b+a}{4b}}\)

From eqn (iv) we have h = a sin 2α 

h = 2a sin a cos α

h = 2a . \(\sqrt{\frac{3b-a}{4b}}\) \(\sqrt{\frac{b+a}{4b}}\)

h = \(\frac{2a}{4b}\)\(\sqrt{(3b-a)(b+a)}\)

h = \(\frac{a}{2b}\)\(\sqrt{(3b-a)(a+b)}\).

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