# 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

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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

by (21.6k points)

(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)}$.