P Q = \mathrm{PQ}= PQ = Tower A = 45 \mathrm{A}=45 A = 45 metre
R S = \mathrm{RS}= RS = Tower B = 15 \mathrm{B}=15 B = 15 metre
Q S = x \mathrm{QS}=x QS = x metre ( ( ( let ) ) )
∠ P S Q = 6 0 ∘ ; ∠ R Q S = θ \angle \mathrm{PSQ}=60^{\circ} ; \angle \mathrm{RQS}=\theta ∠ PSQ = 6 0 ∘ ; ∠ RQS = θ
From Δ P Q S , tan θ 6 0 ∘ = P Q Q S \Delta \mathrm{PQS}, \tan \theta 60^{\circ}=\frac{\mathrm{PQ}}{\mathrm{QS}} Δ PQS , tan θ 6 0 ∘ = QS PQ
⇒ 3 = 45 x ⇒ 3 x = 45 \Rightarrow \sqrt{3}=\frac{45}{x} \Rightarrow \sqrt{3} x=45 ⇒ 3 = x 45 ⇒ 3 x = 45
⇒ x = 45 3 = 15 3 \Rightarrow x=\frac{45}{\sqrt{3}}=15 \sqrt{3} ⇒ x = 3 45 = 15 3 metre
From Δ R S Q , tan θ = R S Q S = 15 15 3 \Delta \mathrm{RSQ}, \tan \theta=\frac{\mathrm{RS}}{\mathrm{QS}}=\frac{15}{15 \sqrt{3}} Δ RSQ , tan θ = QS RS = 15 3 15
⇒ tan θ = 1 3 \Rightarrow \tan \theta=\frac{1}{\sqrt{3}} ⇒ tan θ = 3 1
⇒ tan θ = tan 3 0 ∘ \Rightarrow \tan \theta=\tan 30^{\circ} ⇒ tan θ = tan 3 0 ∘
⇒ θ = 3 0 ∘ \Rightarrow \theta=30^{\circ} ⇒ θ = 3 0 ∘
∴ sin θ = sin 3 0 ∘ = 1 2 \therefore \quad \sin \theta=\sin 30^{\circ}=\frac{1}{2} ∴ sin θ = sin 3 0 ∘ = 2 1