Question

A fire in building B is reported on the telephone to two fire stations P and Q, 20 km apart from each other on a straight road. P observes that the fire is at an angle of ${60}^{\circ }$ to the road and Q observes that it is at an angle of ${45}^{\circ }$ to the road. Which station should send its team and how much will this team have to travel?

Answer 1

According to question.

Let B be the building and A be the fire.

Given : P and Q are 2 fire stations 20 km apart.

Let the height of AB be h km

and PB = x km. then BQ = (20 – x) km

also ∠APB = 60° and ∠AQB = 45°

$In△ABP$

$tan60=\frac{AB}{PB}$

$\sqrt{3}=\frac{h}{x}$

$h=x\sqrt{3}$----(1)

$In△ABQ$

$tan45=\frac{AB}{BQ}$

$1=\frac{h}{20-x}$

$h=20-x$----(2)

Equating (1) and (2),

$x\sqrt{3}=20-x$

$x\sqrt{3}+x=20$

$x(\sqrt{3}+1)=20$

$x=\frac{20}{\sqrt{3}+1}$

$x=\frac{20(\sqrt{3}-1)}{(\sqrt{3}+10)(\sqrt{3}-1)}$

$x=\frac{20(\sqrt{3}-1)}{3-1}$

$x=10(\sqrt{3}-1)$

$x=10(1.732-1)=7.32$

⇒PB = 7.32 km and BQ = (20 – 7.32) km = 12.68 km

since PB < BQ i.e.,

Distance from P < Distance from Q

∴ P station should send its team to travel 7.32 km.