A fire in a building B is reported on telephone to two fire stations P and Q,

Question:

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

Solution:

Let AB be the building of height h. P Observes that the fire is at an angle of 60° to the road and Q observes that the fire is at an angle of 45° to the road.\

Let $Q A=x, A P=y .$ And $\angle B P A=60^{\circ}, \angle B Q A=45^{\circ}$, given $P Q=20$.

Here, clearly $\angle A P B>\angle A Q B$.

$\Rightarrow \quad \angle A B P<\angle A B Q$

$\Rightarrow \quad A P

So stationis near to the building. Hence station P must send its team.

We sketch the following figure

So we use trigonometric ratios.

In ΔPAB

$\tan P=\frac{A B}{A P}$

$\Rightarrow \tan 60^{\circ}=\frac{h}{y}$

$\Rightarrow \sqrt{3}=\frac{h}{y}$

$\Rightarrow h=\sqrt{3} y$

Again in $\triangle Q A B$,

$\Rightarrow \quad \tan Q=\frac{A B}{Q A}$

$\Rightarrow \quad \tan 45^{\circ}=\frac{h}{x}$

$\Rightarrow \quad 1=\frac{h}{x}$

$\Rightarrow \quad x=h$

Now,

$x+y=20$

$\Rightarrow h+y=20 \quad[\because x=h]$

$\Rightarrow \sqrt{3} y+y=20 \quad[\because h=\sqrt{3} y]$

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

Hence, the team from station $\mathrm{P}$ will have to travel $10(\sqrt{3}-1) \mathrm{km}$.

 

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