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?
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 $\angle A P B>\angle A Q B$.
$\Rightarrow \quad \angle A B P<\angle A B Q$
So station
is 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 $10(\sqrt{3}-1) \mathrm{km}$.