Question

A fire in a building B is reported on 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

Let $B$ be the building and $A$ be the fire.

$P$ and $Q$ are $2$ fire station $20km$ apart.

Let the height of $A$ be $hkm$ and $PB=xkm$ then, $BQ=(20-x)km$

Also $\angle APB={60}^o$ and $\angle AQB={45}^o$

$\Rightarrow$ In $△ABP,\tan {60}^o=\frac{AB}{PB}$

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

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

$\Rightarrow$ In $△ABQ,\tan {45}^{\circ }=\frac{AB}{BQ}$

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

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

$\Rightarrow$ $x\sqrt{3}=20-x$ [ Equationg ( 1 ) and ( 2 ) ]

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

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

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

$\Rightarrow$ $x=\frac{20}{2}(\sqrt{3}-1)=10(\sqrt{3}-1)$

$\Rightarrow$ $x=10(1.732-1)=10\times 0.732=7.32km$

$\Rightarrow$ $PB=7.32km$ and $BQ=(20-7.32)km=12.68km$

$\Rightarrow$ Since $PB<BQ$

$\therefore$ $P$ station should send its team to travel $7.32km$.