Question

A fire in the building is reported to 2 diff fire stations p and q.pobserves the fire is on an angle of 60 deg.and q observes it on 45°.if both stations are 20 km apart then find the dist. Which team p will have to cover to reach the tower.

Answer 1

Let AB be the building of height h, Clearly $\angle APB>\angle AQB$.
$\Rightarrow \angle APB>\angle ABQ$
$\Rightarrow AP<AQ$
$\Rightarrow$ Station P is nearer to the building
So, station P must send its team.
In $\Delta APB$, we have
$tan{60}^{\circ }=\frac{AB}{AP}\Rightarrow \sqrt{3}=\frac{h}{AP}\Rightarrow AP=\frac{h}{\sqrt{3}}$
In $\Delta QAB$, we have
$tan{45}^{\circ }=\frac{AB}{AQ}\Rightarrow 1=\frac{h}{AQ}\Rightarrow AQ=h$
Now, PQ=20 km
$\Rightarrow AP+AQ=20\Rightarrow \frac{h}{\sqrt{3}}+H=20\Rightarrow \frac{h+\sqrt{3}h}{\sqrt{3}}=20\Rightarrow h(1+\sqrt{3})=20\sqrt{3}\Rightarrow h=\frac{20\sqrt{3}}{1+\sqrt{3}}\Rightarrow h=\frac{20\sqrt{3}}{1+\sqrt{3}}\times \frac{1-\sqrt{3}}{1-\sqrt{3}}\Rightarrow h=\frac{20\sqrt{3}-60}{1-3}\Rightarrow h=30-10\sqrt{3}\Rightarrow h=10(3-\sqrt{3})\Rightarrow h=12.68km$
Now, in triangle APB,
$sin{60}^{\circ }=AB/BP$
$\frac{\sqrt{3}}{2}=\frac{h}{BP}$
This gives h= 14.64 km