Two observers 1at point and oter is at Q were 500m apart when they observe the flash of am rnemy gun at R. If angle RPQ and angle RQP were 45 & 60respectively then show how far was each observer from the enemygun.

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Please find below the solution to the asked query:

We form our diagram from given information , As :

Here we draw a perpendicular from point " R "  to PQ that intersect at " M  " . 

Let PM  = x  m so QM =  500 - x  m  (  As given P and Q are 500 m apart )

We know : $\mathrm{\theta } = \frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$ , So

In triangle APM we get

$$\begin{gathered} \tan 45° = \frac{\mathrm{RM}}{\mathrm{PM}} \\ \Rightarrow 1 = \frac{h}{x} ( \mathrm{we} \mathrm{know} \tan 45° = 1 ) \\ \Rightarrow h = x --- ( 1 ) \end{gathered}$$\
And
In triangle AQM we get

$$\begin{gathered} \tan 60° = \frac{\mathrm{RM}}{QM} \\ \Rightarrow \sqrt{3} = \frac{h}{500 -x} ( \mathrm{we} \mathrm{know} \tan 60° =\sqrt{3} ) \\ \Rightarrow \sqrt{3} = \frac{h}{500 -h} ( \mathrm{From} \mathrm{equation} 1 ) \\ \Rightarrow 500\sqrt{3} - h\sqrt{3} = h \\ \Rightarrow 500\sqrt{3} = h + h\sqrt{3} \\ \Rightarrow 500\sqrt{3} =h\left(1 + \sqrt{3}\right) \\ \Rightarrow h=\frac{500\sqrt{3}}{\left(1 + \sqrt{3}\right)} \\ \Rightarrow h=\frac{500\sqrt{3}}{\left(1 + \sqrt{3}\right)} \times \frac{\left( \sqrt{3} - 1\right)}{\left( \sqrt{3} - 1\right)} \\ \Rightarrow h=\frac{1500 - 500\sqrt{3}}{\sqrt{3} - 1 +\hspace{0.17em}3 - \sqrt{3}} \\ \Rightarrow h=\frac{500\left(3 - \sqrt{3}\right)}{2} \\ \Rightarrow h=250\left(3 - \sqrt{3}\right) \\ \Rightarrow h=250\left(3 - 1.732\right) ( \mathrm{we} \mathrm{know} \sqrt{3} \approx 1.732 ) \\ \Rightarrow h=250 \times 1.268 \\ \mathbf{\Rightarrow }\mathit{h}\mathbf{=}\mathbf{317} \end{gathered}$$
So,
h  =  x  =  317 m  and QM  =  500 - 317 = 183 m 

Now we apply Pythagoras theorem in triangle APM and get :

PR² =  RM² + PM²  , Substitute values we get

PR² =  317² + 317²

PR² = 100489 + 100489

PR² = 200978

PR = 448.30

And we apply Pythagoras theorem in triangle AQM and get :

QR² =  RM² + QM²  , Substitute values we get

QR² =  317² + 183²

QR² = 100489 + 33489

QR² = 133978

QR = 366.03

Therefore,

P's distance from enemy gun is 448.30 m and Q's distance from enemy gun is 366.03 m                               ( Ans )


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