A man observes the top of a pole with an angle of elevation of 45°. He walks 15 m towards the pole, and the angle of elevation becomes 60°. What is the height of the pole?
\(\dfrac{15(3+\sqrt{3})}{2}\) m
Let the height of the pole be \(h\) and let the nearer point (after walking) be at horizontal distance \(x\) from the foot of the pole. The farther point is then at distance \(x+15\).
From the nearer point the elevation is 60°: \(\tan 60^\circ=\dfrac{h}{x}\), so \(h=x\tan 60^\circ=\sqrt{3}\,x\).
From the farther point the elevation is 45°: \(\tan 45^\circ=\dfrac{h}{x+15}\), so \(h=x+15\) (since \(\tan 45^\circ=1\)).
Equating the two expressions for \(h\): \(\sqrt{3}\,x=x+15\), giving \(x(\sqrt{3}-1)=15\), so \(x=\dfrac{15}{\sqrt{3}-1}\).
Then \(h=\sqrt{3}\,x=\dfrac{15\sqrt{3}}{\sqrt{3}-1}\). Rationalise by multiplying numerator and denominator by \((\sqrt{3}+1)\): \(h=\dfrac{15\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\dfrac{15\sqrt{3}(\sqrt{3}+1)}{3-1}=\dfrac{15\sqrt{3}(\sqrt{3}+1)}{2}\).
Hence, the height of the pole is \(\dfrac{15(3+\sqrt{3})}{2}\) m
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