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Exercise 10(a) · Q11

Q.A man observes the angle of elevation of the top of a tower to be 45∘45^\circ. He walks 3030 m nearer to the tower and observes the angle of elevation to be 60∘60^\circ. Find the height of the tower.

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Step 1. Let hh be the height and dd the initial distance. From tan⁡45∘=hd=1\tan45^\circ=\dfrac hd=1, we get d=hd=h.

Step 2. After walking 3030 m nearer, tan⁡60∘=hd−30\tan60^\circ=\dfrac{h}{d-30}, so d−30=h3d-30=\dfrac{h}{\sqrt3}.

Step 3. Substitute d=hd=h:

h−30=h3 ⟹ h−h3=30 ⟹ h(3−13)=30.h-30=\frac{h}{\sqrt3}\ \Longrightarrow\ h-\frac h{\sqrt3}=30\ \Longrightarrow\ h\left(\frac{\sqrt3-1}{\sqrt3}\right)=30.

Step 4. Solve for hh: …

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