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Exercise 10(a) · Q9

Q.The angle of elevation of the top of a tower from a point on the ground is 30∘30^\circ. On walking 4040 m towards the tower, the angle of elevation becomes 60∘60^\circ. Find the height of the tower.

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Step 1. Let the height of the tower be hh and let the initial distance from the tower's base be dd. From the first observation point, tan⁡30∘=hd\tan30^\circ=\dfrac hd, so d=hcot⁡30∘=h3d=h\cot30^\circ=h\sqrt3.

Step 2. After walking 4040 m towards the tower, the new distance is d−40d-40, and tan⁡60∘=hd−40\tan60^\circ=\dfrac{h}{d-40}, so d−40=hcot⁡60∘=h3d-40=h\cot60^\circ=\dfrac{h}{\sqrt3}.

Step 3. Subtract the second equation from the first: …

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