From the top of a vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be $45^{\circ} a n d 60^{\circ}$ . If the cars are 100 m apart and are on the same side of the tower, find the height of the tower.

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Solution

$L e t A B = x, I n △ A B C, t a n 60^{o} = \frac{A B}{B C} \sqrt{3} = \frac{x}{B C} B C = \frac{x}{\sqrt{3}} I n △ A B D, t a n 45^{o} = \frac{A B}{B C + 100} 1 = \frac{x}{B C + 100} B C = x - 100 \frac{x}{\sqrt{3}} = x - 100 x = x \sqrt{3} - 100 \sqrt{3} x \sqrt{3} - x = 100 \sqrt{3} x (\sqrt{3} - 1) = 100 \sqrt{3} x = \frac{100 \sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} x = \frac{(100) \sqrt{3} \times (\sqrt{3} + 1)}{2} = 236.60 m$

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From the top of a vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be 45∘ and 60∘. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower.

Let AB=x,In △ABC,tan 60o=ABBC√3=xBCBC=x√3 In △ABD,tan 45o=ABBC+1001=xBC+100BC=x−100 x√3=x−100x=x√3−100√3x√3−x=100√3x(√3−1)=100√3x=100√3√3−1×√3+1√3+1x=(100)√3×(√3+1)2=236.60 m

From the top of a vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be $45^{\circ} a n d 60^{\circ}$ . If the cars are 100 m apart and are on the same side of the tower, find the height of the tower.