A TV tower stands vertically on a bank of canal From a

A TV tower stands vertically on a bank of canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60^{\circ}$ . From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $30^{\circ}$ . Find the height of the tower and the width of the canal.

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SOLUTION

In $△ A B C$

$\frac{A B}{B C} = t a n 60^{\circ}$ .

$\frac{A B}{B C} = \sqrt{3}$

$∴ B C = \frac{A B}{\sqrt{3}}$

In $△ A B D$

$\frac{A B}{B D} = t a n 30^{\circ}$ .

$\frac{A B}{B C + C D} = \frac{1}{\sqrt{3}}$

$I n Δ A B D, \frac{A B}{B D} = t a n 30^{o} \frac{A B}{B C + C D} = \frac{1}{\sqrt{3}} \frac{A B}{\frac{A B}{\sqrt{3}} + 20} = \frac{1}{\sqrt{3}} \frac{A B \sqrt{3}}{A B + 20 \sqrt{3}} = \frac{1}{\sqrt{3}} 3 A B = A B + 20 \sqrt{3} 2 A B = 20 \sqrt{3} A B = 10 \sqrt{3} m B C = \frac{A B}{\sqrt{3}} = \frac{10 \sqrt{3}}{\sqrt{3}} = 10 m$

Therefore, the height of the tower is $10 \sqrt{3}$ m and the width of the canal is 10 m

A TV tower stands vertically on a bank of canal From a

Right Answer is:

SOLUTION

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