A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $$60$$°. 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 $$30$$° (see$$ Fig. $$9.12). Find the height of the tower and the width of the canal.

Question

Infinity Learn · Accepted Answer

Let AB be the tower and BC be the canal. C is the point on the other side of the canal directly opposite the tower.Let AB = h, BC = x and CD = 20From right △BAC,

co t 60^{0} = \frac{x}{h} . \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{h} . \Rightarrow x = \frac{h}{\sqrt{3}} .

From right △BAD,

$c o t 30^{0} = \frac{B D}{A B} . \Rightarrow \sqrt{3} = \frac{x + 20}{h} . \Rightarrow x = \sqrt{3} h - 20 .$
Now we have to equate the values of x.

$h \sqrt{3} - 20 = \frac{h}{\sqrt{3}} . \Rightarrow h = 10 \sqrt{3} .$
Replacing, the value of h in x we have, $x = \sqrt{3} (10 \sqrt{3}) - 20 = 10$

Therefore, the width of the canal is 10 m, and the height of the canal is

10 \sqrt{3}

m.

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