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Home » A TV tower stands vertically on a bank of canal. Form a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is . From another point away from the point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is . Find the height of the tower and the width of the canal.
A TV tower stands vertically on a bank of canal. Form 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 \mathrm{~m} away from the 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.
CBSE | Class 10 | Maths | RS Aggarwal | Some Applications Of Trigonometry
A TV tower stands vertically on a bank of canal. Form 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 \mathrm{~m} away from the 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.

Let \mathrm{PQ}=\mathrm{h} \mathrm{m} be the height of the \mathrm{TV} tower and \mathrm{BQ}=\mathrm{x} \mathrm{m} be the width of the canal. We have,

\mathrm{AB}=20 \mathrm{~m}, \angle \mathrm{PAQ}=30^{\circ}, \angle \mathrm{BQ}=\mathrm{x} and \mathrm{PQ}=\mathrm{h}

In \triangle \mathrm{PBQ}

\tan 60^{\circ}=\frac{\mathrm{PQ}}{\mathrm{BQ}}

\Rightarrow \sqrt{3}=\frac{\mathrm{h}}{\mathrm{x}}

\Rightarrow \mathrm{h}=\mathrm{x} \sqrt{3}

Again in \triangle \mathrm{APQ} \tan 30^{\circ}=\frac{\mathrm{PQ}}{\mathrm{AQ}}

\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{\mathrm{AB}+\mathrm{BQ}}

\Rightarrow \frac{1}{\sqrt{3}}=\frac{x \sqrt{3}}{20+3}

\Rightarrow 3 \mathrm{x}=20+\mathrm{x}

\Rightarrow 3 x-x=20

\Rightarrow 2 \mathrm{x}=20

\Rightarrow \mathrm{x}=\frac{20}{2}

\Rightarrow \mathrm{x}=10 \mathrm{~m}

Substituting \mathrm{x}=10 in (i), we get

\mathrm{h}=10 \sqrt{3} \mathrm{~m}

=> the height of the TV tower is 10 \sqrt{3} \mathrm{~m} and the width of the canal is 10 \mathrm{~m}.

i

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