Question

A TV tower stands vertically on a bank of a canal, with a height of 10 √3 m . 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 on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the distance between the opposite bank of the canal and the point with 30° angle of elevation.

Answer 1

Answer:

The distance between the opposite bank of the canal and the point with 30° angle of elevation is 20m

Given :

• The height of the tower is 10√3m
• Angle of elevation from the bank to the top of the tower is 60°
• From another point on the line joining this point of the foot of the tower , the angle of elevation is 30°

Formula to be used :

Here we shall use trigonometric function :

$\sf \tan \theta = \frac{ perpendicular}{base}$

Solition :

Let us consider

• the height of the tower be AB

• the point from which angle of elevation is 60° be C

• the point from which angle of elevation is 30° be D

Applying the function 'tan' in ∆ABC

$\sf{ \tan(C) = \frac{AB}{BC} } \\ \implies \sf \tan(60 \degree) = \frac{10\sqrt{3}m}{BC} \\ \implies \sf \sqrt{3} = \frac{10 \sqrt{3}m }{BC} \\ \implies \sf{BC = \frac{10 \sqrt{3}m }{ \sqrt{3} } } \\ \sf \implies BC = 10m$

And applying in the ∆ABD we have

$\sf \tan(D) = \frac{AB}{BD} \\ \sf{ \implies \tan30 \degree} = \frac{10 \sqrt{3} m}{BD} \\ \implies \sf{ \frac{1}{ \sqrt{3} } = \frac{10 \sqrt{3} m}{BD} } \\ \implies \sf{BD = 10 \times 3m} \\ \implies \sf{BD = 30m}$

Now the distance between the opposite bank of the canal and the point with 30° angle of elevation is CD:

$\sf{BD - BC} \\ = \sf30m - 10m \\ = \boxed{\boxed{ \bold{\sf 20m}}}$

Answer 2

TDW

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QUESTION :

A TV tower stands vertically on a bank of a canal, with a height of 10 √3 m .

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 on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°.

Find the distance between the opposite bank of the canal and the point with 30° angle of elevation.

SOLUTION :

For the entire solution, see the attachments.

ANSWER :

The distance between the opposite bank of the canal and the point with 30° angle of elevation is 20 m.

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