Question

From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60 o and the angle of depression of the foot of the tower is 30 o . find the height of the tower

Answer 1

Answer: 7(√3+1)

Explanation is in the attachment

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Answer 2

Answer:

27.99 m.

Step-by-step explanation:

Refer the attached figure

Height of building = AB = 7 m

Height of tower = EC=ED+DC=ED+7

The angle of elevation of the top of a cable tower i.e.∠EBC =60°

The angle of depression of the foot of the tower i.e. ∠DBC = 30°

Now using alternate interior angle property

∠DBC = ∠BCA=30°

In ΔABC

Using trigonometric ratio

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan30^{\circ} = \frac{AB}{AC}$

$AC= \frac{7}{\frac{1}{\sqrt{3}}}$

$AC=12.124$

AC = BD = 12.124 m

Now In ΔBED

Using trigonometric ratio

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan60^{\circ} = \frac{ED}{BD}$

$Tan60^{\circ} = \frac{ED}{12.124}$

$\sqrt{3} \times 12.124 = ED$

$20.99 = ED$

So, Height of tower = EC=ED+DC=20.99+7=27.99 m.

Hence the height of the tower is 27.99 m.