Question

from the top of a tower of height 50 M the angle of depression of the top and bottom of a pole are 30° and 45° respectively find how far the pole is from the bottom of the tower and the height of the pole

Answer 1

Answer:

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Answer 2

Concept:

This question requires some basic trigonometric equations.

The angle of depression is the angle between the horizontal line and the observation of the object from the horizontal line.

Given:

Height of tower AB= 50m

Angle of depression of the top =30°

Angle of depression of the bottom =45°

To find:

The height of tower CD.

Solution:

From Triangle ΔAOC

We can understand that

OC=50 m     [Since ΔAOC≅ΔABC]

TanA=$\frac{OC}{OA}$

⇒Tan 45°=$\frac{50}{OA}$

⇒$1=\frac{50}{OA}$

⇒OA=50m

Consider ΔAOD,

Tan 30°=$\frac{OD}{OA}$

⇒$\frac{1}{\sqrt{3} }=\frac{OD}{OA}$

⇒$OD=\frac{50}{\sqrt{3} }m$

Therefore,

The length of the pole CD

CD=OC-OD

⇒$CD=50-\frac{50}{\sqrt{3} }$

⇒$CD=\frac{50\sqrt{3}-50 }{\sqrt{3} }$

⇒$CD=\frac{50(\sqrt{3}-1) }{\sqrt{3} }=\frac{50(\sqrt{3}-1) }{\sqrt{3} }*\frac{\sqrt{3} }{\sqrt{3} } =\frac{50(3-\sqrt{3} )}{3} m$

Therefore, length of pole is $\frac{50(3-\sqrt{3} )}{3} m$