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  • A man is on top of tower of 50 meter hieght watching the a small pole, the angle of depression of bottom and top of a pole are A and B respectively, then find the hieght of pole ?<div clas

A man is on top of tower of 50 meter hieght watching the a small pole, the angle of depression of bottom and top of a pole are A and B respectively, then find the hieght of pole ?

Option: 1

50\frac{\cot B+\cot A}{\cot B}]


Option: 2

50[\frac{\cot B- \cot A}{\cot B}]


Option: 3

50[\frac{\cot A- \cot B}{\cot B}]


Option: 4

50[\frac{\cot B- \cot A}{\cot A}]


Answers (1)

best_answer

 

 

Height and Distance -

Height and Distance-

Angle of Elevation

 

The student is looking at the top of the tower. The line AC drawn from the eye of the student to the top of the tower is called the line of sight. The angle BAC, so formed by the line of sight with the horizontal is called the angle of elevation of the top of the tower from the eye of the student.

 

If the object is above the observer’s eye level, the angle between the horizontal line and the line of sight is called the angle of elevation of the object. The horizontal line is taken in the same vertical plane as that of the observer and the object.

Here α is the angle of elevation. 

 

Angle of Depression

If the object is below the observer’s eye level, the angle between the horizontal line and the line of sight is called the angle of depression of the object. The horizontal line is taken in the same vertical plane as that of the observer and the object.

Here β is the angle of depression.

 

\\\mathrm{\tan A=\frac{BC}{AB}\;\;or\;\;\cot A=\frac{AB}{BC}}

 

which on solving would give us BC. By adding AE to BC, you will get the height of the tower.

-

 

 

Diagram-

 

In \Delta RST \\ \cot B=\frac{x}{50-h}\\ In \Delta PQT \\ \cot A=\frac{x}{50}\\ \frac{\cot A}{\cot B}=\frac{50-h}{50}\\ h=50\frac{\cot B- \cot A}{\cot B}

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rishi.raj

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