Question

A vertical pole stands on an horizontal ground. A student positioned due west of the pole observes the angle of elevation to be 40 degrees. When the student moves 10m due south, he observes the angle of elevation to be 35degrees. What is the height of the pole.

Answer 1

Step-by-step explanation:

|S450(2)

(3)

(4)1m3+1AB is a vertical pole with B at the ground level

and A at the top. A man finds that the angle of

elevation of the point A from a certain point C600.He moves away from the pole along the lineCD- 7m

.From D the angle of elevation of the point Aon the ground isBC to a point D such that.Then the height of the pole is (1)7/3

7/3

7/3

7/3V33+1m

Answer 2

Answer:

The height of the pole is 13.02 m.

Step-by-step explanation:

Given,

A student observes a pole at an angle of elevation of 40° by standing west of the pole at 'W'.

He/She then moves 10 m South to point 'S', and observes the pole at an angle of elevation of 35°.

To find,

The height of the pole (h).

Calculation,

Let the pole 'OP' be placed on the ground at point 'O'.

Then, WOP forms a triangle.

From triangle,

tan(40) = OP/WO

⇒ WO = OP/tan(40)

Again if the person has moved to point S then the triangle SOP will be formed,

So, from triangle SOP

tan(35) = OP/OS

⇒ OP = OS tan(35)..(1)

But as OP = height of the pole (h),

And as the triangle, WSO is formed by joining the points W, S, and O is a right-angled triangle with WS = 10, WO = OP/tan(40) = h/tan(40)

Then, OS = $\sqrt{WO^2+WS^2}$

OS = $\sqrt{(\frac{h^2}{\tan^2(40)}) + 10^2}$

$\implies OS^2= (\frac{h^2}{\tan^2(40)}) + 100$..(2)

Substitute equation (2) in equation (1)²

$OP^2 = h^2 = ((\frac{h^2}{\tan^2(40)}) + 100)\times (\tan^2(35))$

But tan(35) = 0.7, tan(40) = 0.83

$\implies \frac{h^2}{\tan^2(35)} = (\frac{h^2}{\tan^2(40)}) + 100$

⇒ h²/(0.7)² = (h²/0.83²) + 100

⇒ h² × 0.589 = 100

⇒ h = 13.02 m

Therefore, the height of the pole is 13.02 m.

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