A man is a watching a bulb glow which is 3 meters above him. If he is standing 3 m away from the point directly below the bulb, what is the angle of elevation in degrees?
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Solution

From the statement we know that the height of the bulb from the man’s head is 3 meters and that he is standing 3 meters away from the bulb.

Let us imagine the bulb and the man is in a 2D plane. Now let us name the position of the bulb as ‘A’ and the position of the man’s eye as ‘C’.

Now imagine line horizontal to C and a vertical line from A. let’s imagine the point where these 2 lines intersect to be ‘B’.

Since, the bulb is 3 meters above the man’s head, AB = 3 meters

And, he is standing 3 meters away from the bulb, i.e. BC = 3 meters

Consider the right angled triangle, $△$ ABC,

$∠$ ABC = $90^{\circ}$

Let, angle of elevation is $θ$ .

Then, $tan$ $θ$ = $\frac{A B}{B C}$ = $\frac{3}{3}$ = 1

Therefore, $θ$ = $45^{\circ}$

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A man is a watching a bulb glow which is 3 meters above him. If he is standing 3 m away from the point directly below the bulb, what is the angle of elevation in degrees? ___

A man is a watching a bulb glow which is 3 meters above him. If he is standing 3 m away from the point directly below the bulb, what is the angle of elevation in degrees?
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