Find the angle of elevation if an object is at a height of $50 m$ on the tower and the distance between the observer and the foot of the tower is $50 m$ .

Open in App

Solution

Let the angle of elevation be $^{'} θ^{'}$
Now, In $△ A B C,$
$\Rightarrow tan θ = \frac{A B}{B C} = \frac{50}{50}$
$\Rightarrow tan θ = 1$

$∴ θ = 45 °$
Hence, the answer is $45 ° .$

Suggest Corrections

Similar questions

Q. Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be

30^{\circ}

and

38^{\circ}

respectively. Find the distance between them, if the height of the tower is 50 m. [2 MARKS]

Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be $30^{\circ}$ and $38^{\circ}$ respectively. Find the distance between them, if the height of the tower is 50 m. [ $tan 38^{\circ} = 0.7813$ ] [2 MARKS]

Question 9
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Join BYJU'S Learning Program

Find the angle of elevation if an object is at a height of 50m on the tower and the distance between the observer and the foot of the tower is 50m.

Let the angle of elevation be ′θ′Now, In △ABC,⇒tanθ=ABBC=5050⇒tanθ=1∴θ=45°Hence, the answer is 45°.

Find the angle of elevation if an object is at a height of $50 m$ on the tower and the distance between the observer and the foot of the tower is $50 m$ .

Let the angle of elevation be $^{'} θ^{'}$
Now, In $△ A B C,$
$\Rightarrow tan θ = \frac{A B}{B C} = \frac{50}{50}$
$\Rightarrow tan θ = 1$

$∴ θ = 45 °$
Hence, the answer is $45 ° .$