Question

What will be the angle of elevation of top of a 50m high tower at apoint 50m away from the football of the tower?

Answer 1

Given: Height of the tower = 50 m and the point of angle of elevation from the tower = 50 m.

To find: The angle of elevation.

Answer:

(Diagram attached below.)

$\sf tan\ C\ =\ \frac{AB}{CB}\\ \\tan\ C\ =\ \frac{50}{50}\\\\tan\ C\ =\ 1\\\\We\ know\ that\ tan\ 45 ^{\circ} \ =\ 1.\\\\\sf Therefore,\ C\ =\ 45 ^{\circ} .\\\\Hence,\ the\ angle\ of\ elevation\ =\ 45 ^{\circ} .$

Answer 2

Given data:-

• Top of tower is 50m high
• football 50m away from base of tower {according to que}

Assumption:-

• Let, AB be the tower and C be the point where football sictuated means distance of football from base of tower with 50m,which {BC = 50m}and angle CAD which is angle of elevation with top of tower.
• Let, tower perpendicular to the base. and hence, according to figure ΔABC is right angle triangle.
• Let, {according to figure} adjacent side of angle of elevation, side AD is parrellel to base BC.

Solution:-

{Accirding to figure}

{Alternate angle property}

—› Angle CAD = Angle ACB ....( 1 )

Now, to find angle of elevation from top of tower, we use trigonometric ratio.

—› tan( θ ) = {opposite side}/{adjacent side}

—› tan( θ ) = {AB}/{BC}

—› tan( θ ) = 50/50

—› tan( θ ) = 1

—› θ = 1/tan ( 1 ) or tan^ - 1 ( 1 )

—› θ = 45°

Hence, angle of elevation from top of tower is 45°.