Question

at a point 50m away from the base of tower an observer measures the angle of elevation of the top of the tower to be 60°.find the height of the tower

Answer 1

For solution prefer to attach file

Answer 2

Given data:-

• At a point 50m away from the base of tower an observer measures the angle of elevation of the top of the tower to be 60°.

Assumption:-

• Let, AB be the tower and BC be the length from base of tower with 50m. and angle CAD is 60° which is angle of elevation.
• 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 height of tower, we use trigonometric ratio.

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

—› tan( 60° ) = {AB}/{BC}

—› √3 = {AB}/50. i.e.

—› AB = 50√3 m or 86.6025 m (approx)

Hence, the height of the tower is 50√3 m or 86.6025 m (approx).