Question

From a point 20m away from the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower.

Answer 1

SOLUTION

Let the distance of the tower (AB)= h m

tower of the point (AC)= 20m

Let theta denotes the angle of elevation.

In ∆ABC,

$tan \theta = \frac{ab}{ac} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{h}{20} \\ \\ = > \sqrt{3} h = 20 \\ \\ = > h = \frac{20}{ \sqrt{3} } \\ \\ = > h = \frac{20}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = > \frac{20 \sqrt{3} }{3} \\ \\ = > h = 6.66 \sqrt{3} \\ \\ = > h =( 6.66 \times 1.732)m \\ \\ = > h = 11.54m$

hope it helps ☺️

Answer 2

Answer:

Let AB be the height of the tower and C be the point .

In rt.△ABC,

tan30° = AB/BC

AB = BC tan30°

= 20/√3 = 20/√3 × √3/√3

= 11.56 m

Therefore the height of the tower is 11.56 m.

#Hope it helps ✨