Question

From a point O on the ground, the angle of elevation of the top of a
tower is 30° and that of the top of the flagstaff on the top of the tower
is 60°. If the length of the flagstaff is 5 metres, find the height of the
tower.

by÷ don ji​

Answer 1

Answer:

10m will be the height. very easy which class questions

Answer 2

Recall:

$\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$

$\sin \theta = \dfrac{\text{opposite}}{\text{hypothenuse}}$

$\cos \theta = \dfrac{\text{adjacent}}{\text{hypothenuse}}$

Define H and D:

Let the height of the tower be H.

let the distance of the point O to the tower be D.

Find the height of the tower in term of D (When the angle is 30):

$\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$

$\tan (30)= \dfrac{\text{H}}{\text{D}}$

$\text{H} = \text{D} \tan(30)$

Find the height of the tower in term of D (When the angle is 60):

$\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$

$\tan (60)= \dfrac{\text{H} + 5}{\text{D}}$

$\text{H} + 5 = \text{D} \tan(60)$

$\text{H} = \text{D} \tan(60) - 5$

Solve D:

$\text{H} = \text{D} \tan(30)$

$\text{H} = \text{D} \tan(60) - 5$

$\text{D} \tan(30) = \text{D} \tan(60) - 5$

$\text{D} \tan(60) - \text{D} \tan(30) = 5$

$\text{D} (\tan(60) - \tan(30)) = 5$

$\text{D} = \dfrac{6}{\tan(60) - \tan(30)}$

$\text{D} = 3\sqrt{3} \text { m}$

Solve H:

$\text{H} = \text{D} \tan(30)$

$\text{H} = 3\sqrt{3} \times \tan(30)$

$\text{H} = 3 \text { m}$

Answer: The height is 3m