Question

From the top of a 9 metres high building ab, the angle of elevation of the top of a tower cd is 30º and the angleof depression of the foot of the tower is 60º. what is the height of the tower?

Answer 1

Height of tower is 12m

Answer 2

Answer:

11.999 ≈ 12 m

Step-by-step explanation:

Height of the building AB = 9 m

The angle of elevation of the top of a tower CD is 30º i.e.∠DBE=30°

The angle of depression of the foot of the tower is 60º .i.e.∠BCA = 60°

AB = CE = 9 m

BE = AC

In ΔABC

$tan \theta = \frac{Perpendicular}{Base}$

$tan 60^{\circ} = \frac{AB}{AC}$

$\sqrt{3}= \frac{9}{AC}$

$AC= \frac{9}{\sqrt{3}}$

$AC=BE=5.196$

In ΔDBE

$tan \theta = \frac{Perpendicular}{Base}$

$tan 30^{\circ} = \frac{DE}{BE}$

$\frac{1}{\sqrt{3}}= \frac{DE}{5.196}$

$\frac{1}{\sqrt{3}} \times 5.196=DE$

$2.999=DE$

Height of tower = CD = CE+ED = 9+2.999 =11.999 ≈ 12 m

Hence the height of the tower is 11.999 ≈ 12 m