Question

Find the angle of depression from the top of 12 m high tower of an object lying at a point 12 m away from the base of the tower.

plsss help

Answer 1

Answer:

24 m

Step-by-step explanation:

Refer the attached image .

The height of the tower i.e. AB = 12 m

The angle of depression from the top of tower at a point on the ground is 30°

i.e. ∠ACB = 30°

Since we are asked to find the distance of the point from the top of the tower i.e. AC

So, in ΔABC, to find AC we will use trigonometric ratio

sin\theta = \frac{Perpendicular}{Hypotenuse}sinθ=

Hypotenuse

Perpendicular

sin 30^{\circ} = \frac{AB}{AC}sin30

∘

=

AC

AB

\frac{1}{2} = \frac{12}{AC}

2

1

=

AC

12

AC = 12*2AC=12∗2

AC = 24AC=24

Thus the distance of the point from the top of the tower is 24 m

Answer 2

Answer:

45⁰

Step-by-step explanation:

in triangle ABC

tan∅=AB/BC

tan∅=12/12

tan∅=1

=45⁰