Question

A 6.0-ft-tall girl stands on level ground. The sun is 23 ∘ above the horizon.

How long is her shadow?

Express your answer using two significant figures.

b) A 5.0-m-wide swimming pool is filled to the top. The bottom of the pool becomes completely shaded in the afternoon when the sun is 25 ∘above the horizon.

How deep is the pool?

Express your answer using two significant figures.

Answer 1

a)Given that

Height of the girl = H = 6 ft ; = 23 degrees.

We need to find the height of girl's shadow. Let it be H'.

The sun is making with thoe ground. The girl herself can be viewed as the perpendicular of the right angled triangle and her shadow as a base. So from trigonometry:(refer to the diag a in the uploaded pic)

tan = H/H'

H' = H/tan = 6 ft / tan 23 = 14.14 feet.

Hence, height of the shadow = H' = 14.14 feets.

(b)Given that,

width = w = 5 m ; (a) = 25 degrees

We need to find the depth of the pool. Let it be "D"

Please refer to the attached diagram (b)

= 90 - (a) = 90 - 25 = 65 deg

From Snell's law we can write:

n(water) sin(w) = n(air) sin

1.33 x sin(w) = 1 x sin65 => sin(w) = 0.9063/1.33 =0.6814

(w) = sin -1 (0.6814) => (w) = 42.96 deg

Now again using trigonometry;

tan (w) = w / D

D = w / tan(w) = 5m / tan(42.96) = 5.37 meters

Hence, Depth of the pool = D = 5.37 meters.