Question
solve these using trig please.
I solved a already and a = 13.04 miles
When the sun is directly overhead, the sun's light is filtered by approximately 10 miles of atmosphere. As the sun sets, the intensity decreases because the light must shown below. Note: 3947 mi is the radius of Earth. 2. pass a greater distance through the atmosphere, as sun at noon Atmosphere 500 G10 mi 3947 mi Find the distance d in the figure when the angle of elevation of the sun is 50° a. b. If the sun is directly overhead at 12 noon, then at what time is the angle of elevation of the sun 50°? The distance from the earth to the sun is 93 million miles. Give your answer in hours and minutes, i.e. 3:35. Find the distance that the sunlight passes through the atmosphere at sunset (angle of elevation 0%).
Explanation / Answer
a )
L = 3947 + 10 = 3957 mi
angle is 90 + 50 = 140
now using sine rule
3957 / sin140 = 3947 / sin C
C = sin-10.6411
C = 39.87o
angle A = 180 - ( 140 + 39.87 )
A = 0.13o
d / sin0.13 = 3957 / sin140
d = 13.96 miles
b )
90 - 50 = 40o
40 X 1/0.004166
= 9601.536 sec
c )
the length is
= ( 39572 - 39472 )1/2
= 281.14 miles