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Anettt [7]
3 years ago
15

A man is flying in a hot-air balloon in a straight line at a constant rate of 4 feet per second, while maintaining it at a const

ant altitude. As he approaches the parking lot of a market, he notices that the angle of depression from his balloon to a friend's car in the parking lot is 35(degrees). A minute and a half later, after flying directly over his friend's car, he looks back to see his friend getting into the car and observes the angle of depression to be 36(degrees). At that time, what is the distance between him and his friend?
Physics
1 answer:
Tanya [424]3 years ago
6 0

Answer: the distance between him and his friend is 218.39 feet

Explanation:

Given the data in the question;

FROM IMAGE A;

distance travelled in minute and a half

⇒ (60 + 30)sec × 4ft/sec = 360 fts

FROM IMAGE B

tan35°= h/x   --- equ 1

and tan36° = h/(360 - x), so h = (360 - x)tan36°

we substitute value of h into euq 1

tan35° = (360 - x)tan36°/x

xtan35° = (360 - x)tan36°

0.7002x = 261.5553 - 0.7265x

0.7002x + 0.7265x = 261.5553

1.4267x = 261.5553

x = 183.32 feet

so

360 - x ⇒ ( 360 - 183.32) = 176.68

FROM IMAGE C

Let the distance between them be d

so

cos36° = 176.68 / d

dcos36° = 176.68

d = 176.68 / 0.809

d = 218.39 feet

Therefore the distance between him and his friend is 218.39 feet        

 

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