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Help needed for year ten trigonometry question (1 Viewer)

atifaj

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I would really appreciate anyone's help with this word problem:

14. The angle of elevation of a vertically rising hot air balloon changes from 27° at 7.00 am to 61° at 7.03 am, according to an observer who is 300 m away from the take-off point.

a. Assuming a constant speed, calculate the speed (in m/s and km/h) at which the balloon is rising, correct to 2 decimal places.

b. The balloon then falls 120 metres. What is the angle of elevation now? Write your answer correct to 1 decimal place.
 

Roy G Biv

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it's rising vertically. so it's a sohcahtoa question.

tan 27 = x / 300
and
tan 61 = y / 300

solve x and y.
now, speed = distance / time

so, speed = (y-x) / (7.03am - 7.00 am)

you should be able to do b now.
 

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