Asked by Angel
A balloon 150 meters from the observer is rising vertically at constant rate. After 1 minute the angle of elevation is 28 degrees and 29 minutes. What is the angle after 3 minutes?
Answers
Answered by
Steve
To figure the height after 1 minute,
h/150 = tan 28°29'
Now, having h, you know that 3h is the altitude after 3 minutes. So, find the angle θ such that
tanθ = 3h/150
Or, in one step,
θ = arctan(3*150*tan28°29')
h/150 = tan 28°29'
Now, having h, you know that 3h is the altitude after 3 minutes. So, find the angle θ such that
tanθ = 3h/150
Or, in one step,
θ = arctan(3*150*tan28°29')
Answered by
mateo
thanks big help
Answered by
mateo
You're so stupid!!!
Answered by
Jyc
tan(28deg59min)= h/150
h= 83.0893 m/min
@3 min
tan(theta)= h(@3min)/150
theta= arctan [(83.0893*3min)/150]
theta= 58.9621 deg
h= 83.0893 m/min
@3 min
tan(theta)= h(@3min)/150
theta= arctan [(83.0893*3min)/150]
theta= 58.9621 deg
Answered by
LukaTheGoat BinjEE
First get the height from its initial position
(Tan28°29') = h / 150
h= 83. 089
Now we will get the angle of the balloon from the observer after 3 mins. (3mins = 3 altitude)
(Tanθ) = (hx3) / 150
(Tanθ) = (83.089x3) / 150
θ = 58.96 deg
(Tan28°29') = h / 150
h= 83. 089
Now we will get the angle of the balloon from the observer after 3 mins. (3mins = 3 altitude)
(Tanθ) = (hx3) / 150
(Tanθ) = (83.089x3) / 150
θ = 58.96 deg
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