SOLUTION: The angle of elevation of a hot air balloon, climbing vertically, changes from 25 degrees at 10:00 am to 60 degrees at 10:02 am. The point of observation of the angle of elevation

Question 210095: The angle of elevation of a hot air balloon, climbing vertically, changes from 25 degrees at 10:00 am to 60 degrees at 10:02 am. The point of observation of the angle of elevation is situated 300 meters away from the take off point. What is the upward speed, assumed constant, of the balloon? Give the answer in meters per second and round to two decimal places.
So, far I have done:
I have drawn the right triangle, and I know that the initial angles is 25 degrees, but how does 60 degrees factor into this problem? It's a right triangle, but 90 + 25 + 60 does not = 180. What is the next step here?

Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Ok, the key here is to find the change in the baloon's height in the two-minute observation period.
You have a good starting point with the right triangle but actually, you would have two right triangles having the same base (300 meters) but different heights.
You can calculate these heights using the tangent function because you are given the two appropriate angles (25 degs. and 60 degs.)
As you probably know, the tangent of the given angles is just the height divided by the base. So we can find the two heights as follows, starting with the lower height which we can call :
...and the upper height is:
Subtracting these two:

Meters.
So the ballon ascended a distance of 379.723 meters in 2 minutes (10:02 - 10:00). Converting this to seconds we have a speed of:
meters per second. Performing the indicated division, we get:
meters per second (rounded to two decimal places.)

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