Trigonometry application question help!

Question

An eagle flying at an altitude of 1000metres uses wind drafts and can be seen oscillating 6 complete cycles of -/+ 50metres within a span of 2 minutes. An ovserver notes at one point (t=0seconds) the eagle begins its cycle flying upwards at 1000metres. The height of the eagle as a function of time is h(t)=50sin[36t]+1000. Determine the first 4 times when the eagle is at a height of 970metres.

Answer 1

Plug in 970 for h(t) and begin to solve for t:

 

970 = 50sin(36t) + 1000

-30 = 50sin(36t)

-3/5 = sin(36t)

sin^-1(-3/5) = 36t

-36.87° = 36t

 

 

 

Think about the first four angles on the unit circle (starting at 0°) that share a sine value with -36.87°. Remember that sine is equivalent to the y-coordinate. So, since -36.87° is in Quadrant IV, sine must be negative. We're therefore looking for angles equivalent to -36.87° in Quadrants III and IV. Those angles are:

 

180 + 36.87 = 216.87 + 360n

360 - 36.87 = 323.13 + 360n

 

The first four of those are 216.87°, 323.13°, 576.87°, and 683.13°.

 

 

 

So:

 

216.87 = 36t

6.02 = t

 

323.13 = 36t

8.98 = t

 

576.87 = 36t

16.02 = t

 

683.13 = 36t

18.98 = t

 

 

 

The eagle will reach a height of 970 meters at approximately 6.02 seconds, 8.98 seconds, 16.02 seconds, and 18.98 seconds.