SOLUTION: Solve this equation; find two values for t (II and III quadrant) 70=55-23cos{{{(2pi/365)}}}(t-30) I've tried to work this out for about an hour, but I keep getting different

Algebra ->  Trigonometry-basics -> SOLUTION: Solve this equation; find two values for t (II and III quadrant) 70=55-23cos{{{(2pi/365)}}}(t-30) I've tried to work this out for about an hour, but I keep getting different      Log On


   

Question 1135889: Solve this equation; find two values for t (II and III quadrant)
70=55-23cos%282pi%2F365%29(t-30)

I've tried to work this out for about an hour, but I keep getting different answers. Any help would be greatly appreciated. Thank you in advance!
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


70+=+55-23cos%28%282pi%2F365%29%28t-30%29%29

15+=+-23cos%28%282pi%2F365%29%28t-30%29%29

cos%28%282pi%2F365%29%28t-30%29%29+=+-15%2F23

%28%282pi%2F365%29%28t-30%29%29+=+arccos%28-15%2F23%29 = 2.281245 (radians) to several decimal places

t-30+=+2.281245%2A%28365%2F%282pi%29%29+=+132.52106 degrees
t+=+162.52106 degrees

That's the answer in degrees for the angle in quadrant II.

For the angle in quadrant III...

%28%282pi%2F365%29%28t-30%29%29+=+2pi+-+arccos%28-15%2F23%29 = 4.00194 (radians)

t-30+=+4.00194%2A%28365%2F%282pi%29%29+=+232.4789 degrees

t+=+262.4789 degrees

ANSWERS (degrees): 162.52106; 262.4789