SOLUTION: Solve for B: tan(B+15°)=cot(2B+30°)

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Question 936968: Solve for B: tan(B+15°)=cot(2B+30°)
Found 2 solutions by lwsshak3, AnlytcPhil:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for B: tan(B+15°)=cot(2B+30°)
for interval:(0,180˚)
tan(B+15°)=cot(2B+30°)
tan(B+15°)=1/tan(2B+30°)
tan(B+15°)=1/tan(2(B+15°))
letx=(B+15˚)
tanx=1/tan(2x)
tanx=1%2F%28%282tanx%29%2F%281-tan%5E2%28x%29%29%29
tanx=%281-tan%5E2%28x%29%29%2F%282tanx%29
2tan^2(x)=1-tan^2(x)
3tan^2(x)=1
tan^2(x)=1/3
tanx=±√(1/3)=±√3/3
x=30˚, 150˚
..
B+15˚=30˚
B=15˚
and
B+15=150˚
B=135˚

Answer by AnlytcPhil(1808) About Me  (Show Source):
You can put this solution on YOUR website!
tan%28B%2B%2215%B0%22%29=cot%282B%2B%2230%B0%22%29

2B+30° = 2(B+15°)

Let A = B+15°, then 2B+30° = 2(B+15°) = 2A

tan%28A%29=cot%282A%29

tan%28A%29=1%2Ftan%282A%29

tan%28A%29tan%282A%29=1

tan%28A%29%282tan%28A%29%2F%281-tan%5E2%28A%29%29%29=1

2tan%5E2%28A%29%2F%281-tan%5E2%28A%29%29=1

2tan%5E2%28A%29=1-tan%5E2%28A%29

3tan%5E2%28A%29=1

tan%5E2%28A%29=1%2F3

tan%28A%29+=+%22%22+%2B-+sqrt%281%2F3%29

tan%28A%29+=+%22%22+%2B-+sqrt%28expr%281%2F3%29%2Aexpr%283%2F3%29%29

tan%28A%29+=+%22%22+%2B-+sqrt%283%29%2F3

A =  30°+ 180°n, 150°+ 180°n

A = B+15°

A-15° = B

B = A - 15° 

B = 30° + 180°n - 15°, 150° + 180°n - 15°

B =  15°+ 180°n, 135°+ 180°n

Edwin