<p>In triangle ABC, 3sin(A) + 4cos(B) = 6 and 4sin(B) + 3cos(A) = 1. What does angle C equal?</p>
<p>If someone can show me how to do this I would greatly appreciate it. Thanks.</p>
<p>A=115.2526835
B=34.747
C approximately = 29.996</p>
<p>First, square both sides of each equation:</p>
<p>9sin^2(A) + 24sin(A)cos(B) + 16cos^2(B) = 36 and
16sin^2(B) + 24sin(B)cos(A) + 9cos^2(A) = 1</p>
<p>Then add the two equations together:</p>
<p>9sin^2(A)+24sin(A)cos(B)+16cos^2(B)+16sin^2(B)+24sin(B)cos(A)+9cos^2(A)=36+1</p>
<p>Next, rearrange and factor the terms on the left side of the equation so you can apply some trig identities:</p>
<p>9(sin^2(A)+cos^2(A))+16(sin^2(B)+cos^2(B))+24(sin(B)cos(A)+sin(A)cos(B)) = 37</p>
<p>Simplify:</p>
<p>9x1 + 16x1 + 24sin(A+B) = 37 </p>
<p>Then, solve for sin(A+B):</p>
<p>sin(A+B) = 12/24 = 1/2, so A+B = 30 degrees and C = 180-30=150 degrees</p>
<p>thanks alot chloe</p>
<p>uhh dude that doesn't make sense.... Angle A must be quite big, > 30 definitely... my solution up top checks</p>
<p>Actually, if sin(A+B) = 1/2, then there is another senario for (A+B) between 0 and 180 degrees and that is A+B = 180 - 30 (reference angle) = 150. Then that would give
C = 180-150=30 degrees. This solution satisfies the original equation whereas the solution I first posted did not. Sorry, I posted late last night while I was sleepy, and am rusty on my trig. Just curious, StarWarsKid, how did you drive your answer? It's very close to the algebraic solution.</p>