Question

In a triangle ABC, 3sinA+4cosB=6 and 4sinB+3cosA=1.Find measure of angle C.

Answer 1

Given that ABC is a triangle.

Then, SinC = Sin(Pi-(A+B)) = Sin(A+B)

Square the 2 given equations.

9 (SinA)^2 + 16(CosB)^2 + 24 SinA Cos B = 36

9(CosA)^2 + 16(SinB)^2 + 24 CosA SinB = 1

Adding the two,

9 + 16 + 24 Sin(A+B) = 37 . . . Or

Sin(A+B) = 0.5

This gives a value of 30 (or 150) degrees for angle C
C=30 DEGREE ONLY

WHY C CAN'T BE 150?

If C is 150 then A +B must be 30

So sinA is maximum when A takes value very close to 30 and cos B is maximum when B takes value very close to 0

In that case 3sinA + 4cosB can take maximum value 3(1/2)+4(1) =5.5 which is less than 6, it's given that 3sinA + 4 cosB=6 ..So c =150 not possible.