Answer:
π/6
Step-by-step explanation:
Let's start to solve by finding the square roots of both functions.
the suqare root of 3sinA+4cosB=6 will be 9sin2A+24sinAcosB+ 16cos2B=36
the square root of 4sinB + 3cosA = 1 will be 16sin2B+24sinBcosA+9cos2A
Then let's sum both functions side by side and group them in appropriate variables:
sin2A+cos2B=1 (in order trigonometric rules)
sinAcosB+cosAsinB= sin(A+B) then:
9(sin2A+cos2B)+24(sinAcosB+cosAsinB)+16(sin2A+cos2B)=37
9+16+24sin(A+B)=37
sin(A+B)=1/2
sin(A+B)=sinπ/6
A+B=π/6 ==> π−π/6=5π/6
If we know A,B and C are angles of a triangle,
Thus, A+B+C=π
⟹π/6+C=π==>5π/6+C=π ==>π/6
