In a triangle `ABC, 3 sinA+ 4 cos B=6 and 4 sinB+3cosA=1` hold. Then the angle C equals (measured in degree)

In a triangle ABC,3sin A+4cos B=6 and 4sin B+3cos A=1 hold.Then the angle C equals (measured in degree)

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

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

In triangle ABC, we are given that 3 sin A +4 cos B=6 and 4 sin B+3 cos A=1 . Then the measure of the angle C is -

in a Delta ABC,3sin A+4cos B=6 and 3cos A+4sin B=1 then angle C is

In triangle ABC, if sinA cos B=1/4 and 3 tanA = tanB, then cot^(2)A is equal to

In a triangle ABC, cos 3A + cos 3B + cos 3C = 1 , then then one angle must be exactly equal to

If in a triangle ABC ,(sin A+sin B +sin C)(sinA+sinB-sin C)=3sinAsinB then angle C is equal to

If 3sinA +4cosA=4, " then find " 4sinA -3cosA .