cos(a) + 1.341cos(b) = 2 and sin(a) + 1.341sin(b) = 0 are given. What are the a and b ?

Question

the answer should be degree type

Answer 1

As you know Sin^2(b) + Cos^2(b) =1

Solve the given two equations for Sinb and Cosb as follows:

Sinb = -sina/1.341 (from second equation)

Cosb = (2-cosa)/1.341

Now, use these two values for Sinb and Cosb in first equation:

(-Sin(a)/1.34)^2 + ((2-Cosa)/1.341)^2 = 1

solving this we get

Sin^a + 4 - 4 Cosa + Cos^2a = (1.34)^2

1+4 - 4 Cos a = (1.34)^2

5- (1.34)^2 = 4 Cosa

Cosa = (5- (1.34)^2)/4

a = arcCos(((5- (1.34)^2)/4))

Use this value of Cosa in any of the above equation (like Cosb = (2-cosa)/1.341) to get value of b.

Hope this helps.