Question

Question #6f11c

Answer 1

tan b = 3 (Q. III); sin a = 2/3 (Q.I)

cos^2 b = 1/(1 + tan^2 b) = 1/(1+ 9) = 1/10 -> cos b = - 0.32
sin^2 b = 1 - 1/10 = 9/10
cos ^a = 1 - sin^2 a = 1 - 4/9 = 5/9 --> cos a = 0.75
tan a = sin a/cos a = 0.89
cos (a + b) = cos a.cos b - sin a.sin b
= -0.32(0.75) - 0.90(0.67) = -0.24 - 0.6 = - 0.84
`2cos^2 (b/2) = 1 + cos b = 1 - 0.32 = 0.68 -> cos^2 (b/2) = 0.34 -> cos (b/2) = 0.58`
`tan 2b = (2tan b)/(1 - tan^2 b) = 2/(1 - 9) = -1/4`