Find the exact value under the given conditions. sin a =7/25, 0 < a < 3.14/2 ;cos B = 12/13 , 0<B< 3.14/2 find tan(a+B)

Answer 1

sina = 7/25 and angle a is acute, so cosa = √(1 - sin²a) = 24/25 and tana = sina/cosa = 7/24

cosB = 12/13 and angle B is acute, so sinB = √(1 - cos²B) = 5/13 and tanB = sinB/cosB = 5/12

Tan(a + B) = [Tana + TanB] / (1 - TanaTanB) = (7/24 + 5/12) / (1 - 35/288)

Multiply the top and bottom by 288 to get (84 + 120) / (288 - 35) = 204 / 253

Answer 2

sinα=.28

α=16.26°

If the value of π/2 is 180/2.

cosβ=.923

β=22.61°

If the value of π/2 is 180/2.

tan(16.26°+22.61°)=.806