If sinalpha 2129 where 0 Solutionsin alpha 2129 opposite21

Solution

sin alpha= 21/29

opposite=21 and hypotenuse=29

adjacent=sqrt(29²-21²) = sqrt(400)=20

cos alpha=20/29

cos beta=4/5

adjacent=4   and hypotenuse=5

opposite=sqrt(5²-4²)= 3

sin beta= 3/5

a. sin (alpha + beta) =sin alpha cos beta + cos alpha sin beta

      sin(alpha + beta) = (21/29)(4/5) + (20/29)(3/5)

      sin(alpha + beta) = 144/145

b. cos(alpha-beta)= cos alpha cos beta + sin alpha sin beta

     cos(alpha-beta) = (20/29)(4/5) + (21/29)(3/5)

      cos(alpha-beta) = 143/145

sin(alpha-beta)= sin alpha cos beta -cos alpha sin beta

                            = (21/29)(4/5) - (20/29)(3/5)

sin(alpha-beta) = 24/145

c. tan(alpha-beta)= sin (alpha-beta)/cos(alpha-beta)

                                =24/145/143/145=24/143