Please help me find the answer

Question

Given sin(x) = 1/8 with 0 deg < x < 90 deg and sin(y) = 7/9 with 0 deg < y < 90 deg Find sin(x + y)

Answer 1

sinx = 1/8, 0<x<90 cosx = sqr(8^2-1^2)/8 = sqr(63)/8

siny = 7/9 0<y<90 cosy =sqr(9^2-7^2)= sqr(81-49)/9= sqr(32)/9

find sin(x+y) = sinxcosy +sinycosx

=(1/8)(sqr32)/9 +(7/9)(sqr63)/8

= sqr32/72 + 7sqr63/72

=(sqr32+7sqr63)/72

=about .85

or use a calculator with an inverse trig function

x = arcsin(1/8) = 7.18 degrees

y= arcsin(7/9) = 51.06

x+y = 58.24

sin58.24 = about .85