this can be solved in the following manner.
solve for the angles as if they were in quadrant 1.
this means make them all positive when you solve for them.
you will be using a calculator for this method.
arcsin(20/29) = 43.60281897 degrees.
arctan(15/8) = = 61.92751306 degrees.
now find the equivalent angle in the quadrant these angles are supposed to be in.
43 degrees in quadrant 4 is equivalent to 360 - 43 degrees in quadrant 4.
that angle becomes 316.397181 degrees to be more exact.
61 degrees in quadrant 2 is equivalent to 180 - 61 degrees in quadrant 2.
that angle becomes 118.0724869 degrees to be more exact.
now you want to find sin (A + B)
you can use the formula for sin (A + B) which is equal to sin(A)cos(B) + cos(A)sin(B).
use your calculator to get sin(A+B+ = .9634888438 which is equivalent to 475/493 in fraction form.
you can also solve this in a similar but different way using fractions.
start with sin(A) = -20/29 in fourth quadrant.
this means opposite side is -20 and hypotenuse is 29
you can solve for the adjacent side by using pythagorus to get x^2 + 20^2 = 29^2 which results in x^2 = 29^2 - 20^2 which results in x = 21.
your triangle in the fourth quadrant has opposite side = -20 and hypotenuse = 29 and adjacent side = 21.
now start with tan(B) = -15/8 in quadrant 2.
use pythagorus again to get 8^2 + 15^2 = x^2 which results in x = 17.
your triangle in the second quadrant has opposite side = 15 and adjacent side = -8 and hypotenuse = 17.
now you want to find sin(A+B), so use the formula for sin(A+B).
sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
now, however you can find these directly from the triangles formed.
sin(A) = -20/29
cos(B) = -8/17
sin(B) = 15/17
cos(A) = 21/29
formula of sin(A+B) = sin(A)cos(B) + cos(A)sin(B) becomes:
sin(A+B) = -20/29 * -8/17 + 21/29 * 15/17
simplify this to get:
sin(A+B) = 160 / 29*17 + 315 / 29*17
simplify further to get:
sin(A+B) = (160 + 315) / (29*17)
simplify finally to get:
sin(A+B) = 475 / 493
that equals selection c.
a picture of your triangles is shown below.
