SOLUTION: Given sin B=5/13 in QII,and(6,-8) is on the terminal side of a, find the exact value of sin (A+B)

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Question 1201490: Given sin B=5/13 in QII,and(6,-8) is on the terminal side of a, find the exact value of sin (A+B)

Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

Answer: 63/65

Explanation:

sine = opposite/hypotenuse
cosine = adjacent/hypotenuse
tangent = opposite/adjacent

we can rewrite those formulas into this shorter notation
sin(theta) = y/r
cos(theta) = x/r
tan(theta) = y/x
where (x,y) is the terminal point's location and x^2+y^2 = r^2
x = adjacent
y = opposite
r = hypotenuse = distance from origin to terminal point
Using this notation will avoid having to memorize things like "cosine is negative in Q2, cosine is positive in Q4, etc".
All you'll need to worry about is the terminal point's coordinates.

sin(B) = 5/13
opposite = y = 5
hypotenuse = r = 13
angle B is in quadrant QII, aka Q2
This is the northwest quadrant

The adjacent side of 12 is found through the pythagorean theorem (translating to )
Plug in y = 5 and r = 13 to find x = 12 or x = -12
We will go for x = -12 since we're in Q2.

From that drawing, we can determine:
cos(B) = adjacent/hypotenuse
cos(B) = x/r
cos(B) = -12/13

The terminal side of angle A is at (6,-8)

Angle A is in quadrant Q4, in the southeast.
The coordinates of the point (6,-8) lead directly to the adjacent and opposite sides in that order.
The hypotenuse is determined through the pythagorean theorem.
Plug in x = 6 and y = -8 to find r = 10.

sin(A) = opp/hyp = y/r = -8/10 = -4/5
cos(A) = adj/hyp = x/r = 6/10 = 3/5

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Let's recap what we have so far

sin(A) = -4/5
cos(A) = 3/5
and
sin(B) = 5/13
cos(B) = -12/13

We'll then use a trig identity to finish up the problem.

A reference sheet can be found here
https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
sin(A+B) = (-4/5)*(-12/13) + (3/5)*(5/13)
sin(A+B) = 48/65 + 15/65
sin(A+B) = (48 + 15)/65
sin(A+B) = 63/65

Answer by greenestamps(13203) (Show Source):
You can put this solution on YOUR website!

sin(B)=5/13 in quadrant II, so cos(B)=-12/13. (5-12-13 is a Pythagorean Triple)

The point (6,-8) is on the terminal side of angle A, so angle A is in quadrant IV, and tan(A)=-4/3. That makes sin(A)=-4/5 and cos(A)=3/5.

sin(A+B) = sin(A)cos(B)+cos(A)sin(B) = (-4/5)(-12/13)+(3/5)(5/13) = 48/65+15/65 = 63/65.

ANSWER: 63/65