Recall the sum identity for cosine:
cos(a + b) = cos(a) cos(b) - sin(a) sin(b)
so that
cos(a + b) = 12/13 cos(a) - 8/17 sin(b)
Since both a and b terminate in the first quadrant, we know that both cos(a) and sin(b) are positive. Then using the Pythagorean identity,
cos²(a) + sin²(a) = 1 ⇒ cos(a) = √(1 - sin²(a)) = 15/17
cos²(b) + sin²(b) = 1 ⇒ sin(b) = √(1 - cos²(b)) = 5/13
Then
cos(a + b) = 12/13 • 15/17 - 8/17 • 5/13 = 140/221
Hey.
acc'g to the conditions :
Let Catherine has x
and James has 5 + 3x
Then,

Hence
• Catherine has = x = $9,
• and James has = 5 +3x = 5 + 3(9) = $32
Thanks.
Is there anymore to the question?
Answer:
If your vertex is at the orgin you can't have a point on (6,0)
Step-by-step explanation: