Point a is found at (-3, -3). find the sin at that point.
a. √2/2
b. -√2/2
c. √3/3
d. -√3/3

B. -√2/2.

Step-by-step explanation:

This is in the third quadrant  so the sine will be negative.

The hypotenuse =  √ [( -3)^2 + (-3)^2]

= √18 = 3√2.

Therefore the sine =  -3 / 3√2.

= -1 / √2

= -√2/2..

any options?

step-by-step explanation:

place r′ so that qr′ is parallel to the tangent.

now we get that ∠r′py=∠pr′q, where y is on the tangent, on the opposite side of p from x.

also, because of the symmetry, ∠r′py=∠qpx.

finally, note that no matter where we place r, as long as it's on the correct arc, we have ∠prq=∠pr′q.

stringing all these equalities together, we end up with ∠prq=∠qpx.

this picture will show how its going to look like.

hope this by jack monhen