we know that θ is in the III Quadrant, and let's recall that on the III Quadrant sine and cosine are both negative, and since tangent = sine/cosine, that means that tangent is positive. Let's also keep in mind that tan(π) = sin(π)/cos(π) = 0/-1 = 0.
well, the hypotenuse is just a radius unit, so is never negative, since we know sin(θ) = -(5/13), well, the negative number must be the 5, so is really (-5)/13.
The roots of the equation are 2 and k substituting x=2 in the equation (x+4)(x-p) = 2-x (2+4)(2-p) = 2-2 6(2-p) = 0 dividing 6 both side, the answer is p = 2
substituting x in the equation we (x-4)(x-p) = 2-x x² - 2x + 4x- 8 = 2-x x² - 2x + 4x -8 -2 -x = 0 x² + 3x - 10 = 0 x² + 5x -2x - 10 = 0 x(x + 5) -2( x + 5) = 0 (x + 5)( x - 2) =0 (x + 5) = 0 or (x - 2) = 0 x = -5 and x = 2 therefor k = -5 and p = 2