Angle between the tangents to the curve $y=x^2-5x+6$ at the points $(2,0)$ and $(3,0)$ is:

$y=x^2-5x+6$ Slope of tangent $m=\frac{dy}{dx}=2x-5$ At $(2,0)$: $m_1=2(2)-5=-1$ At $(3,0)$: $m_2=2(3)-5=1$ The product of slopes $m_1\cdot m_2=(-1)(1)=-1$.Since the product of slopes is $-1$, the tangents are perpendicular.The angle is $\pi /2$.