Solution

Given curves are:

$y=(x-2)^2$

and

$y=-4+6x-x^2$

At the point of intersection,

$(x-2)^2=-4+6x-x^2$

$x^2-4x+4=-4+6x-x^2$

$2x^2-10x+8=0$

$x^2-5x+4=0$

$(x-1)(x-4)=0$

So,

$x=1$ or $x=4$

Now, slopes of the curves are:

For $y=(x-2)^2$,

$m_1=\frac{dy}{dx}=2(x-2)$

For $y=-4+6x-x^2$,

$m_2=\frac{dy}{dx}=6-2x$

At $x=1$,

$m_1=2(1-2)=-2$

$m_2=6-2(1)=4$

Angle between two curves is given by:

$\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|$

So,

$\tan\theta=\left|\frac{4-(-2)}{1+(-2)(4)}\right|$

$=\left|\frac{6}{1-8}\right|$

$=\left|\frac{6}{-7}\right|$

$=\frac{6}{7}$

Therefore,

$\theta=\tan^{-1}\left(\frac{6}{7}\right)$