What is the angle at which these two curves intersect?
$$x^2 + y^2 = 12x$$
$$y = ((x-6)^{2})^{1/3}$$
I have a book where I found this problem. It's even solved there but I think there're two mistakes in the solution. So I think the answer given there is not correct.
I solved it myself too. Here is what I got. The curves intersect at two points.
$A(6+3\sqrt{3}, 3)$
$B(6-3\sqrt{3}, 3)$
Is this correct?
Also, I got that the angles of intersection of the two curves are equal at these two points (ignoring the orientation/sign of the angles of course). And I got that the angle is $$\arctan\ {\frac{11}{\sqrt{3}}}$$ Is this correct?
The answer in the book is
$$\arctan\ {\frac{5\sqrt{3}}{7}}$$
but I think this one is not correct.
The intersection points are correct by inspection.
For the angles we need to consider derivatives at that points, that is for point $A$
$x^2 + y^2 = 12x \implies y=\sqrt{12x-x^2} \implies y'(x)=\frac{6-x}{\sqrt{12x-x^2}}\implies y'(6+3\sqrt 3)=-\sqrt 3$
$y = (x-6)^{\frac{2}{3}} \implies y'(x)=\frac{2}{3(x-6)^\frac13}\implies y'(6+3\sqrt 3)=\frac{2\sqrt 3}9$
therefore, using that
$$\arctan{u}+\arctan{v}=\arctan\left(\frac{u+v}{1-uv}\right)$$
we obtain
$$\theta = \arctan\left(\frac{2\sqrt 3}9\right)-\arctan(-\sqrt 3)=$$ $$= \arctan\left(\frac{2\sqrt 3}9\right)+\arctan(\sqrt 3)=\arctan \frac{11}{\sqrt 3}$$
which corresponds to your solution.
Same solution holds also for point $B$ by symmetry.