Given that $\theta=45°$, find $\angle AOC$.

Question

I only know that $AB=OB$, and that the two right triangles $AOC$ and $AOB$ share a same line $OA$. Does $\angle BOC=45°$ hint something? The answer from the book is $60°$

Answer 1

I think, it means $AB\perp(OBC)$, otherwise it's undefined.

The hint:

Prove that $$\measuredangle OCB=90^{\circ}.$$

Indeed, $AB\perp(OBC)$, which says $AB$ is perpendicular to any line in the plane,

which says $AB\perp OC$.

Now, we see that $$OC\perp AB$$ and from the given $$OC\perp AC.$$ Also, $$AC\cap AB=\{A\}.$$ Thus, $$OC\perp(ABC)$$ and from here $$OC\perp BC.$$

From here: let $AB=a$.

Thus, $$\tan\measuredangle AOC=\frac{AC}{OC}=\frac{a\sqrt{1+\left(\frac{1}{\sqrt2}\right)^2}}{\frac{a}{\sqrt2}}=\sqrt3.$$