Nice Coordination with angles

First, I have created an image to represent all the information given in the question. Let point X lie on the x-axis where x > 0.

Now,

The interval AB has a distance of 1 as the radius of the circle is 1.

The interval AC has a distance of 1 as the radius of the circle is 1.

Let the interval BC have a length of h.

Let the angle $\theta$ be $\angle BCX$

Now using that information,

$\angle ACB$ is equal to 90 - $\theta$

As AC = AB, triangle ABC is an isosceles triangle so $\angle ACB$ = $\angle ABC$

Now the sum of a triangle equals 180,

So $\angle ACB$ + $\angle ABC$ + $\angle CAB$ = 180,

Simplifying this we get $\angle CAB$ = $2\theta$

Now using cosine rule,

$h^2 = 1^2 + 1^2 - 2*1*1*\cos 2\theta$

So $h^2 = 2 - 2\cos 2\theta$

Now as $\cos 2\theta = 1 - 2(\sin \theta)^2$

Then $h^2 = 2 - 2 + 2*2(\sin \theta)^2$

Thus $h = 2\sin \theta$

Letting $\theta$ = 60,

We get that h = 1.73