How do you find the length of the arm of an inscribed angle?

Question

How do you find the length of an arm of an angle inscribed in a circle with its vertex on the circle itself? The circle has radius $1$ and the triangle that is formed is always an isosceles, as in the picture below.

Answer 1

If you connect the center to the vertex you will get an isosceles triangle whit two sides being the radii of the circle and the angle between the radii is $\pi-2\alpha$

Using law of cosines, you will get $$ L^2 = 1+1 -2\cos (\pi -2\alpha)$$ Thus you find your $$L = \sqrt {2+2\cos (2\alpha)}$$

Answer 2

There are two ways to arrive at the correct answer.

First, we can approach the problem the same way as Mohammad Riazi-Kermani using the law of cosines.

\begin{eqnarray} L^2&=&1^2+1^2-2(1)(1)\cos(\pi-2\alpha)\\ &=&2-2(\cos\pi\,\cos(2\alpha)+\sin\pi\,\sin(2\alpha))\\ &=&2+2\cos(2\alpha)\\ &=&4\left(\frac{1+\cos(2\alpha)}{2}\right)\\ &=&4\cos^2\alpha\\ L&=&2\cos\alpha \end{eqnarray}

A second way to arrive at the same result is to notice that if we drop a perpendicular from the vertex to the base of the isosceles triangle we obtain a right triangle with height $h=1+\cos(2\alpha)$, base $b=\sin(2\alpha)$, and hypotenuse $L$. Applying the Pythagorean theorem we get a different sign

\begin{eqnarray} L^2&=&\left(\sin(2\alpha)\right)^2+\left(1+\cos(2\alpha)\right)\\ &=&\sin^2(2\alpha)+1+2\cos(2\alpha)+\cos^2(2\alpha)\\ &=&2+2\cos(2\alpha) \end{eqnarray}

From there we proceed as in the first solution.

But, in either case, the correct solution is

\begin{equation} L=2\cos\alpha \end{equation}

Answer 3

Another way would be to use two expressions for the area $S_{\triangle ABC}$:

\begin{align} S_{\triangle ABC}&= \tfrac12b^2\sin2\alpha ,\\ S_{\triangle ABC}&= 2R^2\sin2\alpha\sin(90-\alpha)^2 =2R^2\sin2\alpha\cos^2\alpha ,\\ b&=2R\cos\alpha . \end{align}

Or, just using the sine law,

\begin{align} |AB|&=2R\sin\angle BCA=2R\sin(90-\alpha)=2R\cos\alpha . \end{align}

Answer 4

We have several ways to arrive at a solution. I think the following is especially simple.

Let $L$ be the length of the side that is to be found.

Due to the symmetry of the triangle, the radius from the center of the circumscribed circle to the apex of the triangle bisects the angle at the apex. That is, this radius bisects the angle $2\alpha$ into two angles of measure $\alpha.$ Construct that radius as in the figure below, and also drop a perpendicular from the radius to one leg of the triangle, which bisects that leg into two segments of length $L/2.$

Recall the formula for $\cos \alpha$ in terms of the sides of a right triangle with angle $\alpha$ at one vertex ("adjacent divided by the hypotenuse"), and you can almost read the answer directly from the figure. That is, you immediately have $$ \cos \alpha = L/2, $$ and the value of $L$ in terms of $\alpha$ is obvious.

Answer 5

The law of sines: $\frac{a}{\sin \alpha} = 2R.$

By the law of sines: $$ \frac{x}{\sin{\frac{180^\circ-2\alpha}{2}}}=2\cdot 1 \iff x=2\cos \alpha.$$