Question

Circumscribing a TriangleShow that $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2r}$

where $r$is the radius of the circle circumscribing the triangle PQR whose sides are a, b, and c, as shown in the figure.

Short answer

It is proved that$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2r}$

Step-by-step solution

Step 1. Given Information

We have

Let P' be a point on the circumference such that$PP\text{'}=2r$

Step 2. Identifying angles

First, notice that angles $\angle PQR$and $\angle PP\text{'}R$are both inscribed angles.

Also, they both originate from the same two points P and R.

This means that they are equal.$\angle PQR=\angle PP\text{'}R$

Since$\angle PQR=B,thenalso\angle PP\text{'}R=B$

Step 3. Proving the result

Since $$\begin{gathered} \sin \left(\angle PP\text{'}R\right)=\frac{b}{2r} \\ \frac{\sin \left(\angle PP\text{'}R\right)}{b}=\frac{1}{2r} \\ \frac{\sin B}{b}=\frac{1}{2r}\left[Since\angle PP\text{'}R=B\right] \end{gathered}$$

Now, role="math" localid="1646590875136" $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}and\frac{\sin B}{b}=\frac{1}{2r}$

then$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2r}$