Chapter 8.2, Problem 62AYU

To determine

Circumscribing a triangle, show that, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2r}$ .

Answer to Problem 62AYU

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

Explanation of Solution

Given information:

Circumscribing a triangle, show 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 $a,bandc$ as shown in the figure.[hint: draw the diameter $PP\text{'}$ . Then $B=\angle PQR=PP\text{'}R$ and angle $PRP\text{'}=90°$ .

  

Calculation:

Let us a consider a circle with radius $r$ circumscribing a triangle $\Delta PQR$ having angles $A,B,C$ correspond to the opposite sides $a,b,c$ .

Now, drawing a diameter $PP\text{'}$ and making a chord $P\text{'}R$ by joining the points $P\text{'}$ and $R$ lying on the circle.

We have,

  $\Delta RPQ=A,\Delta PQR=B,\Delta QRP=C$ and $QR=a,RP=b,PQ=c$

By the law of sine in the triangle, $\Delta PQR$ .

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

We know that $pp\text{'}=2r$ since a chord of a circle makes the same angle with minor or major segment.

  $\angle PP\text{'}R=\angle PQR=B$

  $\angle PRP\text{'}=90°$ , Because diameter always make a right angle with any point on the circle. so $\Delta PRP\text{'}$ is a right triangle, now by using trigonometric ratio function, we get.

  $\begin{array}{l}\sin p\text{'}=\frac{RP}{PP\text{'}},\sin B=\frac{b}{2r}\\ \frac{\sin B}{b}=\frac{1}{2r}\\ \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2r}\end{array}$

Hence $\boxed{\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2r}}$ proved.