Chapter 8.2, Problem 62AYU

To determine

To find: Show that $\frac{\sin \text{A}}{a}=\frac{\sin \text{B}}{b}=\frac{\sin C}{c}=\frac{1}{2r}$ of a circumscribing a triangle.

Answer to Problem 62AYU

The answer is

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

Explanation of Solution

Given information:

The given radius of the circle is $r$

The sides and opposite angles are $a,b,c$ , $P,Q,R$ of triangle $\Delta \text{PQR}$

Concept used:

Law of Sine:

If ABC is a triangle with sides $a$ , $b$ and $c$ , then

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

The given radius of the circle is $r$

The given angles and sides of circumscribing triangle $\text{ΔPQR}$ are

  $\begin{array}{l}\text{ΔRPQ=A}\\ \Delta \text{PQR=B}\\ \Delta \text{QRP=C}\end{array}$ ,

  $\text{QR=a,}\text{RP=b,}\text{PQ=c }\mathrm{....}\left(1\right)$

By using the law of sine in the circumscribing triangle $\text{ΔPQR}$

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\mathrm{....}\left(2\right)$

The diameter of circle is $\text{PP'}$

  $\text{PP'=2r}\mathrm{...}\left(3\right)$

By using theorem-Subtended angles in the same segment of a circle are equal.

  $\angle \text{PP'R=}\angle \text{PQR=B}\mathrm{....}\left(4\right)$

From figure $\angle {\text{PRP'=90}}^{\circ }\mathrm{....}\left(5\right)$

In the circumscribing triangle $\Delta \text{PP'R}$

By using the trigonometric ratio for right triangle $\Delta \text{PP'R}$

  $\text{sinP'=}\frac{\text{opposite}}{\text{hypotenuse}}\text{=}\frac{\text{RP}}{\text{PP'}}$

According to equations $\mathrm{..}\left(1\right),\mathrm{...}\left(2\right),\mathrm{..}\left(3\right)$

  $\begin{array}{l}\sin \text{B}=\frac{b}{2r}\\ \frac{\sin \text{B}}{b}=\frac{1}{2r}\mathrm{...}\left(6\right)\end{array}$

Put the values of equation $\mathrm{....}\left(6\right)$ in the equation $\mathrm{....}\left(2\right)$

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

Hence, the law of Tangent is $\frac{a-b}{a+b}=\frac{\tan \left[\frac{1}{2}\left(A-B\right)\right]}{\tan \left[\frac{1}{2}\left(A+B\right)\right]}$ .