SATHEE: Properties of Triangles 2 Question 11

Properties of Triangles 2 Question 11

11. Consider the following statements concerning a $△ A B C$ (i) The sides $a, b, c$ and area of triangle are rational.

(ii) $a, \tan \frac{B}{2}, \tan \frac{C}{2}$ are rational.

(iii) $a, \sin A, \sin B, \sin C$ are rational.

$Prove that (i) \Rightarrow (ii) \Rightarrow (iii) \Rightarrow (i)$

(1994, 5M)

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Solution:

It is given that $a, b, c$ and area of triangle are rational.

We have, $\tan \frac{B}{2} = \sqrt{\frac{(s - c) (s - a)}{s (s - b)}}$

$= \frac{\sqrt{s (s - a) (s - b) (s - c)}}{s (s - b)}$

Again, $a, b, c$ are rational given, $s = \frac{a + b + c}{2}$ are rational, Also, $(s - b)$ is rational, since triangle is rational, therefore we get

$\tan \frac{B}{2} = \frac{Δ}{s (s - b)} is rational.$

Similarly, $\tan \frac{C}{2} = \frac{Δ}{s (s - c)}$ is rational.

Therefore $a, \tan \frac{B}{2}, \tan \frac{C}{2}$ are rational.

which shows that, (i) $\Rightarrow$ (ii).

Again, it is given that, $a, \tan \frac{B}{2}, \tan \frac{C}{2}$ are rational, then

$\tan \frac{A}{2} = \tan \frac{π}{2} - \frac{B + C}{2}$

$\begin{aligned} = \cot \frac{B + C}{2} = \frac{1}{\tan \frac{B}{2} + \frac{C}{2}} \\ = \frac{1 - \tan \frac{B}{2} \cdot \tan \frac{C}{2}}{\tan \frac{B}{2} + \tan \frac{C}{2}} \end{aligned}$

Since, $\tan (B / 2)$ and $\tan (C / 2)$ are rational, hence $\tan (A / 2)$ is a rational.

Now, $\sin A = \frac{2 \tan A / 2}{1 + \tan^{2} A / 2}$ as $\tan (A / 2)$ is a rational number, $\sin A$ is a rational. Similarly, $\sin B$ and $\sin C$ are. Thus, $a, \sin A, \sin B, \sin C$ are rational, therefore (ii) $\Rightarrow$ (iii).

Again, $a, \sin A, \sin B, \sin C$ are rational.

By the sine rule,

$\begin{aligned} \frac{a}{\sin A} & = \frac{b}{\sin B} = \frac{c}{\sin C} \\ \Rightarrow b & = \frac{a \sin B}{\sin A} and c = \frac{a \sin C}{\sin A} \end{aligned}$