Properties Of Triangles Ques 27
- Consider the following statements concerning a $\triangle A B C$ (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.
$$ \text { Prove that (i) } \Rightarrow \text { (ii) } \Rightarrow \text { (iii) } \Rightarrow \text { (i) } $$
(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
$$ \tan \frac{B}{2}=\frac{\Delta}{s(s-b)} \text { is rational. } $$
Similarly, $\quad \tan \frac{C}{2}=\frac{\Delta}{s(s-c)}$ is rational.
Therefore $\quad a, \tan \frac{B}{2}, \tan \frac{C}{2}$ 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
$$ \tan \frac{A}{2}=\tan \frac{\pi}{2}-\frac{B+C}{2} $$
$$ \begin{aligned} & =\cot \frac{B+C}{2}=\frac{1}{\tan \frac{B}{2}+\frac{C}{2}} \\ & =\frac{1-\tan \frac{B}{2} \cdot \tan \frac{C}{2}}{\tan \frac{B}{2}+\tan \frac{C}{2}} \end{aligned} $$
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,
$$ \begin{aligned} \frac{a}{\sin A} & =\frac{b}{\sin B}=\frac{c}{\sin C} \\ \Rightarrow \quad b & =\frac{a \sin B}{\sin A} \text { and } c=\frac{a \sin C}{\sin A} \end{aligned} $$
Since $a, \sin A, \sin B$ and $\sin C$ are rational,
Hence, $b$ and $c$ are also rational.
Also, $\quad \Delta=\frac{1}{2} b c \sin A$
As $b, c$ and $\sin A$ are rational, so triangle is rational number. Therefore, $a, b, c$ and triangle are rational.
Therefore, (iii) $\Rightarrow$ (i).
BETA
