Question

If ADB be the $\Delta^{le}$ with the length of side as Integer's the which of the following will be rational number

• A. Circum radius
• B. hradius
• C. Area of $\Delta^{le}$
• D. COS (A-B), Cos (B-C), COS (C-A).

Answer 1

Answer: (D)

(iv)

Answer 2

Let the side lengths of the triangle be a, b, and c, where a, b, c are integers.

1. Area ($\Delta$): $\Delta^2 = s(s-a)(s-b)(s-c) = \frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}$. Since $a,b,c$ are integers, $\Delta^2$ is always rational. However, $\Delta = \sqrt{\Delta^2}$ is not always rational (e.g., for sides 5,6,7, $\Delta = 6\sqrt{6}$).

2. Circumradius (R): $R = \frac{abc}{4\Delta}$. Since $\Delta$ can be irrational, $R$ can be irrational (e.g., for sides 5,6,7, $R = \frac{35\sqrt{6}}{24}$).

3. Inradius (r): $r = \frac{\Delta}{s}$. Since $\Delta$ can be irrational and $s$ is rational, $r$ can be irrational (e.g., for sides 5,6,7, $r = \frac{2\sqrt{6}}{3}$).

4. $\cos(X-Y)$: Consider $\cos(A-B) = \cos A \cos B + \sin A \sin B$.

• $\cos A = \frac{b^2+c^2-a^2}{2bc}$. Since $a,b,c$ are integers, $\cos A$ is rational. Similarly, $\cos B$ and $\cos C$ are rational.
• $\sin A = \frac{2\Delta}{bc}$ and $\sin B = \frac{2\Delta}{ac}$.
• $\cos(A-B) = \cos A \cos B + \left(\frac{2\Delta}{bc}\right)\left(\frac{2\Delta}{ac}\right) = \cos A \cos B + \frac{4\Delta^2}{abc^2}$.

Since $\cos A \cos B$ is a product of rationals (hence rational), and $\frac{4\Delta^2}{abc^2}$ is a ratio of a rational ($4\Delta^2$) and an integer ($abc^2$) (hence rational), their sum $\cos(A-B)$ is always rational. Similarly, $\cos(B-C)$ and $\cos(C-A)$ are always rational.