Question

In $\Delta ABC$ if $\frac{\cos A}{a}=\frac{\cos B}{b}$ then it is an isosceles triangle. (With the standard notations.)

• A. True
• B. False

Answer 1

The correct option is A True

We have $\Delta$ ABC, $\cos A=\frac{b^2+c^2-a^2}{2bc},\cos B=\frac{a^2+c^2-b^2}{2ac}$.

Now $\frac{\cos A}{a}=\frac{\cos B}{b}$

or, $\frac{b^2+c^2-a^2}{2abc}=\frac{a^2+c^2-b^2}{2abc}$.

or, $b^2+c^2-a^2=a^2+c^2-b^2$

or, $b^2=a^2$

$\Rightarrow b=a$ [ Since $b$ and $a$ are length of sides of $\Delta$ ABC]

Now in $\Delta$ ABC, two sides are of equal length , so the triangle is isosceles.