Question

In a triangle $ABC$, $\sin A\cos B=\frac{1}{4}$ and $3\tan A=\tan B$ , the triangle is

• A. a right angled
• B. an equilateral
• C. an isosceles
• D. none of these

Answer 1

The correct option is A a right angled

We have, $3\tan A=\tan B$ ...(1)
$\Rightarrow \frac{3\cos B}{\sin B}=\frac{\cos A}{\sin A}$
$\Rightarrow 3\left(\frac{a^2+c^2-b^2}{2ac}\right)\frac{1}{b}=\left(\frac{b^2+c^2-a^2}{2bc}\right)\frac{1}{a}$ ..{ Using Sine and Cosine rule}
$\Rightarrow a^2+\frac{c^2}{2}=b^2$ ...(2)

Given that: $\sin A\cos B=\frac{1}{4}$

$\Rightarrow \sin A\left(\frac{a^2+c^2-b^2}{2ac}\right)=\frac{1}{4}$ ...{ cosine rule }

$\Rightarrow \sin A\left(\frac{c^2-\frac{c^2}{2}}{2ac}\right)=\frac{1}{4}$ ...{ From (2)}

$\Rightarrow \frac{a}{\sin A}=\frac{c}{1}=\frac{c}{\sin \frac{\pi }{2}}$

$\therefore C=\frac{\pi }{2}\{Sinerule:\frac{a}{\sin A}=\frac{c}{\sin C}\}$

$\therefore △ABC$ is right angled triangle.

Ans: A