Question

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

• A. Angle C is more than ${90}^{\circ }$
• B. Right angled at C
• C. Angle C is less than ${90}^{\circ }$
• D. Information is insufficient

Answer 1

The correct option is B Right angled at C

$3tanA=tanB\Rightarrow 3sinAcosB=sinBcosA$
$\Rightarrow \frac{3}{4}=sinBcosA$
$\Rightarrow sinAcosB+cosAsinB=\frac{3}{4}+\frac{1}{4}=1$
$\Rightarrow sin(A+B)=1\Rightarrow A+B={90}^{\circ }\Rightarrow \angle C={90}^{\circ }$