Question

In a $△ABC$, if $\cot A\cot B\cot C>0$. then the $△$ is

• A. Acute angled
• B. Right angled
• C. Abtuse angles
• D. Does not exist.

Answer 1

The correct option is A Acute angled

$\cot A\cot B\cot C>0$
$\cot \theta$ is positive in first and third quadrant only, since no angle of a triangle can be greater than ${180}^o$. So we will not consider third quadrant
possible only when
i) $\cot A,\cot B$ and $\cot C$ all lie in first quadrant i.e. $A,B,C\in (0,\frac{\pi }{2})$
In this case all angles become acute angles
ii)when say, $\cot A,\cot B$ is negative and $\cot C$ is positive i.e. any two value is negative and one is positive
Thus,$(A,B),(B,C),(C,A)\in (\frac{\pi }{2},\pi )$...which is not possible in a triangle as only one angle can be greater than ${90}^o$
So, In a triangle $A,B,C\in (0,\frac{\pi }{2})$ hence $cotA,cotB,cotC>0\Rightarrow cotA.cotB.cotC>0$
Option A