Question

In a $△ABC$, if $\cot A\cdot \cot C=\frac{1}{2}$ and $\cot B\cdot \cot C=\frac{1}{18}$, then which of the following is/are true?

• A. $2\sec A=\sqrt{3}$
• B. $\sec C=\sqrt{17}$
• C. $△ABC$ is an acute angled triangle
• D. $△ABC$ is an obtuse angled triangle

Answer 1

Answer: (B), (C)

The correct options are
B $\sec C=\sqrt{17}$
C $△ABC$ is an acute angled triangle

Given : $\cot A\cdot \cot C=\frac{1}{2}$ and $\cot B\cdot \cot C=\frac{1}{18}$
So, $\tan A\cdot \tan C=2\cdots \left(1\right)$
$\tan B\cdot \tan C=18\cdots \left(2\right)$

Let $\tan C=x.$ Then
$\tan A=\frac{2}{x},\tan B=\frac{18}{x}$

In a triangle, we know that
$\tan A+\tan B+\tan C=\tan A\cdot \tan B\cdot \tan C\Rightarrow \frac{2}{x}+\frac{18}{x}+x=\frac{2}{x}\cdot \frac{18}{x}\cdot x\Rightarrow 20+x^2=36\Rightarrow x=\pm 4$
When $x=-4$, then
$\tan A=-\frac{1}{2},\tan B=-\frac{9}{2},\tan C=-4$
All angles are greater than ${90}^{\circ }$, which is not possible, so
$x=4\Rightarrow \tan A=\frac{1}{2},\tan B=\frac{9}{2},\tan C=4$
So, $△ABC$ is acute angled triangle.

Now, $\sec A=\sqrt{1+{\tan }^{2}A}=\frac{\sqrt{5}}{2}$
$\sec C=\sqrt{1+{\tan }^{2}C}=\sqrt{17}$