Question

If $A(-2,1),B(2,3)$ and $C(-2,-5)$ are the vertices of an acute angled $△ABC,$ then the value of $\tan \angle B$ is

• A. $\frac{3}{4}$
• B. $\frac{2}{3}$
• C. $\frac{3}{2}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A $\frac{3}{4}$

Given : $A(-2,1),B(2,3)$ and $C(-2,-5)$

Clearly, $\angle B$ is the angle between line $BA$ and $BC$, then

Now,
$m_1=\text{slope of}BA=\frac{3-1}{2+2}=\frac{1}{2}{m}_2=\text{slope of}BC=\frac{-5-3}{-2-2}=2$

So,
$\tan \angle B=\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|\Rightarrow \tan \angle B=\left|\frac{2-\frac{1}{2}}{1+2\times \frac{1}{2}}\right|\therefore \tan \angle B=\frac{3}{4}$