Question

If A(-2, 1), B(2, 3) and C(-2, -4) be the vertices of a \[\Delta ABC\] , show that \[\tan B=\dfrac{2}{3}\] .

Answer 1

Hint:
 Use the fact that the slope of the line joining the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$. Substitute the value of ${{x}_{1}},{{x}_{2}},{{y}_{1}},{{y}_{2}}$ in each case and hence find the slopes of the lines.
Also, another important fact that would be used in this question would be as follows
The formula to calculate the angle between two line having slopes as x and y is as follows
\[\theta ={{\tan }^{-1}}\left[ \left| \dfrac{x-y}{1-x\cdot y} \right| \right]\]
(Where \[\theta \] is the angle between the two lines with slopes as x and y)
So, in this question, we will first calculate the slopes of the lines AB and BC(using the formula that is given in the hint), and then we will use the formula to find the angle between these two lines which would be the angle B.

Complete step-by-step answer:
Now, as mentioned in the question, we have points as A(-2, 1), B(2, 3) and C(-2, -4).
We know that the slope of the line joining the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
In the first case, for AB, we have ${{x}_{1}}=-2,{{x}_{2}}=2,{{y}_{1}}=1$ and ${{y}_{2}}=3$
Hence, we have
${{m}_{1}}=\dfrac{3-1}{2-(-2)}=\dfrac{2}{4}=\dfrac{1}{2}$
Hence the slope of the line AB is$\dfrac{1}{2}$ .
In the second case, for BC, we have ${{x}_{1}}=-2,{{x}_{2}}=2,{{y}_{1}}=-4$ and ${{y}_{2}}=3$
Hence, we have
${{m}_{2}}=\dfrac{3-4}{2-(-2)}=\dfrac{-1}{4}$
Hence the slope of the line BC is $\dfrac{-1}{4}$.
Now, using the formula that is given in the hitn, we can find the angle between the two line AB and BC as follows
The formula to calculate the angle between two line having slopes as x and y is as follows
\[\theta ={{\tan }^{-1}}\left[ \left| \dfrac{x-y}{1-x\cdot y} \right| \right]\]
(Where \[\theta \] is the angle between the two lines with slopes as x and y)
Now, we can write as follows
\[\begin{align}
  & B={{\tan }^{-1}}\left[ \left| \dfrac{\dfrac{1}{2}-\dfrac{\left( -1 \right)}{4}}{1-\dfrac{1}{2}\cdot \dfrac{\left( -1 \right)}{4}} \right| \right] \\
 & B={{\tan }^{-1}}\left[ \left| \dfrac{\dfrac{3}{4}}{1+\dfrac{1}{8}} \right| \right] \\
 & B={{\tan }^{-1}}\left[ \left| \dfrac{\dfrac{3}{4}}{\dfrac{8+1}{8}} \right| \right] \\
 & B={{\tan }^{-1}}\left[ \left| \dfrac{2}{3} \right| \right] \\
 & \tan B=\dfrac{2}{3} \\
\end{align}\]
Hence, the result is proved.



Note: Another approach to the question is draw a triangle ABC then from that find the length of each side and use the cosine rule to find the angle B , Now once we have angle B we can manipulate the cosine to term to get the sine term as well , Now use tangent identity in terms of sine in cosine to get the required result.