Question

if side BC is parallel to an equilateral $\Delta $ABC is parallel to the x-axis. What are the slopes of its other sides?

Answer 1

Hint: In this problem, we are given that triangle $\Delta $ABC is an equilateral triangle that has side BC parallel to the x-axis which is figured as follows and the side AC and AB are extended to O and E respectively. Then, by using the concept that on the straight-line sum of the angles is ${{180}^{\circ }}$, we get the value of x. Then, by using another concept given by the slope s of any line by the formula where $\theta $ is the angle made by the line as $s=\tan \theta $.

Complete step by step answer:
We are supposed to find the slope of other sides of the triangle when side BC is parallel to an equilateral $\Delta $ABC is parallel to the x-axis.
Now given that triangle $\Delta $ABC is an equilateral triangle which has side BC parallel to the x-axis which is figured as follows and the side AC and AB are extended to O and E respectively as:

Now, by using the concept of the parallel lines that line EO is parallel to BC and we get the same angles using the property of equilateral triangles as:
$\begin{align}
  & \angle AEO={{60}^{\circ }} \\
 & \angle AOE={{60}^{\circ }} \\
\end{align}$
Now, by using the concept that on the straight line sum of the angles is ${{180}^{\circ }}$.
So, by using the above property, we get the value of x as:
$\begin{align}
  & {{60}^{\circ }}+x={{180}^{\circ }} \\
 & \Rightarrow x={{180}^{\circ }}-{{60}^{\circ }} \\
 & \Rightarrow x={{120}^{\circ }} \\
\end{align}$
Then, by using another concept given by the slope s of any line by the formula where $\theta $ is the angle made by the line as:
$s=\tan \theta $
So, we get the slope of side AB is given by the formula as:
$\begin{align}
  & \tan \left( \angle AEO \right)=\tan {{60}^{\circ }} \\
 & \Rightarrow \sqrt{3} \\
\end{align}$
So, we get the slope of the side AB is $\sqrt{3}$.
Then, we get the slope of side AC is given by the formula as:
$\begin{align}
  & \tan x=\tan {{120}^{\circ }} \\
 & \Rightarrow -\sqrt{3} \\
\end{align}$
So, we get the slope of the side AC is $-\sqrt{3}$.
Then, we get the slope of side BC which is parallel gives the angle as zero is given by the formula as:
$\begin{align}
  & \tan {{0}^{\circ }} \\
 & \Rightarrow 0 \\
\end{align}$
So, we get the slope of the side BC is $0$.
 Hence, the slope of the side AB is $\sqrt{3}$, the slope of the side AC is $-\sqrt{3}$, and the slope of the side BC is $0$.

Note: In this problem, we should know the all side of the equilateral triangle has equal length and it’s all angle are of ${{60}^{\circ }}$. Moreover, any line subtended between parallel lines gives all the angles as equal.