Question 186·Medium·Geometry and Trigonometry

The equal sides $RS$ and $RT$ meet at vertex $R$. That makes the angle at $R$ the vertex angle. Which side is the base, and where are the base angles located?

Let $x$ represent the measure of each base angle. Write an equation using the fact that all three angles in a triangle add up to $180^\circ$, then solve for $x$.

In Desmos, type the expression (180 - 68) / 2 on a new line. The value that Desmos outputs is the measure of each base angle, including $\angle RST$.

In an isosceles triangle, the angles opposite the equal sides are equal.

Here, sides $RS$ and $RT$ are congruent and meet at vertex $R$, so $R$ is the vertex of the isosceles triangle. That means the base is side $ST$, and therefore the base angles are the ones at $S$ and $T$:

• $\angle RST$ (at $S$)
• $\angle RTS$ (at $T$)

These two base angles are congruent (have the same measure).

The sum of the interior angles of any triangle is $180^\circ$.

We are given that $\angle SRT = 68^\circ$. This is the angle at $R$ (the vertex angle).

Let $x$ be the measure of each base angle. Then the three angles are:

• Vertex angle at $R$: $68^\circ$
• Base angle at $S$: $x$
• Base angle at $T$: $x$

So we write the equation:

Simplify:

Now solve for $2x$:

We found that $2x = 112$, so divide both sides by 2 to get one base angle:

So $\angle RST$, one of the base angles, measures $56^\circ$.