# Finding the angles of a triangle using algebraic expressions

I think I have solved this question but want to check if I am correct with my formula

I used the theorem (rule) : the measure of the exterior angle of a triangle is equal to the sum of the remote (opposite) interior angles.

My workings as follows:

I am given one interior angle as $25^{\circ}$.

Another interior as $x + 15^{\circ}$ and the other is missing.

The final angle given is an exterior angle $3x - 10^{\circ}$ that is opposite the two interior angles.

$x+ 15^{\circ} + 25^{\circ}$ (opposite interior angles) $= 3x - 10^{\circ}$ (exterior angle). \begin{align*} x + 15^{\circ} + 25^{\circ} & = 3x - 10^{\circ}\\ x - x + 15^{\circ} + 25^{\circ} + 10^{\circ} & = 3x - 10^{\circ} + 10^{\circ} - x\\ 50^{\circ} & = 2x\\ \frac{50^{\circ}}{2} & = \frac{2x}{2}\\ 25^{\circ} & = x\\ x & = 25^{\circ} \end{align*}

So one of the interior angles is a Ready given in this question which is: $25$ degrees

And the other given is: $x + 15^{\circ}$. So if $x=25^{\circ}$ then this angle $= 40$ degrees.

And the exterior angle is $3 \times 25 - 10 = 65$ degrees.

Am I correct in this working?

Yes, it's correct. $$x+15^{\circ}+25^{\circ}=3x-10^{\circ}$$ or $$x=25^{\circ},$$ which gives measured angles of the triangle: $25^{\circ}$, $40^{\circ}$ and $115^{\circ}$.