We are asked to find the size of angle $q$ in the given triangle. We are given the angles $47^\circ$ and $24^\circ$ at two vertices of the triangle, and the exterior angle $235^\circ$ at the third vertex.

1. Problem Description

We are asked to find the size of angle $q$ in the given triangle. We are given the angles $47^\circ$ and $24^\circ$ at two vertices of the triangle, and the exterior angle $235^\circ$ at the third vertex.

2. Solution Steps

First, we need to find the interior angle adjacent to the exterior angle $235^\circ$. Since the sum of an interior and exterior angle on a straight line is $180^\circ$, the interior angle is $180^\circ - 235^\circ = -55^\circ$. This does not make sense, so we need to consider the angles around the point where the three sides meet. The angles $47^\circ$ and $24^\circ$ are shown as exterior angles and are thus not angles inside the triangle.

We are given exterior angles of $235^\circ, 47^\circ, 24^\circ$.

Let the interior angles of the triangle be $a, b, c$, where $a$ is the angle adjacent to $235^\circ$, $b$ is the angle adjacent to $47^\circ$, and $c$ is the angle adjacent to $24^\circ$. Note that $q$ in the figure is simply $a$.

We have:

$a + 235^\circ = 360^\circ$, which means $a = 360^\circ - 235^\circ = 125^\circ$.

We have:

$b = 180^\circ - 47^\circ = 133^\circ$.

$c = 180^\circ - 24^\circ = 156^\circ$.

Since $a, b, c$ are the interior angles of a triangle and hence should add up to $180^\circ$, we made a mistake.

Let us find the interior angles that are adjacent to the exterior angles of $47^\circ$ and $24^\circ$ at the base. Let these interior angles be $x$ and $y$.

Then $x = 180^\circ - 47^\circ = 133^\circ$.

And $y = 180^\circ - 24^\circ = 156^\circ$.

This is still not right.

The angles inside the triangle, say $a, b, c$, form angles of $235^\circ$, $47^\circ$ and $24^\circ$ *outside* the triangle at the respective corners of the triangle. So $235^\circ + a = 360^\circ \Rightarrow a = 360^\circ - 235^\circ = 125^\circ$. But this still implies exterior angles $47^\circ, 24^\circ$.

We are given that the exterior angle at one vertex is $235^\circ$. Thus the interior angle is $360^\circ - 235^\circ = 125^\circ$. The other two angles are $47^\circ$ and $24^\circ$. The sum of angles in a triangle is $180^\circ$. So we must calculate $180^\circ - (47^\circ + 24^\circ)$ = $180^\circ - 71^\circ = 109^\circ$.

Thus $q = 109^\circ - a - b = 109^\circ - a - b$ cannot be the case since a and b are unknown.

Instead let us deduce the interior angles at the bottom of the triangle. The angles at the base are $180^\circ - 47^\circ = 133^\circ$ and $180^\circ - 24^\circ = 156^\circ$. Then this is impossible since sum of the angles in a triangle must be $180^\circ$.

Instead, let the angles inside the triangle at the base be $47^\circ$ and $24^\circ$. The angle adjacent to $235^\circ$ must be $360^\circ - 235^\circ = 125^\circ$. The interior angles of the triangle must add up to $180^\circ$. Let the angle $q$ be unknown.

$q + 47^\circ + 24^\circ = 180^\circ$.

$q + 71^\circ = 180^\circ$.

$q = 180^\circ - 71^\circ = 109^\circ$.

3. Final Answer

$q = 109^\circ$