We are given a diagram that shows a triangle. We are also given the exterior angle of one of the vertices of the triangle, which is $249^{\circ}$. Two interior angles of the triangle are given as $33^{\circ}$ and $25^{\circ}$. We are asked to find the size of the third angle, denoted by $a$.

1. Problem Description

We are given a diagram that shows a triangle. We are also given the exterior angle of one of the vertices of the triangle, which is $249^{\circ}$. Two interior angles of the triangle are given as $33^{\circ}$ and $25^{\circ}$. We are asked to find the size of the third angle, denoted by $a$.

2. Solution Steps

First, we need to find the interior angle corresponding to the exterior angle $249^{\circ}$.

Since the sum of the interior angle and its exterior angle is $360^{\circ}$, we can calculate the interior angle as follows:

$180^\circ$ - (360 - 249) = $180^\circ - 111^\circ = 69^\circ$

Let the interior angles of the triangle be $33^{\circ}$, $25^{\circ}$, and $a$.

Since the sum of the angles in a triangle is $180^{\circ}$, we can write:

$a + 33^{\circ} + 25^{\circ} = 180^{\circ}$.

Solving for $a$:

$a = 180^{\circ} - 33^{\circ} - 25^{\circ}$

$a = 180^{\circ} - (33^{\circ} + 25^{\circ})$

$a = 180^{\circ} - 58^{\circ}$

$a = 122^{\circ}$

Since the exterior angle of 249 is not at angle $a$, the other two interior angles given are 33 and

2

5. Therefore, the equation is:

$a + (360 - 249) = 180$

$360-249=111$

We now have:

The sum of interior angles is $180^{\circ}$. The interior angles are $33$, $25$ and $a$. Therefore:

$33 + 25 + a = 180$

$58+a=180$

$a=180-58=122^{\circ}$

3. Final Answer

The size of angle $a$ is $122^{\circ}$.