The problem asks to find the size of angle $k$ in the triangle. We are given an exterior angle of $244^\circ$, and two interior angles of the smaller triangles formed by the lines extending from the vertices where angle $k$ sits; these are $29^\circ$ and $35^\circ$.

1. Problem Description

The problem asks to find the size of angle $k$ in the triangle. We are given an exterior angle of $244^\circ$, and two interior angles of the smaller triangles formed by the lines extending from the vertices where angle $k$ sits; these are $29^\circ$ and $35^\circ$.

2. Solution Steps

First, find the two interior angles adjacent to the exterior angle of $244^\circ$.

Since the sum of angles on a straight line is $180^\circ$, these two angles can be found by subtracting $29^\circ$ and $35^\circ$ from $180^\circ$ respectively. However, since we know the exterior angle we calculate the adjacent interior angle by $360 - 244 = 116$.

The sum of angles around a point is $360^\circ$. The interior angle adjacent to $244^\circ$ is $360^\circ - 244^\circ = 116^\circ$.

Then, calculate the remaining interior angles of the main triangle by finding the supplementary angles to $29^\circ$ and $35^\circ$.

Supplementary angle to $29^\circ = 180^\circ - 29^\circ = 151^\circ$. But since we know the supplementary angle to the exterior angle that forms with the $244^\circ$ we proceed.

Let the two remaining angles in the triangle be $a$ and $b$.

$a = 180^\circ - 29^\circ = 151^\circ$, we want the other angle.

$b = 180^\circ - 35^\circ = 145^\circ$, we want the other angle.

So instead use the $180 - (180-a)= a = 29$, $b = 35$ to construct small angles from $180- a$ and $180 - b$

Let the two angles be $x = 180-a$ and $y=180-b$.

We know that $a+b+116=180$. Then $a+b=180 - 116= 64$.

We know that the sum of angles in a triangle is $180^\circ$.

$k + (180^\circ - 29^\circ) + (180^\circ - 35^\circ) = 180$ doesn't work.

Let's try another approach. The adjacent angles of $29$ and $35$ are $a$ and $b$ such that the given exterior angle is $244$.

The angle at the bottom is therefore $360 - 244= 116$. Therefore the sum of the other two angles is

$180-116 = 64$.

Let's find the angles in the two smaller triangles near the bottom vertices.

The small angles are $180-116-29 = 35$ and $180 - 116 - 35 = 29$. We know therefore that $k$ is an exterior angle that can be found from $29+35 = 64$

However, if the two smaller angles were inside the main triangle, $29$ and $35$, $k+29+35 =180$ becomes a problem.

The two inside angles are not formed by small triangles, therefore the angle we have been given is $360-244 = 116$.

Thus we now only need the angles of $29+35 =180-64 =116$.

Final angles are now $64$.

Since the sum of angles in a triangle is $180^\circ$, we have $k+29+35 =180$ so $180 - 64=116$. This can't be either since this can't be $244 =29+35+2(90)=

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4.

Looking closely we realize that the $29$ and $35$ do not make up the inside angles, but the exterior angles to them. So $A=180-29=151$, and the other exterior angle = $180-35=145$. So $k +151 + 145=180=k+296$. This does not add up therefore

Let us draw a diagram and see where that gets us

3. Final Answer

64