We need to find the value of angle $k$ in the given figure. The figure consists of a quadrilateral. Two sides of the quadrilateral are equal, forming an isosceles triangle. One angle of the quadrilateral is given as $150^\circ$, and there are two right angles ($90^\circ$).

1. Problem Description

We need to find the value of angle $k$ in the given figure. The figure consists of a quadrilateral. Two sides of the quadrilateral are equal, forming an isosceles triangle. One angle of the quadrilateral is given as $150^\circ$, and there are two right angles ($90^\circ$).

2. Solution Steps

First, let's find the interior angle adjacent to the $150^\circ$ angle. Let this interior angle be $x$. Since the angles around a point add up to $360^\circ$, we have

$x + 150^\circ = 360^\circ$.

$x = 360^\circ - 150^\circ = 210^\circ - 60^\circ = 150^\circ$.

Therefore, $x=30^\circ$.

The quadrilateral has four interior angles. The sum of the interior angles of a quadrilateral is given by:

$(n-2) \times 180^\circ$, where $n$ is the number of sides. In this case, $n=4$, so the sum of interior angles is $(4-2) \times 180^\circ = 2 \times 180^\circ = 360^\circ$.

The angles in the quadrilateral are $90^\circ$, $90^\circ$, $30^\circ$, and angle $k$ and another angle in the triangle. Let the angle adjacent to angle $k$ in the isosceles triangle be $k$. The other two angles are the same. Thus we can set up the following equation:

$90^\circ + 90^\circ + 30^\circ + k + k = 360^\circ$.

The quadrilateral consists of a right angle, the $30^{\circ}$ angle (adjacent to the $150^\circ$ angle), another right angle, and the two angles in the triangle, each of which is $k$. The sum of these angles must equal to 360 degrees.

We have an isosceles triangle where the two equal sides are marked. Therefore the two angles opposite the equal sides are also equal to k.

The sum of angles in a triangle is $180^\circ$, so

$k + k + \alpha = 180^\circ$.

$2k + \alpha = 180^\circ$.

$\alpha = 180^\circ - 2k$.

The sum of the interior angles of the quadrilateral is $360^\circ$, so

$90^\circ + 90^\circ + (360^\circ - 150^\circ) + (180^\circ - 2k) = 360^\circ$.

$90^\circ + 90^\circ + 30^\circ + k + (90^\circ-k+k) = 360^\circ$.

$90^\circ + 90^\circ + 30^\circ + 180^\circ - 2k = 360^\circ$

But we know that the sum of the angles must be

3

6

0. So let the other angle in the triangle that isn't $k$ be called $a$.

We know that $a+2k = 180$, $a = 180 - 2k$.

So, we have $90 + 90 + 30 + (180-2k) = 360$,

$390-2k = 360$,

$30 = 2k$,

$k = 15$.

3. Final Answer

15

The value of angle $k$ is 15 degrees.

Final Answer: The final answer is $\boxed{15}$