The figure below is composed of a square and a regular pentagon.

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-14.jpg?height=517&width=517&top_left_y=455&top_left_x=678)

Calculate the sum of the angles $a^{\circ}$ and $b^{\circ}$.

Facts that can help. (You can use them!). The sum of the interior angles of a triangle is always equal to $180^{\circ}$. Additionally, the sum of the interior angles of a quadrilateral is always equal to $360^{\circ}$. To see this, simply divide the quadrilateral into two triangles by connecting two opposite vertices.

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-14.jpg?height=340&width=348&top_left_y=1338&top_left_x=754)

Each triangle has $180^{\circ}$ as the sum of its interior angles, hence we get $180^{\circ} \times 2 = 360^{\circ}$ as the sum of the interior angles of the quadrilateral. And the sum of the interior angles of a pentagon is equal to $180^{\circ} \times 3 = 540^{\circ}$, since we can divide any pentagon into three triangles as shown in the following figure.

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-14.jpg?height=366&width=369&top_left_y=1987&top_left_x=752)

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Solution

Consider the triangle $A P M$ in the following figure:

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-42.jpg?height=554&width=532&top_left_y=1236&top_left_x=662)

Since $A \hat{M} P=(180-a)^{\circ}$, and the sum of the internal angles of a triangle is $180^{\circ}$, then $A \hat{P} M=(a-90)^{\circ}$. Applying the same argument to triangle $B Q N$, we get that $B \hat{Q} N=(b-90)^{\circ}$. Note that $S \hat{P} Q=A \hat{P} M=(a-90)^{\circ}$ and that $P \hat{Q} R=B \hat{Q} N=(b-90)^{\circ}$ since $S \hat{P} Q$ and $A \hat{P} M$ are vertically opposite angles at vertex $P$ and, similarly, $P \hat{Q} R$ and $B \hat{Q} N$ are vertically opposite angles at vertex $Q$.

Let's focus on the quadrilateral $S P Q R$ illustrated below:

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-42.jpg?height=229&width=688&top_left_y=2170&top_left_x=593)

Since the pentagon is regular, all five of its internal angles are equal, so the sum of its internal angles is equal to $5 \times P \hat{S} R$. Since the sum of the internal angles of the pentagon is $540^{\circ}$, we have that $5 \times P \hat{S} R=540^{\circ}$, from which we obtain that $P \hat{S} R=108^{\circ}$. Since the measure of $Q \hat{R} S$ is equal to that of $P \hat{S} R$, we have that $Q \hat{R} S=108^{\circ}$.

The sum of the internal angles of the quadrilateral $P Q R S$ is equal to $360^{\circ}$. Therefore,

$$
(a-90)^{\circ}+(b-90)^{\circ}+108^{\circ}+108^{\circ}=360^{\circ}
$$

Thus, we have that $(a+b)^{\circ}-180^{\circ}+108^{\circ}+108^{\circ}=360^{\circ}$, from which we conclude that $a^{\circ}+b^{\circ}=$ $324^{\circ}$.