4. On the figure, lines $a, b, c, d, e$ and $f$ as well as angles $\alpha, \beta$ and $\gamma$ are marked. For the lines on that figure, it holds that $a\|b, c\| d, e \| f$ and $a \perp d$. The measure of angle $\alpha$ is $50^{\circ} 24^{\prime}$. Determine the measures of angles $\beta$ and $\gamma$ from the figure.

![](https://cdn.mathpix.com/cropped/2024_06_03_6e1b38aefafca10e6d1eg-5.jpg?height=808&width=848&top_left_y=1840&top_left_x=635)

The use of a pocket calculator or any reference materials is not allowed.

Turn the page!
4. On the figure, lines $a, b, c, d, e$ and $f$ as well as angles $\alpha, \beta$ and $\gamma$ are marked. For the lines on that figure, it holds that $a\|b, c\|$ $d, e \| f$ and $a \perp d$. The measure of angle $\alpha$ is $50^{\circ} 24^{\prime}$. Determine the measures of angles $\beta$ and $\gamma$ from the figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_0b00c6f52e60bdba44a4g-14.jpg?height=820&width=851&top_left_y=384&top_left_x=637)

## Solution.

Angles $\alpha$ and $\gamma$ are angles with parallel sides ($c \| d$ and $e \| f$).

$\alpha$ is an acute angle, $\gamma$ is an obtuse angle, so they are supplementary, i.e., $\alpha+\gamma=180^{\circ}$.

1 POINT

From this expression, we calculate $\gamma=180^{\circ}-50^{\circ} 24^{\prime}=179^{\circ} 60^{\prime}-50^{\circ} 24^{\prime}=129^{\circ} 36^{\prime}$.

1 POINT

Let's denote the auxiliary angles $\delta$ and $\delta^{\prime}$ as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_0b00c6f52e60bdba44a4g-14.jpg?height=914&width=854&top_left_y=1573&top_left_x=607)

Angles $\alpha$ and $\delta$ are acute angles formed by the transversal $e$ of the parallel lines $c$ and $d$, so their measures are equal.

(It can also be said that $\alpha$ and $\delta$ are angles with parallel sides $c \| d, e$ is the common side.)

Thus, $\delta=\alpha=50^{\circ} 24^{\prime} \quad 1$ POINT

From the relations $a \perp d$ and $d \| c$ it follows that $a \perp c$, so $\delta+\delta^{\prime}=90^{\circ}$. 1 POINT

$\delta+\delta^{\prime}=90^{\circ}$, so $50^{\circ} 24^{\prime}+\delta^{\prime}=90^{\circ}$, from which it follows

$\delta^{\prime}=90^{\circ}-50^{\circ} 24^{\prime}=89^{\circ} 60^{\prime}-50^{\circ} 24^{\prime}=39^{\circ} 36^{\prime}$.

1 POINT

Angles $\delta^{\prime}$ and $\beta$ are acute and obtuse angles formed by the transversal $e$ of the parallel lines $a$ and $b$, so they are supplementary, i.e., $\delta^{\prime}+\beta=180^{\circ}$.

From this, it follows that $\beta=180^{\circ}-\delta^{\prime}=179^{\circ} 60^{\prime}-39^{\circ} 36^{\prime}=140^{\circ} 24^{\prime}$.

1 POINT

TOTAL 6 POINTS

Note: If the student solves the problem using the sum of the measures of the angles in a triangle, such a solution should be evaluated in accordance with the provided distribution of points.