Question

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Explanation:

Step1: Find \( m\angle1 \)

In the first triangle (with angles \( 85^\circ \), \( 40^\circ \), and the angle adjacent to \( \angle1 \)), the sum of angles in a triangle is \( 180^\circ \). So the angle adjacent to \( \angle1 \) is \( 180 - 85 - 40 = 55^\circ \). Since \( \angle1 \) and this angle are vertical angles? No, wait, \( \angle1 \) and the angle we just found are supplementary? Wait, no, the two triangles are connected by vertical angles. Wait, the first triangle has angles \( 85^\circ \), \( 40^\circ \), so the third angle (let's call it \( \angle x \)) is \( 180 - 85 - 40 = 55^\circ \). Then \( \angle1 \) is equal to \( \angle x \)? Wait, no, \( \angle1 \) and the angle we just calculated are vertical angles? Wait, no, the two triangles are connected at the vertex, so \( \angle1 \) and \( \angle2 \) are vertical angles, so they are equal. Wait, let's re - calculate.

In the left - hand triangle, the sum of interior angles is \( 180^{\circ} \). So the angle opposite to \( \angle1 \) (the angle inside the left - hand triangle) is \( 180-(85 + 40)=180 - 125 = 55^{\circ} \). Since \( \angle1 \) and this angle are vertical angles? No, \( \angle1 \) and the angle we just found are actually the same as vertical angles? Wait, no, the two triangles are formed by intersecting lines, so \( \angle1=\angle2 \) (vertical angles). And in the left - hand triangle, the angle adjacent to \( \angle1 \) is \( 55^{\circ} \), so \( \angle1 = 55^{\circ} \) (because the sum of angles in a triangle is \( 180^{\circ} \), so \( 85+40 + \text{angle adjacent to } \angle1=180 \), so that angle is \( 55^{\circ} \), and \( \angle1 \) is equal to that angle? Wait, no, the angle adjacent to \( \angle1 \) and \( \angle1 \) are supplementary? No, I think I made a mistake. Let's start over.

The sum of angles in a triangle is \( 180^{\circ} \). For the left triangle with angles \( 85^{\circ} \) and \( 40^{\circ} \), the third angle (let's call it \( \alpha \)) is \( 180-(85 + 40)=55^{\circ} \). Now, \( \angle1 \) and \( \alpha \) are supplementary? No, \( \angle1 \) and \( \alpha \) are actually vertical angles? Wait, no, the two triangles are connected at the vertex, so the angle \( \alpha \) and \( \angle1 \) are vertical angles? Wait, no, when two lines intersect, vertical angles are equal. The two triangles are formed by two intersecting lines, so the angle inside the left triangle ( \( \alpha = 55^{\circ} \)) and \( \angle1 \) are vertical angles? Wait, no, \( \angle1 \) is equal to \( \alpha \), so \( m\angle1 = 55^{\circ} \).

Step2: Find \( m\angle2 \)

Since \( \angle1 \) and \( \angle2 \) are vertical angles, vertical angles are equal. So if \( m\angle1 = 55^{\circ} \), then \( m\angle2=m\angle1 = 55^{\circ} \).

Step3: Find \( m\angle3 \)

In the right - hand triangle, we know \( m\angle2 = 55^{\circ} \) and \( m\angle\text{(the other angle)}=55^{\circ} \) (wait, no, the right - hand triangle has an angle of \( 55^{\circ} \) ( \( \angle2 \)) and another angle of \( 55^{\circ} \)? Wait, no, the right - hand triangle has angles \( \angle2 \), \( 55^{\circ} \), and \( \angle3 \). The sum of angles in a triangle is \( 180^{\circ} \). So \( m\angle3=180-(m\angle2 + 55^{\circ}) \). Since \( m\angle2 = 55^{\circ} \), then \( m\angle3=180-(55 + 55)=180 - 110 = 70^{\circ} \).

Answer:

\( m\angle1 = 55^{\circ} \), \( m\angle2 = 55^{\circ} \), \( m\angle3 = 70^{\circ} \)