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Question
which statements are always true regarding the diagram? select three options. m∠5 + m∠3 = m∠4 m∠3 + m∠4 + m∠5 = 180° m∠5 + m∠6 = 180° m∠2 + m∠3 = m∠6 m∠2 + m∠3 + m∠5 = 180°
Step1: Analyze the angle - sum property of a triangle
In a triangle, the sum of the interior angles is \(180^{\circ}\). But here, we are dealing with exterior - angle relationships.
The exterior - angle theorem states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.
Step2: Check each option
- Option 1: \(m\angle5 + m\angle3=m\angle4\)
This is not correct. By the exterior - angle theorem, \(m\angle2 + m\angle3=m\angle6\) (not related to this equation).
- Option 2: \(m\angle3 + m\angle4 + m\angle5 = 180^{\circ}\)
This is not correct. The sum of angles in a triangle is \(180^{\circ}\), but \(\angle3,\angle4,\angle5\) do not form a triangle.
- Option 3: \(m\angle5 + m\angle6=180^{\circ}\)
Since \(\angle5\) and \(\angle6\) are adjacent angles forming a linear pair (they are on a straight line), by the linear - pair postulate, \(m\angle5 + m\angle6 = 180^{\circ}\).
- Option 4: \(m\angle2 + m\angle3=m\angle6\)
By the exterior - angle theorem (the exterior angle \(\angle6\) of a triangle is equal to the sum of the two non - adjacent interior angles \(\angle2\) and \(\angle3\)), \(m\angle2 + m\angle3=m\angle6\).
- Option 5: \(m\angle2 + m\angle3 + m\angle5=180^{\circ}\)
Since \(m\angle2 + m\angle3=m\angle6\) (from the exterior - angle theorem) and \(m\angle5 + m\angle6=180^{\circ}\) (linear pair), substituting \(\angle6\) gives \(m\angle2 + m\angle3 + m\angle5 = 180^{\circ}\).
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C. \(m\angle5 + m\angle6 = 180^{\circ}\), D. \(m\angle2 + m\angle3=m\angle6\), E. \(m\angle2 + m\angle3 + m\angle5=180^{\circ}\)