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Question
interactive practice: describe angle relationships in triangles
use the diagram to show that the sum of the measures of the exterior angles of
a triangle, one at each vertex, is 360.
Step1: Use the linear - pair relationship
For each vertex of the triangle, if an interior angle is \(x\) and the corresponding exterior angle is \(y\), then \(x + y=180^{\circ}\).
Let the interior angles of the triangle be \(\angle1,\angle2,\angle3\) and the exterior angles be \(\angle4,\angle5,\angle6\).
We have \(\angle1+\angle4 = 180^{\circ}\), \(\angle2+\angle5=180^{\circ}\), \(\angle3+\angle6 = 180^{\circ}\).
So \((\angle1+\angle4)+(\angle2+\angle5)+(\angle3+\angle6)=180^{\circ}\times3 = 540^{\circ}\).
Step2: Use the sum of interior angles of a triangle
We know that the sum of the interior angles of a triangle \(\angle1+\angle2+\angle3=180^{\circ}\) (by the triangle - angle sum theorem).
Let \(S=\angle4+\angle5+\angle6\).
From \((\angle1+\angle4)+(\angle2+\angle5)+(\angle3+\angle6)=540^{\circ}\), we can rewrite it as \((\angle1+\angle2+\angle3)+S = 540^{\circ}\).
Step3: Solve for the sum of exterior angles
Substitute \(\angle1+\angle2+\angle3 = 180^{\circ}\) into \((\angle1+\angle2+\angle3)+S = 540^{\circ}\).
We get \(180^{\circ}+S=540^{\circ}\).
Subtract \(180^{\circ}\) from both sides: \(S=\angle4+\angle5+\angle6=540^{\circ}-180^{\circ}=360^{\circ}\).
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The sum of the measures of the exterior angles of a triangle, one at each vertex, is \(360^{\circ}\)