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8th period 2 - 3 mathxl for school: practice & problem - solving copy 1…

Question

8th period 2 - 3 mathxl for school: practice & problem - solving copy 1 due sep 24 - 11:45 pm part 2 of 2 for the figure shown, find m∠1 and m∠2. m∠1 = 119° m∠2 = □° (the figure is not drawn to scale.)

Explanation:

Step1: Use angle - sum property of a triangle

The sum of interior angles of a triangle is 180°. In the outer - most triangle, one angle is 146° and another is 21°. Let's first find the third angle of this triangle. Let this angle be \(x\). Then \(x=180-(146 + 21)=180 - 167=13^{\circ}\).

Step2: Use linear - pair property

\(\angle1\) and the angle adjacent to it (the one we just found, \(x = 13^{\circ}\)) form a linear pair. So \(m\angle1=180 - 13=119^{\circ}\) (as given).

Step3: Use angle - sum property of the inner triangle

In the inner triangle, one angle is \(36^{\circ}\) and \(\angle1 = 119^{\circ}\). Let \(m\angle2=y\). Then, by the angle - sum property of a triangle (\(y+36 + 119=180\)). Solving for \(y\), we get \(y=180-(36 + 119)=180 - 155 = 25^{\circ}\).

Answer:

\(m\angle2 = 25^{\circ}\)