Question

example 7 if m∠2 = 41°, m∠5 = 94°, and m∠10 = 109°, find each measure. a. m∠1 = b. m∠3 = c. m∠4 = d. m∠6 = e. m∠7 = f. m∠8 = g. m∠9 = © gina wilson (all things algebra®, llc), 2014 - 2019

Explanation:

Step1: Use linear - pair angles property

$\angle1$ and $\angle2$ form a linear - pair. Since the sum of angles in a linear - pair is $180^{\circ}$, and $m\angle2 = 41^{\circ}$, then $m\angle1=180 - 41=139^{\circ}$

Step2: Use linear - pair angles property for $\angle3$

$\angle3$ and $\angle2$ form a linear - pair. So $m\angle3 = 180 - 41=139^{\circ}$

Step3: Use angle - sum property of a triangle

In the triangle with $\angle4$, $\angle5$ and an unknown angle, we know that the sum of angles in a triangle is $180^{\circ}$. Given $m\angle5 = 94^{\circ}$, and assuming the third angle in that triangle is part of a linear - pair with $\angle1$ or $\angle3$. Let's use the fact that the sum of angles around a point is $360^{\circ}$. But a more straightforward way is to note that if we consider the large triangle formed by $\angle4$, $\angle5$ and an angle adjacent to $\angle2$. Since the sum of angles in a triangle is $180^{\circ}$, and we know one angle related to $\angle2$ and $\angle5$, we have $m\angle4=180-(94) = 86^{\circ}$

Step4: Use vertical - angles property for $\angle6$

$\angle6$ and $\angle2$ are vertical angles. Vertical angles are equal. So $m\angle6 = m\angle2=41^{\circ}$

Step5: Use angle - sum property of a triangle for $\angle7$

In the triangle with $\angle6$, $\angle7$ and $\angle10$. The sum of angles in a triangle is $180^{\circ}$. Given $m\angle6 = 41^{\circ}$ and $m\angle10 = 109^{\circ}$, then $m\angle7=180-(41 + 109)=30^{\circ}$

Step6: Use linear - pair angles property for $\angle8$

$\angle8$ and $\angle10$ form a linear - pair. So $m\angle8=180 - 109 = 71^{\circ}$

Step7: Use vertical - angles property for $\angle9$

$\angle9$ and $\angle2$ are vertical angles. So $m\angle9 = m\angle2=41^{\circ}$

Answer:

a. $m\angle1 = 139^{\circ}$
b. $m\angle3 = 139^{\circ}$
c. $m\angle4 = 86^{\circ}$
d. $m\angle6 = 41^{\circ}$
e. $m\angle7 = 30^{\circ}$
f. $m\angle8 = 71^{\circ}$
g. $m\angle9 = 41^{\circ}$