Question

find m∠1.
9.
10.
11.
see problem 1.

Explanation:

Step1: Recall angle - sum property of a triangle

The sum of the interior angles of a triangle is $180^{\circ}$.

Step2: For problem 9

Let the three angles of the triangle be $\angle1$, $117^{\circ}$, and $33^{\circ}$. Then, using the angle - sum property, we have $\angle1+117^{\circ}+33^{\circ}=180^{\circ}$.
So, $\angle1 = 180^{\circ}-(117^{\circ}+33^{\circ})$.
First, calculate the sum inside the parentheses: $117^{\circ}+33^{\circ}=150^{\circ}$.
Then, $\angle1=180^{\circ}-150^{\circ}=30^{\circ}$.

Step3: For problem 10

Let the three angles of the triangle be $\angle1$, $44.7^{\circ}$, and the angle supplementary to $52.2^{\circ}$. The angle supplementary to $52.2^{\circ}$ is $180^{\circ}-52.2^{\circ}=127.8^{\circ}$.
Using the angle - sum property of a triangle $\angle1 + 44.7^{\circ}+127.8^{\circ}=180^{\circ}$.
So, $\angle1=180^{\circ}-(44.7^{\circ}+127.8^{\circ})$.
First, calculate the sum inside the parentheses: $44.7^{\circ}+127.8^{\circ}=172.5^{\circ}$.
Then, $\angle1 = 180^{\circ}-172.5^{\circ}=7.5^{\circ}$.

Step4: For problem 11

Let the three angles of the triangle be $\angle1$, $33^{\circ}$, and $57^{\circ}$. Using the angle - sum property of a triangle $\angle1+33^{\circ}+57^{\circ}=180^{\circ}$.
So, $\angle1=180^{\circ}-(33^{\circ}+57^{\circ})$.
First, calculate the sum inside the parentheses: $33^{\circ}+57^{\circ}=90^{\circ}$.
Then, $\angle1 = 90^{\circ}$.

Answer:

1. $m\angle1 = 30^{\circ}$
2. $m\angle1 = 7.5^{\circ}$
3. $m\angle1 = 90^{\circ}$