Question

m∠1 =
m∠2 =
m∠3 =
m∠4 =
m∠5 =
m∠6 =
m∠7 =
m∠8 =

Explanation:

Step1: Find $m\angle1$

In a triangle, the sum of interior - angles is $180^{\circ}$. In the triangle with angles $42^{\circ}$, $47^{\circ}$ and $\angle1$, we have $m\angle1=180-(42 + 47)=91^{\circ}$.

Step2: Find $m\angle2$

$\angle2$ and the $71^{\circ}$ angle are vertical angles. Vertical angles are equal, so $m\angle2 = 71^{\circ}$.

Step3: Find $m\angle3$

In the triangle with $\angle2 = 71^{\circ}$ and $\angle1 = 91^{\circ}$, we find $m\angle3$. Since the sum of angles in a triangle is $180^{\circ}$, $m\angle3=180-(71 + 91)=18^{\circ}$.

Step4: Find $m\angle4$

In the triangle with $\angle3 = 18^{\circ}$ and the $36^{\circ}$ angle, we find $m\angle4$. Using the angle - sum property of a triangle, $m\angle4=180-(18 + 36)=126^{\circ}$.

Step5: Find $m\angle5$

$\angle5$ and $\angle2$ are supplementary (linear - pair). Since $\angle2 = 71^{\circ}$, $m\angle5=180 - 71=109^{\circ}$.

Step6: Find $m\angle6$

In the triangle with $\angle5 = 109^{\circ}$ and $\angle1 = 91^{\circ}$, we find $m\angle6$. Using the angle - sum property of a triangle, $m\angle6=180-(109 + 91)= - 20^{\circ}$, which is wrong. Let's use another way. In the large triangle with angles $36^{\circ}$, $42^{\circ}$ and $\angle6$, $m\angle6=180-(36 + 42)=102^{\circ}$.

Step7: Find $m\angle7$

In the triangle with $\angle6 = 102^{\circ}$ and the $71^{\circ}$ angle, $m\angle7=180-(102 + 71)=7^{\circ}$.

Step8: Find $m\angle8$

In the triangle with $\angle7 = 7^{\circ}$ and the $36^{\circ}$ angle, $m\angle8=180-(7 + 36)=137^{\circ}$.

Answer:

$m\angle1 = 91^{\circ}$
$m\angle2 = 71^{\circ}$
$m\angle3 = 18^{\circ}$
$m\angle4 = 126^{\circ}$
$m\angle5 = 109^{\circ}$
$m\angle6 = 102^{\circ}$
$m\angle7 = 7^{\circ}$
$m\angle8 = 137^{\circ}$