Question

if necessary, state your solution as an improper fraction.
40°
20°
m∠1 =
m∠2 =
m∠3 =
m∠4 =
m∠5 =
m∠6 =

Explanation:

Step1: Find $\angle6$

In the small - right triangle with the $40^{\circ}$ angle, since the sum of the interior angles of a triangle is $180^{\circ}$ and one angle is $90^{\circ}$, then $m\angle6=180^{\circ}-90^{\circ}-40^{\circ}=50^{\circ}$.

Step2: Find $\angle1$

$\angle1$ and the $40^{\circ}$ angle are vertical angles. Vertical angles are equal, so $m\angle1 = 40^{\circ}$.

Step3: Find $\angle2$

In the small - right triangle with $\angle1$, since the sum of the interior angles of a triangle is $180^{\circ}$ and one angle is $90^{\circ}$ and $\angle1 = 40^{\circ}$, then $m\angle2=180^{\circ}-90^{\circ}-40^{\circ}=50^{\circ}$.

Step4: Find $\angle3$

$\angle3$ and $\angle2$ are vertical angles. Vertical angles are equal, so $m\angle3 = 50^{\circ}$.

Step5: Find $\angle4$

$\angle4$ and $\angle1$ are vertical angles. So $m\angle4 = 40^{\circ}$.

Step6: Find $\angle5$

In the large triangle on the right, one angle is $90^{\circ}$ and another non - right angle is composed of a $20^{\circ}$ angle and part of the right - angle from the other triangle. The sum of the interior angles of a triangle is $180^{\circ}$. The non - right angle of the large right - triangle is $20^{\circ}+(90^{\circ}-50^{\circ}) = 60^{\circ}$. Then $m\angle5=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$.

Answer:

$m\angle1 = 40^{\circ}$
$m\angle2 = 50^{\circ}$
$m\angle3 = 50^{\circ}$
$m\angle4 = 40^{\circ}$
$m\angle5 = 30^{\circ}$
$m\angle6 = 50^{\circ}$