In the diagram provided for problem 5, you are given the following information:

• $m \angle A = 76^\circ$,
• $m \angle B = 119^\circ$,
• $m \angle ADE = 53^\circ$, and you are asked to find $m \angle C$.

Steps to Solve:

1. Identify the Quadrilateral: The diagram shows quadrilateral $ABCD$ with angles at $A$, $B$, $C$, and $D$.

2. Angle Sum in a Quadrilateral: The sum of the interior angles in any quadrilateral is always $360^\circ$. Therefore, we can write the equation: $m \angle A + m \angle B + m \angle C + m \angle D = 360^\circ$

3. Find $m \angle D$:

   • From the diagram, $\angle ADE$ is an exterior angle to $\triangle ABD$.
   • The exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles. So, $m \angle ADE = m \angle A + m \angle B$ Substituting the known values: $53^\circ = 76^\circ + 119^\circ$ Therefore, this relationship holds true, but we don't directly solve for $m \angle D$ using this method.

4. Finding $m \angle C$: Now, using the angle sum property of a quadrilateral: $76^\circ + 119^\circ + m \angle C + 127^\circ = 360^\circ$ Simplifying this: $322^\circ + m \angle C = 360^\circ$ Solving for $m \angle C$: $m \angle C = 360^\circ - 322^\circ = 38^\circ$

Therefore, $m \angle C = 38^\circ$.

This result matches what is shown in the diagram for $\angle C$, so the solution is consistent.

Would you like more details on any of these steps?

Related Questions:

1. What is the sum of the exterior angles of any polygon, including quadrilaterals?
2. How does the exterior angle theorem apply to other polygons?
3. Can the angle sum property be extended to polygons with more than four sides? If yes, how?
4. How do you classify quadrilaterals based on their angles?
5. How can you use the properties of parallel lines to deduce angles in polygons?

Tip: Remember that the sum of interior angles for any polygon can be found using the formula $180^\circ(n - 2)$, where $n$ is the number of sides.