https://file.aiquickdraw.com/mathbot/file/172546091787251yf2id0.jpg

Let's solve each of the problems in the image one by one:

### 1. Triangle ABC
- Given: \( \angle B = 80^\circ \) and triangle \( ABC \) is isosceles (since \( AB = AC \)).
- We need to find \( \angle A \) (denoted as \( x \)).

**Solution:**
In an isosceles triangle, the base angles are equal. Therefore, \( \angle C = \angle B = 80^\circ \).

Using the angle sum property of a triangle:
\[
\angle A + \angle B + \angle C = 180^\circ
\]
\[
x + 80^\circ + 80^\circ = 180^\circ
\]
\[
x + 160^\circ = 180^\circ
\]
\[
x = 20^\circ
\]

So, \( \angle A = 20^\circ \).

### 2. Triangle NML
- Given: \( \angle N = 20^\circ \), and the triangle is isosceles (\( NL = ML \)).
- We need to find \( \angle M \) (denoted as \( b \)).

**Solution:**
Since triangle \( NML \) is isosceles, the base angles \( \angle NML = \angle MNL \).

Using the angle sum property of a triangle:
\[
\angle N + \angle M + \angle L = 180^\circ
\]
Let \( \angle NML = \angle MNL = b \). Therefore:
\[
20^\circ + 2b = 180^\circ
\]
\[
2b = 160^\circ
\]
\[
b = 80^\circ
\]

So, \( \angle M = 80^\circ \).

### 3. Triangle PRQ
- Given: \( \angle PQR = 50^\circ \), \( \angle QRP = x \), and the triangle is isosceles (\( PQ = PR \)).
- We need to find \( \angle PRQ \).

**Solution:**
Since triangle \( PRQ \) is isosceles, the base angles \( \angle PQR = \angle PRQ = x \).

Using the angle sum property of a triangle:
\[
\angle P + 2x = 180^\circ
\]
\[
50^\circ + 2x = 180^\circ
\]
\[
2x = 130^\circ
\]
\[
x = 65^\circ
\]

So, \( \angle PRQ = 65^\circ \).

### 4. Triangle PQS
- Given: \( \angle QPR = 59^\circ \), \( \angle PRS = 32^\circ \), and \( QR = RS \).
- We need to find \( \angle QRS = x \) and \( \angle RQS = y \).

**Solution:**
Since \( QR = RS \), triangle \( QRS \) is isosceles.

We first find \( \angle PRQ \) using the angle sum property in triangle \( PRQ \):
\[
\angle PRQ = 180^\circ - \angle QPR - \angle PRS = 180^\circ - 59^\circ - 32^\circ = 89^\circ
\]

Now in triangle \( QRS \), since it is isosceles:
\[
\angle RQS = \angle QRS = y
\]

Using the angle sum property:
\[
89^\circ + 2y = 180^\circ
\]
\[
2y = 91^\circ
\]
\[
y = 45.5^\circ
\]

So, \( \angle RQS = \angle QRS = 45.5^\circ \).

### 5. Triangle CDE
- Given: \( \angle D = 63^\circ \), \( CD = DE \), and \( \angle CED = x \).
- We need to find \( \angle BCA \).

**Solution:**
Since \( CD = DE \), triangle \( CDE \) is isosceles. The base angles \( \angle DCE = \angle EDC \).

Using the angle sum property in triangle \( CDE \):
\[
x + 63^\circ + x = 180^\circ
\]
\[
2x = 117^\circ
\]
\[
x = 58.5^\circ
\]

So, \( \angle CED = 58.5^\circ \).

If you have any questions or need further details, feel free to ask!

### Related Questions:
1. What is the general formula for the sum of the angles in any polygon?
2. How do you determine the type of triangle given three side lengths?
3. Can you prove that the base angles of an isosceles triangle are always equal?
4. What is the relationship between the exterior angle of a triangle and the interior opposite angles?
5. How can you use the properties of an equilateral triangle to solve for unknown angles?

### Tip:
Always double-check your triangle's properties (isosceles, equilateral, scalene) as they often provide helpful clues to solving for unknown angles.

Find the lettered angle in each of the following triangles.

Learn how to find the lettered angle in various isosceles triangles using key geometric concepts and theorems. This problem involves the Angle Sum Property and the Isosceles Triangle Theorem. Suitable for Grade 7 students.

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