 ## Sum of Angles in a Triangle

The sum of all three angles in a triangle is always 180°. In the given figure ABC is a triangle.

The sum of its all three angles is always 180° i.e. ∠ABC + BAC + ACB = 180°.

### Theoretical proof:

Given: ABC is a triangle. BAC, ABC and ACB are interior angles of triangle ABC.
To prove: ABC + BAC + ACB = 180°
Construction: Through A, a line XAY parallel to BC is drawn.

Proof:
Statements                                          Reasons
1.    XAB = ABC -------------------------> Alternate angles are equal.
2.    YAC = ACB -------------------------> Alternate angles are equal.
3.    XAB + BAC + YAC = 180° -----> Straight angle.
4.    ABC + BAC + ACB = 180° -----> From statements 1, 2 and 3.
Proved.

### Alternative method:

Given: ABC is a triangle. BAC, ABC and ACB are interior angles of triangle ABC.
To prove: ABC + BAC + ACB = 180°
Construction: A circle through A, B and C i.e. a circumcircle ABC is drawn.

Proof:
Statements                                    Reasons
1.    ABC = ½ A͡C -------------------> Inscribed angle is half of opposite arc.
2.    BAC = ½ B͡C -------------------> Inscribed angle is half of opposite arc.
3.    ACB = ½ A͡C -------------------> Inscribed angle is half of opposite arc.
4.    ABC + BAC + ACB = ½ (A͡C + B͡C + A͡C) ---> Adding statements 1, 2 and 3.
= ½ ʘABC
= ½ × 360°
= 180°         Proved.

### Workout Examples

Example 1: Find the unknown angle in the given figure.
Solution: From the figure,
x + 50° + 70° = 180° ----------> Sum of angles in a triangles.
or,     x + 120° = 180°
or,     x = 180° – 120°
or,     x = 60°

Example 2: Find the unknown angles in the given figure.
Solution: From the figure,
x + 40° + 80° = 180° ----------> Sum of angles in a triangles.
or,     x + 120° = 180°
or,     x = 180° – 120°
or,     x = 60°

y = 40° + 80° ------> exterior angle of a Δ is equal to the sum of opposite interior angles.
or,     y = 120°

x = 60° and y = 120°

Example 3: Find the values of x, y and z in the given figure.
Solution: From the figure,
2x + 9° + x + x + 7° = 180° ----------> Sum of angles in a triangles.
or,     4x + 16° = 180°
or,     4x = 180° – 16°
or,     4x = 164°
or,     x = 164°/4
or,     x = 41°

y = x + 7°------> Alternate angles.
= 41° + 7°
= 48°

z = 2x + 9°------> Alternate angles.
= 2×41° + 9°
= 82° + 9°
= 91°

x = 41° and y = 48° and z = 91°

Example 4: Find the values of x and y in the given figure.
Solution: From the figure,
x = y – 20° ----------> base angles of an isosceles Δ.

x + y – 20° + 120° = 180° ------> sum of angles of a Δ.
or,     y – 20° + y – 20° + 120° = 180° ------> putting x = y – 20°.
or,     2y + 80° = 180°
or,     2y = 180° – 80°
or,     2y = 100°
or,     y = 100°/2
or,     y = 50°

x = y – 20°
= 50° – 20°
= 30°

x = 30° and y = 50°

Example 5: Find the values of a, and b in the given figure.
Solution: From the figure,
a + 50° + 60° = 180° ----------> Sum of angles in ΔABC.
or,     a + 110° = 180°
or,     a = 180° – 110°
or,     a = 70°

Again,
b + 60° + 30° = 180° ----------> Sum of angles in ΔADE.
or,     b + 90° = 180°
or,     b = 180° – 90°
or,     b = 90°

a = 70° and b = 90°

You can comment your questions or problems regarding the angles of a triangle here.