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