Sum of angles of a triangle is 180°

Here,

∠A + ∠B + ∠C = 180°

For proof, please check Theorem 6.7 (link)

Let’s solve some examples

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Show that sum of angles of the triangle is 180°
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Sum of all angles of triangle = ∠A + ∠B + ∠C

= 30° + 80° + 70°

= 110° + 70°

= 180°

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Show that sum of angles of the triangle is 180°
**

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Sum of all angles of triangle = ∠P + ∠Q + ∠R

= 40° + 120° + 20°

= 180°

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**
Show that sum of angles of the triangle is 180°
**

**
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Sum of all angles of triangle = ∠X + ∠Y + ∠Z

= 45° + 90° + 45°

= 180°

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Find the missing angle of following triangles
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In ∆ABC,

∠A + ∠B + ∠C = 180°
*
(Angle sum property of triangle)
*

55° + 75° + ∠C = 180°

130° + ∠C = 180°

∠C = 180° − 130°

∠C = 50°

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Find the missing angle of following triangles
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In ∆PQR,

∠P + ∠Q + ∠R = 180°
*
(Angle sum property of triangle)
*

30° + ∠Q + 50° = 180°

∠Q + 80° = 180°

∠Q = 180° − 80°

∠Q = 100°

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**
Find the missing angle of following triangles
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In ∆XYZ,

∠X + ∠Y + ∠Z = 180°
*
(Angle sum property of triangle)
*

20° + 90° + ∠Z = 180°

110° + ∠Z = 180°

∠Z = 180° − 110°

∠Z = 70°