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Exterior angle is sum of interior opposite angles
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**
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For ∆ABC

∠1 = ∠ABC + ∠ACB

∠2 = ∠BAC + ∠ACB

∠3 = ∠BAC + ∠ABC

Let’s solve some questions

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Proof of exterior angle property
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In ∆ABC,

Also,

∠1 + ∠4 = 180°
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(Linear Pair)
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(180° − ∠2 − ∠3) + ∠4 = 180°
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(From (1
*
*
)
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∠4 = 180° − 180° + ∠2 + ∠3

∠4 = 0° + ∠2 + ∠3

**
∠4 = ∠2 + ∠3
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∴ Exterior angle is equal to sum of interior opposite angles

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Find exterior
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angle ∠ 1
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Here,

∠1 = ∠B + ∠C (
*
Exterior angle property)
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∠1 = 45° + 60°

∠1 = 105°

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Find exterior angle
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**
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In ∆ABC,

∠BCD = ∠CAB + ∠ABC
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(Exterior angle property)
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∠BCD = 85° + 25°

**
∠BCD = 110°
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Find exterior angle
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**
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In ∆PQR,

∠SQR = ∠P + ∠R
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(Exterior angle property)
*

∠SQR = 30° + 15°

**
∠SQR = 45°
**

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Find exterior angle
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**
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In ∆XYZ,

∠XZO = ∠X + ∠Y
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(Exterior angle property)
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∠XZO = 20° + 90°

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∠XZO = 110°
**

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Find ∠ACB
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In ∆ABC,

∠DAC = ∠B + ∠C
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(Exterior angle property)
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120° = 40° + ∠C

120° − 40° = ∠C

80° = ∠C

∠C = 80°

i.e.
**
∠ACB = 80°
**

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Find
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∠PQR
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**
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In ∆PQR,

∠PRS = ∠PQR + ∠QPR
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(Exterior angle property)
*

160° = ∠PQR + 50°

160° − 50° = ∠PQR

110° = ∠PQR

**
∠PQR = 110°
**

###
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Find
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∠XZY
**

**
**

In ∆XYZ,

∠OXZ = ∠XZY + ∠XYZ
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(Exterior angle property)
*

140° = ∠XZY + 90°

140° − 90° = ∠XZY

50° = ∠XZY

**
∠XZY = 50°
**