###
**
Exterior angle is sum of interior opposite angles
**

**
**

For ∆ABC

∠1 = ∠ABC + ∠ACB

∠2 = ∠CAB + ∠CBA

∠3 = ∠BAC + ∠BCA

Let’s solve some questions

###
**
Proof of exterior angle property
**

**
**

In ∆ABC,

Also,

∠1 + ∠4 = 180°
*
(Linear Pair)
*

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

∠4 = 180° − 180° + ∠2 + ∠3

∠4 = 0° + ∠2 + ∠3

**
∠4 = ∠2 + ∠3
**

∴ Exterior angle is equal to sum of interior opposite angles

###
**
Find exterior
**
**
angle ∠ 1
**

**
**

Here,

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

∠1 = 45° + 60°

∠1 = 105°

###
**
Find exterior angle
**

**
**

In ∆ABC,

∠BCD = ∠CAB + ∠ABC
*
(Exterior angle property)
*

∠BCD = 85° + 25°

**
∠BCD = 110°
**

###
**
Find exterior angle
**

**
**

In ∆PQR,

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

∠SQR = 30° + 15°

**
∠SQR = 45°
**

###
**
Find exterior angle
**

**
**

In ∆XYZ,

∠XZO = ∠X + ∠Y
*
(Exterior angle property)
*

∠XZO = 20° + 90°

**
∠XZO = 110°
**

###
**
Find ∠ACB
**

**
**

In ∆ABC,

∠DAC = ∠B + ∠C
*
(Exterior angle property)
*

120° = 40° + ∠C

120° − 40° = ∠C

80° = ∠C

∠C = 80°

i.e.
**
∠ACB = 80°
**

###
**
Find
**
**
∠PQR
**

**
**

In ∆PQR,

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

160° = ∠PQR + 50°

160° − 50° = ∠PQR

110° = ∠PQR

**
∠PQR = 110°
**

###
**
Find
**
**
∠XZY
**

**
**

In ∆XYZ,

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

140° = ∠XZY + 90°

140° − 90° = ∠XZY

50° = ∠XZY

**
∠XZY = 50°
**