Exterior angle is sum of interior opposite angles
For ∆ABC
∠1 = ∠ABC + ∠ACB
∠2 = ∠BAC + ∠ACB
∠3 = ∠BAC + ∠ABC
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°
