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Finding Exterior Angles
Last updated at July 21, 2026 by Teachoo
Class 7 (Ganita Prakash 1, 2 & old NCERT)
Chapter 7 Class 7 - A tale of three Intersecting Lines (Ganit Prakash)
- Definition
- Constructing Equilateral Triangle
- Constructing a Triangle when its Sides are given
- Figure it out - Page 150, 151
- Are Triangles Possible for any Lengths?
- Figure it out - Page 154
- Visualising the Construction of Circles
- Figure it out - Page 156
- Triangle Inequality & Construction of Circles
- Figure it out - Page 159
- Constructing Triangle when 2 Sides and the Included Angle are given
- Constructing Triangle when 2 Angles and the Included Side are given
- Do triangles always exist?
- Sum of angles of a triangle
- Angle Sum Property
-
Exterior Angles
- Constructions Related to Altitudes of Triangles
- Figure it out - Page 170, 171
Transcript
Finding Exterior Angles In ∆ABC, ∠BCD = ∠CAB + ∠ABC ∠BCD = 85° + 25° ∠BCD = 110° (Exterior angle property) Find exterior angle In ∆PQR, ∠SQR = ∠P + ∠R ∠SQR = 30° + 15° ∠SQR = 45° (Exterior angle property) Find exterior angle In ∆XYZ, ∠XZO = ∠X + ∠Y ∠XZO = 20° + 90° ∠XZO = 110° (Exterior angle property) Find ∠ACB In ∆ABC, ∠DAC = ∠B + ∠C 120° = 40° + ∠C 120° − 40° = ∠C 80° = ∠C ∠C = 80° i.e. ∠ACB = 80° (Exterior angle property) Find ∠PQR In ∆PQR, ∠PRS = ∠PQR + ∠QPR 160° = ∠PQR + 50° 160° − 50° = ∠PQR 110° = ∠PQR ∠PQR = 110° (Exterior angle property) Find ∠XZY In ∆XYZ, ∠OXZ = ∠XZY + ∠XYZ 140° = ∠XZY + 90° 140° − 90° = ∠XZY 50° = ∠XZY ∠XZY = 50° (Exterior angle property)
