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About this video
One Grade 8 exam question worked in full: use the exterior angle of a triangle to set up an equation with only one unknown, solve it, then use a straight line for the second unknown and classify the triangle.
What it covers
- Exterior angle = the sum of the two interior opposite angles
- Choosing the property that gives an equation with ONE unknown
- Substituting the answer back, in brackets, to find each angle
- Angles on a straight line for the second unknown
- Classifying the triangle — read the whole question
Extra worked examples
✍️ Extra example: in △ABC, BC is extended to D. The exterior angle AĈD = 3x + 10°, Â = x + 30° and B̂ = x + 20°. Find x, then classify △ABC.
- 13x + 10° = (x + 30°) + (x + 20°)Ext ∠ of △ABC = the two interior opposite angles — the only equation with one unknown.
- 23x + 10° = 2x + 50° → x = 40°Variables to the left, numbers to the right.
- 3Â = 70°, B̂ = 60°, AĈD = 3(40°) + 10° = 130°Check: 70° + 60° = 130° ✓
- 4AĈB = 180° − 130° = 50°∠s on a str line.
- 570°, 60° and 50°: no two equal, so △ABC is scalene.The question also asked you to classify — always read the full question.
✍️ Extra example: in △PQR, QR is extended to S. PR̂S = 2y + 16°, PR̂Q = 3y − 6° and P̂ = 40°. Find y and Q̂.
- 1(2y + 16°) + (3y − 6°) = 180°∠s on a str line.
- 25y + 10° = 180° → 5y = 170° → y = 34°Group like terms, then the inverses.
- 3PR̂S = 2(34°) + 16° = 84° and PR̂Q = 3(34°) − 6° = 96°Substitute in brackets.
- 4Q̂ = 84° − 40° = 44°Ext ∠ = sum of the int opp ∠s. Check: 40° + 44° + 96° = 180° ✓
Exercises to try
Work each one out, then click to reveal the answer.
- 1In △KLM, LM is extended to N. KM̂N = 125° and K̂ = 70°. Find L̂.
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125° = 70° + L̂ (ext ∠ of △) → L̂ = 55° - 2In △ABC, BC is extended to D. AĈD = 4x, Â = x + 25° and B̂ = 2x + 5°. Find x.
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4x = 3x + 30° (ext ∠ of △) → x = 30° - 3An exterior angle of a triangle is 142° and one interior opposite angle is 67°. Find the other interior opposite angle and the angle next to the exterior angle.
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142° − 67° = 75°; 180° − 142° = 38° (∠s on a str line); check: 67° + 75° + 38° = 180° ✓ - 4Two angles on a straight line are 3y − 20° and y + 40°. Find y.
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(3y − 20°) + (y + 40°) = 180° → 4y = 160° → y = 40° - 5A triangle has angles of 50°, 65° and 65°. Classify it by its angles and by its sides.
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All three are less than 90°, so acute; two equal angles mean two equal sides, so isosceles
