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Answer :

(i) \(\angle \)x + 50° = 120° (Exterior angle of a triangle)

\(\therefore \) \(\angle \)x = 120°- 50° = 70°

\(\angle \)x + \(\angle \)y + 50° = 180° (Angle sum property of a triangle)

70° + \(\angle \)y + 50° = 180°

\(\angle \)y + 120° = 180°

\(\angle \)y = 180° – 120°

\(\therefore \) \(\angle \)y = 60°

Thus \(\angle \)x = 70 and \(\angle \)y – 60°

(ii) \(\angle \)y = 80° (Vertically opposite angles are same)

\(\angle \)x + \(\angle \)y + 50° = 180° (Angle sum property of a triangle)

\(\therefore \) \(\angle \)x + 80° + 50° = 180°

\(\therefore \) \(\angle \)x + 130° = 180°

\(\therefore \)\(\angle \)x = 180° – 130° = 50°

Thus, \(\angle \)x = 50° and \(\angle \)y = 80°

(iii) \(\angle \)y + 50° + 60° = 180° (Angle sum property of a triangle)

\(\angle \)y + 110° = 180°

\(\therefore \)\(\angle \)y = 180°- 110° = 70°

\(\angle \)x + \(\angle \)y = 180° (Linear pairs)

Or, \(\angle \)x + 70° = 180°

\(\therefore \) \(\angle \)x = 180° – 70° = 110°

Thus, \(\angle \)x = 110° and y = 70°

(iv) \(\angle \)x = 60° (Vertically opposite angles)

\(\angle \)x + \(\angle \)y + 30° = 180° (Angle sum

Or, \(\angle \)y + 90° = 180°

Or, \(\angle \)y = 180° – 90° = 90°

Thus, \(\angle \)x = 60° and \(\angle \)y = 90°

(v) \(\angle \)y = 90° (Vertically opposite angles)

\(\angle \)x + \(\angle \)x + \(\angle \)y = 180° (Angle sum property of a triangle)

Or, 2 \(\angle \)x + 90° = 180°

Or, 2\(\angle \)x = 180° – 90°

Or, 2\(\angle \)x = 90°

\(\therefore \) \(\angle \)x=90°/2=45°

Thus, \(\angle \)x = 45° and \(\angle \)y = 90

(vi) From the given figure, we have

\(\angle \)y = \(\angle \)x

\(\angle \)1 = \(\angle \)x

\(\angle \)2 = \(\angle \)x

Adding both sides, we have
Adding both sides, we have
Adding both sides, we have
\(\angle \)y + \(\angle \)1 + \(\angle \)2 = 3\(\angle \)x

Or, 180° = 3\(\angle \)x [Angle sum property of a triangle]

\(\therefore \) \(\angle \)x=180°/3=60°

\(\angle \)x = 60°, \(\angle \)y = 60°

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