Q. 2 Find the values of the unknowns x and y in the following diagrams:

(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, 60° + \(\angle \)y + 30° = 180°
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°