CBSE Class 8 Mathematics Lines and Angles MCQs Set 32

Answer: (d) 80

Explanation:
1. The sum of the interior angles of a hexagon is (6 − 2) × 180° = 720°. So 9x + 0 = 720, which gives x = 80.

Teacher's Note:
1. Use the angle sum formula (n − 2) × 180° for an n-sided polygon.
2. Add the x terms and the constants separately.

Answer: (b) 40°

Explanation:
1. ∠AOC = 90°, so ∠COD = ∠AOD − ∠AOC = 130° − 90° = 40°. Check: ∠BOD = 50° and ½ (130° − 50°) = 40°.

Teacher's Note:
1. Split ∠AOD into the right angle AOC and the angle COD.
2. The result equals half the difference of ∠AOD and ∠BOD.

Answer: (b) 90°

Explanation:
1. Interior angles on the same side add up to 180°, so ∠BMN + ∠DNM = 180°. Their halves add up to 90°. In triangle MPN, ∠MPN = 180° − 90° = 90°.

Teacher's Note:
1. Co-interior angles between parallel lines are supplementary.
2. Half of 180° is 90°, and the triangle sum then leaves 90° for the angle at P.

Answer: (d) 20

Explanation:
1. The three angles lie on a straight line, so 60° + 4x° + 40° = 180°. Then 4x = 80 and x = 20.

Teacher's Note:
1. Angles on a straight line add up to 180°.
2. Do not forget to divide by 4 at the end.

Answer: (c) 50

Explanation:
1. The angles lie on the straight line PQ, so (x + 10) + x + (x + 20) = 180. Hence 3x + 30 = 180, 3x = 150 and x = 50.

Teacher's Note:
1. Opposite rays form a straight line, so the angles add up to 180°.
2. Collect like terms before solving.

Answer: (c) 71°

Explanation:
1. The exterior angle ∠CBD of triangle ABC = 34° + 30° = 64°. In triangle BED, ∠EBD = 64° and ∠D = 45°, so ∠BED = 180° − 64° − 45° = 71°.

Teacher's Note:
1. An exterior angle equals the sum of the two opposite interior angles.
2. Then use the angle sum of triangle BED.

Answer: (b) 180°

Explanation:
1. In a regular five-pointed star each tip angle is 36°, so the five tips add up to 5 × 36° = 180°. The same total holds for any such star, because each tip angle can be written using exterior angles of the small triangles.

Teacher's Note:
1. Think of the star as a pentagon with five triangles on its sides.
2. In a regular star each tip angle is 36°, and 5 × 36° = 180°.

Answer: (a) 60°

Explanation:
1. ∠QPR = 180° − 110° = 70° and ∠QRP = 180° − 130° = 50°. In triangle PQR, ∠PQR = 180° − 70° − 50° = 60°.

Teacher's Note:
1. Use linear pairs to get the angles of the triangle at P and R.
2. Then apply the angle sum of a triangle.

Answer: (d) 16°

Explanation:
1. ∠QPR = 180° − 65° − 33° = 82°, so ∠QPS = 41°. In right triangle PTQ, ∠QPT = 90° − 65° = 25°. Hence ∠TPS = 41° − 25° = 16°.

Teacher's Note:
1. Find the angle QPR first, then half of it.
2. Use the right triangle PTQ to get ∠QPT.

Answer: (d) 4 : 11

Explanation:
1. ∠POR + ∠POS = 180°, so ∠POA + ∠POB = 90°. With ratio 2 : 7, ∠POA = 20° and ∠POB = 70°. Then ∠POR = 40°, so ∠SOQ = 40° (vertically opposite). ∠BOQ = 180° − 70° = 110°. The ratio is 40 : 110 = 4 : 11.

Teacher's Note:
1. Half of a linear pair adds up to 90°.
2. Vertically opposite angles are equal, and ∠BOQ and ∠POB form a linear pair.

Answer: (b) 35

Explanation:
1. Complementary angles add up to 90°. So 2x − 10 + x − 5 = 90, 3x = 105 and x = 35. (The angles are 60° and 30°.)

Teacher's Note:
1. Complementary means the sum is 90°.
2. Check: 2(35) − 10 = 60 and 35 − 5 = 30.

Answer: (a) x = 108°, y = 72°

Explanation:
1. Interior angles on the same side of the transversal are supplementary: x + y = 180°. With ratio 3 : 2, x = 3/5 × 180° = 108° and y = 2/5 × 180° = 72°.

Teacher's Note:
1. Co-interior angles add up to 180°.
2. Divide 180° in the ratio 3 : 2.

Answer: (d) x = 65°, y = 60°, z = 55°

Explanation:
1. y = ∠PRC = ∠QPR = 60° (alternate angles). ∠PQR = 180° − 125° = 55°, and z = ∠QRD = ∠PQR = 55° (alternate angles). Then x = 180° − 60° − 55° = 65°.

Teacher's Note:
1. Alternate angles between parallel lines are equal.
2. The three angles at R on the line CD add up to 180°.

Answer: (c) 45°

Explanation:
1. Let the angle be x. Then x = (180° − x)/3, so 3x = 180° − x, 4x = 180° and x = 45°.

Teacher's Note:
1. Write the supplement as 180° − x.
2. Check: the supplement is 135° and one third of 135° is 45°.

Answer: (b) 70°

Explanation:
1. Vertically opposite angles are equal, so 3x + 10 = 5x − 30, which gives x = 20. The angle is 3(20) + 10 = 70°.

Teacher's Note:
1. Vertically opposite angles are equal.
2. Substitute x back to get the angle itself.

Answer: (a) 53° and 62°

Explanation:
1. Complement = 90° − 37° = 53°. Supplement = 180° − 118° = 62°.

Teacher's Note:
1. Complement uses 90° and supplement uses 180°.
2. Do each subtraction separately.

Answer: (c) 70°

Explanation:
1. Interior angles on the same side of the transversal are supplementary, so the other angle is 180° − 110° = 70°.

Teacher's Note:
1. Co-interior angles add up to 180°.
2. They are not equal unless each is 90°.

Answer: (c) 95°

Explanation:
1. An exterior angle of a triangle equals the sum of the two interior opposite angles: 55° + 40° = 95°.

Teacher's Note:
1. The exterior angle theorem avoids finding the third angle.
2. Check: third angle = 85°, and 180° − 85° = 95°.

Answer: (a) 1080°

Explanation:
1. Sum = (n − 2) × 180° = 6 × 180° = 1080°.

Teacher's Note:
1. Formula: (n − 2) × 180°.
2. An octagon has 8 sides.

Answer: (d) 110°

Explanation:
1. Let the angles be x and x − 40. Then 2x − 40 = 180, so x = 110°. The other angle is 70°.

Teacher's Note:
1. Write one angle in terms of the other.
2. Their sum must be 180°.