Worksheet Solutions: Angles | Mathematics Class 6 ICSE PDF Download

A. Multiple Choice Questions

Q1. What type of angle is 74°?
(a) Right 
(b) Obtuse 
(c) Acute 
(d) Straight
Ans: (c) Acute
Sol: Acute angles are more than 0° and less than 90°. 74° fits this range.

Q2. Around a point, three angles are 120°, 95°, and x°. What is x?
(a) 145° 
(b) 155° 
(c) 165° 
(d) 175°
Ans: (a) 145°
Sol: Sum around a point = 360°. x = 360 − (120 + 95) = 360 − 215 = 145°.

Q3. The complement of 38° is:
(a) 38° 
(b) 42° 
(c) 52° 
(d) 142°
Ans: (c) 52°
Sol: Complementary angles sum to 90°. 90 − 38 = 52°.

Q4. Two lines intersect. If one angle is 128°, the vertically opposite angle is:
(a) 52° 
(b) 128° 

(c) 232° 
(d) 180°
Ans: (b) 128°
Sol: Vertically opposite angles are equal.

Q5. Two adjacent angles form a straight line. One is 63°. The other is:
(a) 27° 
(b) 90° 
(c) 117° 

(d) 123°
Ans: (c) 117°
Sol: Linear pair sum = 180°. 180 − 63 = 117°.

Part B — Short Answer (5)

Q6. A line segment AB is 14 cm long. Point C is on AB such that AC = 9 cm. Find CB.
Ans: 5 cm
Sol:

A line segment AB is divided into two parts: AC and CB.

By property: AB = AC + CB.

Substitute: 14 = 9 + CB.

Rearranging: CB = 14 − 9 = 5.

Therefore, CB = 5 cm.

Q7. Find the supplement of 71°. Also say what type of angle it is.
Ans: 109°, obtuse
Sol: 

Supplementary angles add up to 180°.

So, supplement of 71° = 180° − 71° = 109°.

Since 90° < 109° < 180°, the angle is obtuse.

Q8. In triangle XYZ, sides XY = 6 cm, YZ = 8 cm, XZ = 10 cm.
(a) Is this a valid triangle (closed figure)?
(b) Check if XY + YZ > XZ.
Ans: 
A triangle is formed if the sum of any two sides is greater than the third side (Triangle Inequality Rule).

Check: XY + YZ = 6 + 8 = 14 > 10 = XZ. Condition holds.

Similarly, XY + XZ = 6 + 10 = 16 > 8 = YZ (holds).

And YZ + XZ = 8 + 10 = 18 > 6 = XY (holds).

Since all conditions are satisfied, it is a valid triangle.

Q9. Three angles around a point are 90°, 110°, and k°. Find k and state the angle type.
Ans: k = 160°, 
.The sum of all angles around a point is always 360°.

So, 90 + 110 + k = 360.

Add known angles: 200 + k = 360.

Subtract: k = 360 − 200 = 160.

Since 90° < 160° < 180°, k is an obtuse angle.

Q10. Two adjacent angles form a linear pair. One angle is 30° more than the other. Find both angles.
Ans: 75° and 105°
Sol 

Linear pair means two angles are adjacent and their sum = 180°.

Let the smaller angle = x. Then larger angle = x + 30.

Equation: x + (x + 30) = 180.

Simplify: 2x + 30 = 180 → 2x = 150 → x = 75.

Larger angle = 75 + 30 = 105.

So, the two angles are 75° and 105°.

C . Long Answer Questions

Q11. Angles around a point
At point O, five angles are placed in order: AOB = 55°, BOC = 70°, COD = 90° (right angle), DOE = p°, and EOA = q°.
(a) Find p + q.
(b) If the region EOA equals the complement of AOB, find q and then p.
(c) Classify DOE by type.

Ans:
The property of angles around a point states: ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.
Substitute known values: 55 + 70 + 90 + p + q = 360.
Add: 215 + (p + q) = 360.
So, p + q = 360 − 215 = 145°.
Given that ∠EOA = complement of ∠AOB. Complement of 55° = 90 − 55 = 35°.
So, q = 35°.
From step 4: p + q = 145 ⇒ p + 35 = 145 ⇒ p = 110.

Now classify: ∠DOE = 110°. Since 90° < 110° < 180°, it is obtuse.

Q12. 

Two angles are such that one is twice the other.
(a) If they are complementary, find both angles.
(b) If they are supplementary, find both angles.
Also state the type of each angle in both cases.

Ans:
(a) Complementary: 30° and 60° (acute and acute)
(b) Supplementary: 60° and 120° (acute and obtuse)

Sol (Stepwise):

1. Let the smaller angle = x. Then the larger angle = 2x.

Case (a): Complementary
2. Complementary means sum = 90°.
Equation: x + 2x = 90.
3. Simplify: 3x = 90 ⇒ x = 30.
4. Larger = 2x = 60.
5. So, the angles are 30° and 60° (both acute).

Case (b): Supplementary
6. Supplementary means sum = 180°.
Equation: x + 2x = 180.
7. Simplify: 3x = 180 ⇒ x = 60.
8. Larger = 2x = 120.
9. So, the angles are 60° (acute) and 120° (obtuse).

Q13. Two lines AB and CD intersect at O. One of the angles is 112°.
(a) Find the vertically opposite angle.
(b) Find the two adjacent angles that form linear pairs with 112°.
(c) Verify the linear-pair property.

Ans:
(a) 112°
(b) 68° and 68°
(c) 112° + 68° = 180° (twice), verified.

Sol:

1. Vertically opposite angles are equal ⇒ opposite to 112° is 112°.

2. Adjacent angles on a straight line form a linear pair: they sum to 180°. So each adjacent angle = 180 − 112 = 68°.

3. Check: 112 + 68 = 180 (holds for both adjacent angles).