**Q1: Find the complement of each of the following angles:**

**(i) \( 35^{\circ}\)**

**(ii) \( 47^{\circ}\)**

**(iii) \( 60^{\circ}\)**

**(iv) \( 73^{\circ}\)**

**Sol:**

(i) The given angle measures \(35^{\circ}\).

Let the measure of its complement be \( x^{\circ} \).

x + \(35^{\circ}\) = \(90^{\circ}\)

or, x = \( 90^{\circ} – 35^{\circ} = 55^{\circ} \)

Hence, the complement of the given angle will be \( 55^{\circ} \)

(ii) The given angle measures \( 47^{\circ}\).

Let the measure of its complement be \( x^{\circ} \).

x + \(47^{\circ}\) = \(90^{\circ}\)

or, x = \( 90^{\circ} – 47^{\circ} = 43^{\circ} \)

Hence, the complement of the given angle will be \( 43^{\circ} \)

(iii) The given angle measures \( 47^{\circ}\).

Let the measure of its complement be \( x^{\circ} \).

x + \( 60^{\circ}\) = \(90^{\circ}\)

or, x = \( 90^{\circ} – 60^{\circ} = 30^{\circ} \)

Hence, the complement of the given angle will be \( 30^{\circ} \)

(iv) The given angle measures \( 73^{\circ}\).

Let the measure of its complement be \( x^{\circ} \).

x + \( 73^{\circ}\) = \(90^{\circ}\)

or, x = \( 90^{\circ} – 73^{\circ} = 17^{\circ} \)

Hence, the complement of the given angle will be \( 17^{\circ} \)

**Q2: Find the supplement of each of the following angles:**

**(i) \( 80 ^{\circ}\)**

**(ii) \( 54 ^{\circ}\)**

**(iii) \( 105 ^{\circ}\)**

**(iv) \( 123 ^{\circ}\)**

**Sol:**

(i) The given angle measures \( 80 ^{\circ}\).

Let the measure of its complement be \( x^{\circ} \).

x + \( 80^{\circ}\) = \(180^{\circ}\)

or, x = \( 180^{\circ} – 80^{\circ} = 100^{\circ} \)

Hence, the supplement of the given angle will be \( 100^{\circ} \)

(ii) The given angle measures \( 54 ^{\circ}\).

Let the measure of its complement be \( x^{\circ} \).

x + \( 54^{\circ}\) = \(180^{\circ}\)

or, x = \( 180^{\circ} – 54^{\circ} = 126^{\circ} \)

Hence, the supplement of the given angle will be \( 126^{\circ} \)

(iii) The given angle measures \( 105 ^{\circ}\).

Let the measure of its complement be \( x^{\circ} \).

x + \( 105^{\circ}\) = \(180^{\circ}\)

or, x = \( 180^{\circ} – 105^{\circ} = 75^{\circ} \)

Hence, the supplement of the given angle will be \( 75^{\circ} \)

(iv) The given angle measures \( 123 ^{\circ}\).

Let the measure of its complement be \( x^{\circ} \).

x + \( 123^{\circ}\) = \(180^{\circ}\)

or, x = \( 180^{\circ} – 123^{\circ} = 57^{\circ} \)

Hence, the supplement of the given angle will be \( 57^{\circ} \)

**Q3: Among two supplementary angles, the measure of the larger angle is \( 36 ^{\circ}\) more than the measure of the smaller. Find their measures.**

**Sol:**

Let the two supplementary angles be \( x^{\circ} \). and \( (180 – x)^{\circ} \).

Since it is given that the measure of the larger angle is \( 36 ^{\circ} \) more than the smaller angle, let the larger angle be \( x^{\circ} \).

Therefore, \((180 – x)^{\circ} + 36^{\circ} = x^{\circ}\)

or 216 = 2x

Or x = 108

Thus larger angle = \(108^{\circ}\)

Smaller angle = \( (108 – 36)^{\circ}\)

= \( 72^{\circ}\)

**Q4:Find the angle which is equal to its supplement.**

**Sol:**

Let the measure of the required angle be x.

Since it is its own supplement.

x + x = \(180^{\circ}\)

or 2x = \(180^{\circ}\)

or x = \( 90^{\circ}\)

Therefore the required angle is \( 90^{\circ}\).

**Q5: Can two angles be supplementary if both of them are: **

**(i) acute? (ii) Obtuse? (iii) right?**

**Sol:**

(i) No. If both the angles are acute i.e. less than \( 90^{\circ}\), they cannot be supplement as their supplement will always be less than \(180 ^{\circ}\).

(ii) No. If both the angles are obtuse i.e. more than \( 90^{\circ}\), they cannot be supplement as their sum will always be more than \(180 ^{\circ}\).

(iii) Yes. If both the angles are right i.e. they both measure \( 90^{\circ}\), then they form a supplementary pair.

\( 90^{\circ} + 90^{\circ} = 180^{\circ}\)**Q6: In the given figure, AOB is a straight line and the ray OC stands on it. If \(\angle AOC = 64^{\circ}\) and \(\angle BOC = x^{\circ}\), find the value of x.**

**Sol:**

By linear pair property,

\(\angle AOC + \angle COB = 180^{\circ}\) \(64^{\circ} + \angle COB = 180^{\circ}\) \(\angle COB = 180^{\circ} – 64^{\circ} = 116^{\circ}\)Therefore x = \( 116^{\circ} \)

**Q7: In the given figure, AOB is a straight line and the ray OC stands on it. If \(\angle AOC = (2 x – 10)^{\circ}\) and \(\angle BOC = (3 x + 20)^{\circ}\), find the value of x. Also, find \(\angle AOC \;\; and \;\; \angle BOC\).**

**Sol:**

By linear property:

\(\angle AOC + \angle BOC = 180^{\circ}\)\((2 x – 10)^{\circ} + (3x + 20) ^{\circ} = 180^{\circ}\) (Given)

Or 5x + 10 = 180

Or x = \( 34^{\circ}\)

Therefore \( \angle AOC = (2 x – 10) ^{\circ} = (2 \times 34 – 10) ^{\circ} = 58^{\circ} \)

\( \angle BOC = (3 x + 20) ^{\circ} = (3 \times 34 + 20) ^{\circ} = 122^{\circ} \)**Q8: In the given figure, AOB is a straight line and the ray OC and OD stand on it. If \(\angle AOC = 65^{\circ}\) , \(\angle BOC = 70^{\circ}\) and \(\angle COD = x^{\circ}\), find the value of x.**

**Sol: **

Since AOB is a straight line, we have:

\(\angle AOC + \angle BOD + \angle COD = 180^{\circ}\)or, \(65^{\circ} + 70 ^{\circ} + x ^{\circ} = 180^{\circ}\) (Given)

Or, \(135 ^{\circ} + x ^{\circ} = 180^{\circ}\)

Or, \( x ^{\circ} = 45^{\circ}\)

Thus, the value of x is \( 45^{\circ} \)

**Q9: In the given figure, two straight lines AB and CD intersect at a point O. **

**If \(\angle AOC = 42^{\circ}\), find the measure of each of the angles:**

**(i) \(\angle AOD \)**

**(ii) \(\angle BOD \)**

**(iii) \(\angle COB \)**

**Sol:**

**Sol:**

AB and CD intersect at O and CD is a straight line.

(i) \(\angle COA + \angle AOD = 180^{\circ}\) (Linear pair)

\(42^{\circ} + \angle AOD = 180^{\circ}\) \( \angle AOD = 138^{\circ}\)(ii) \(\angle COA \;\; and \;\; \angle BOD \) are vertically opposite angles.

Therefore, \( \angle COA = \angle BOD = 42^{\circ}\) ( from (i) )

(iii) \(\angle COB \;\; and \;\; \angle AOD \) are vertically opposite angles.

Therefore, \(\angle COB = \angle AOD = 138^{\circ}\) ( from (i) )

**Q10: In the given figure, two straight lines PQ and RS intersect at O. If \( \angle POS = 114^{\circ}\), find the measure of each of the angles:**

**(i) \(\angle POR \)**

**(ii) \(\angle ROQ \)**

**(iii) \(\angle QOS \)**

** **

**Sol:**

(i) \( \angle POS + \angle PQR = 180^{\circ}\) (Linear pair)

Or \(114^{\circ} + \angle PQR = 180^{\circ}\)

Or \(\angle PQR = 180^{\circ} – 114^{\circ} = 66^{\circ}\)

(ii) Since \( \angle POS \;\; and \;\; \angle QOR \) are vertically opposite angles, they are equal.

Therefore, \(\angle QOR = 114^{\circ}\)

(iii) Since \( \angle POR \;\; and \;\; \angle QOS \) are vertically opposite angles, they are equal.

Therefore, \(\angle QOS = 66^{\circ}\)

**Q11: In the given figure, ray OA, OB, OC and OD are such that \(\angle AOB = 56^{\circ}\) , \(\angle BOC = 100^{\circ}\), \(\angle COD = x^{\circ}\) and \(\angle DOA = 74^{\circ}\), find the value of x.**

**Sol:**

Sum of all the angles around a point is \(360^{\circ}\)

Therefore, \(\angle AOB + \angle BOC + \angle COD + \angle DOA = 360^{\circ}\)

or, \(56^{\circ} + 100^{\circ} + x^{\circ} + 74^{\circ} = 360^{\circ}\) (Given)

or, \(230^{\circ} + x^{\circ} = 360^{\circ}\)

or, \(x^{\circ} = 360^{\circ} – 230^{\circ}\)

or, x = \(130^{\circ}\)