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

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 + \(47^{circ}\)

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 + \( 60^{circ}\)

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 + \( 73^{circ}\)

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 + \( 80^{circ}\)

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 + \( 54^{circ}\)

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 + \( 105^{circ}\)

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 + \( 123^{circ}\)

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 \( x^{circ} \)

Since it is given that the measure of the larger angle is \( 36 ^{circ} \)

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 \( 90^{circ}\)

(ii) No. If both the angles are obtuse i.e. more than \( 90^{circ}\)

(iii) Yes. If both the angles are right i.e. they both measure \( 90^{circ}\)

\( 90^{circ} + 90^{circ} = 180^{circ}\)

**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:

\(angle AOC + angle BOC = 180^{circ}\)

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

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, \(65^{circ} + 70 ^{circ} + x ^{circ} = 180^{circ}\)

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

\(42^{circ} + angle AOD = 180^{circ}\)

\( angle AOD = 138^{circ}\)

(ii) \(angle COA ;; and ;; angle BOD \)

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

(iii) \(angle COB ;; and ;; angle AOD \)

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

**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) \( angle POS + angle PQR = 180^{circ}\)

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

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

(ii) Since \( angle POS ;; and ;; angle QOR \)

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

(iii) Since \( angle POR ;; and ;; angle QOS \)

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

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

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

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