**Q1: Define the following terms**

**(i)angle**

**(ii) Interior of an angle.**

**(iii) Obtuse angle**

**(iv) Reflex angle**

**(v) Complementary angles**

**(vi) Supplementary angles**

Answer :

(i) Two rays OA and OB, with a common end-point 0, form an angle AOB that is represented as \(\angle AOB\) .

(ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A.

– ► B

(iii) An angle greater than 90° but less than 180° is called an obtuse angle.

(iv) An angle greater than 180° but less than 360° is called a reflex angle.

Reflex angle

(v) Two angles are said to be complementary if the sum of their measures is 90°.

(vi) Two angles are said to be supplementary if the sum of their measures is 180°.

**Q2: If \(\angle A=36^{\circ}{27}'{46}”\) and \(\angle B=28^{\circ}{43}'{39}”\), find \(\angle A+\angle B\).**

Ans:

angle A + angle B = [(36^{o}27’46”) + (28^{o}43’39”)]

We know that 1^{o} = 60 mins or 60’

1’ = 60 sec or 60”

Deg | Min | Sec |

36 | 27 | 46 |

28 | 43 | 39 |

65 | 11 | 25 |

Hence, the measure of the required angle = 65^{o} 11’ 25”

**Q3: Find the difference between two angles measuring \(36^{\circ}\; and \; 24^{\circ}{28}'{30}”\).**

Ans:

We know that 36^{o} = 35^{o} 59’ 60”

Deg | Min | Sec |

35 | 59 | 60 |

24 | 28 | 30 |

11 | 31 | 30 |

Hence, the measure of the required angle = 11^{o} 31’ 60”

**Q4: Find the complement of each of the following angles.**

**(i) 58 ^{o}**

**(ii) 16 ^{o}**

**(iii) \(\frac{2}{3}\) of a right angle**

**(iv) 46 ^{o} 30’**

**(v) 52 ^{o} 43’ 20”**

**(vi) 68 ^{o} 35’ 45”**

Ans:

(i) Complement of 58^{o} = (90 -58)^{o} = 32^{o}

(ii) Complement if 16^{o} = (90 – 16)^{o} = 74^{o}

(iii) \(\frac{2}{3}\) of a right angle = \((90 \times \frac{2}{3})^{o} = 60^{o}\)

Complement of \(\frac{2}{3}\) of a right angle = (90 – 60)^{o} = 30^{o}

(iv) We know that 90^{o} = 89^{o} 60’

Complement of 46^{o} 30’ = (89^{o} 60’ – 46^{o} 30’)^{o} = 43^{o} 30’

(v) We know that 90^{o} = 89^{o} 60’

Complement of 52^{o} 43’ 20” = [(89^{o} 59’ 60”) – (52^{o} 43’ 20”)]

Deg | Min | Sec |

89 | 59 | 60 |

52 | 43 | 20 |

37 | 16 | 40 |

Hence, the measure of angle = 37^{o} 16’ 40”

(vi) We know that 90^{o} = 89^{o} 59’ 60”

Complement of 68^{o} 35’ 45” – [(89^{o} 59’ 60”) – (68^{o} 35’ 45”)]

Deg | Min | Sec |

89 | 59 | 60 |

68 | 35 | 45 |

21 | 24 | 15 |

Hence, the measure of the required angle = 21^{o} 24’ 15”

**Q5: Find the supplement of each of the following angles.**

**(i) 63°**

**(ii) 138°**

**(iii) \(\frac{3}{5}\) of a right angle**

**(iv) 75° 36′**

**(v) 124° 20′ 40″**

Ans:

(i) Supplement of 63° = (180 – 63)^{o} =117^{o}

(ii) Supplement of 138° = (180 – 138)° =42^{o}

(iii) \(\frac{3}{5}\) of a right angle = \((\frac{3}{5} \times 90)\) = 54°

Supplement of \(\frac{3}{5}\) of a right angle = (180 – 54)^{o} = 126°

(iv) We know that 180° = 179° 60′

Supplement of 75° 36′ = (179° 60′ – 75° 36′) = 104° 24′

(v) We know that 180° = 179° 59′ 60″

Supplement of 124° 20′ 40″ = (179′ 59′ 60″ —124° 20′ 40″)

Deg | Min | Sec |

179 | 59 | 60 |

124 | 20 | 40 |

55 | 39 | 20 |

Hence, the measure of the required angle = 55^{o }39’ 20”

**Q6: Find the measure of an angle which is**

**(i) Equal to its complement.**

**(ii) Equal to its supplement.**

Ans:

(i) Let the measure of the required angle be x°.

Then, in case of complementary angles:

x + x = 90°

= 2x = 90°

= x = 45°

Hence, measure of the angle that is equal to its complement is 45°.

(ii) Let the measure of the required angle be x°.

Then, in case of supplementary angles:

x—x=180°

= 2x = 180°

= x = 90°

Hence, measure of the angle that is equal to its supplement is 90°.

**Q7:** **Find the measure of an angle which is \(36^{\circ}\) more than its complement.**

Ans:

Let the measure of the required angle be x°.

Then, measure of its complement = (90 – x)^{o }.

Therefore,

X – (90° – x) = 36°

2x = 126°

X = 63°

Hence, the measure of the required angle is 63°.

**Q8: Find the measure of an angle which is \(25^{\circ}\) less than its supplement.**

Ans:

Let the measure of the required angle be x.

Then, measure of its supplement = (180° – x).

Therefore, (180° – x) – x = 25°

2x = 155°

x = 77.5°

Hence, the measure of the required angle is 77.5°.

**Q9:** **Find the angle which is four times its complement.**

Ans:

Let the measure of the required angle be x.

Then, measure of its complement = (90° – x).

Therefore,

x = (90° – x) 4

x = 360° – 4x

5x = 360°

x = 72°

Hence, the measure of the required angle is 72°.

**Q10:** **Find the angle which is five times its supplement.**

Ans:

Let the measure of the required angle be x.

Then, measure of its supplement = 180° – x.

Therefore, x = (180° – x) 5

X = 900° – 5x

6x = 900°

x = 150°

Hence, the measure of the required angle is 150°.

**Q11: Find the angle whose supplement is four times its complement.**

Sol:

Let the measure of the required angle be x°.

Then, measure of its complement = (90 – x)^{o}

And, measure of its supplement= (180 – x)^{o}

Therefore,

(180 – x) = 4(90 – x)

180 – x = 360 – 4x

3x = 180

x = 60

Hence, the measure of the required angle is 60°.

**Q12: Find the angle whose complement is one-third of its supplement.**

Answer :

Let the measure of the required angle be x°.

Then, the measure of its complement = (90 – x).

And the measure of its supplement = 180 – x°.

Therefore, 90 – x =\(\frac{1}{3}\)(180 – x)

3(90 – x) = (180 – x)

270 – 3x = 180 – x

2x = 90

x = 45

Hence, the measure of the required angle is 45°.

**Q13: Two supplementary angles are in the ratio 3:2. Find the angles.**

Answer :

Let the two angles be 3x and 2x, respectively.

Then,

3x + 2x = 180

5x = 180

X = 36°

Therefore, the two angles are

3x = 3 x 36° = 108° and 2x = 2 x 36° = 72°.

**Q14: Two complementary angles are in the ratio 4:5. Find the angles.**

Answer :

Let the two angles be 4x and 5x, respectively.

Then, 4x + 5x = 90

9x = 90

X = 10°

Hence, the two angles are

4x = 4 x 10° = 40° and 5x = 5 x 10° = 50°.

**Q15:** **Find the measure of an angle, if 7 times its complement is \(10^{\circ}\) less than three times its supplement.**

Answer :

Let the measure of the required angle be x°.

Then, the measure of its complement = (90 – x)°.

And the measure of its supplement = (180 – x)°.

Therefore, 7(90 – x) = 3(180-x) – 10

= 630 – 7x = 540 – 3x-10

= 4x = 100

= x = 25

Hence, the measure of the required angle is 25°.

**Question 16: In the adjoining figure, AOB is a straight line. Find the value of x.**

** **

Ans: We know that the sum of angles in a linear pair is 180^{o}.

Therefore,

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

Hence, the value of x is 118^{o}

**Question 17: In the adjoining figure, AOB is a straight line. Find the value of x. Hence find \(\angle AOC\) and \(\angle BOD\).**

** **

Ans: As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180^{o}

Therefore,

\(\angle AOC + \angle COD + \angle BOD = 180^{o}\) \(\Rightarrow (3x-6)^{o} +55^{o} +(x+20)^{o} = 180^{o}\) \(\Rightarrow 4x=111^{o}\) \(\Rightarrow x = 27.5^{o} \)Hence,

\(\angle AOC = 3x-6\)=3 x 27.5-6

=76.5^{o}

And \(\angle BOD = x + 20\)

= 27.5+20

= 47.5^{o}

**Question 18: In the adjoining figure, AOB is a straight line. Find the value of x. Hence find \(\angle AOC\), \(\angle COD\) and \(\angle BOD\).**

** **

Ans: AOB Is a straight line. Therfore,

\(\angle AOC + \angle COD + \angle BOD\) \(\Rightarrow (3x+7)^{o} + (2x -19)^{o} +x^{o} =180^{o}\) \(\Rightarrow 6x=192^{o}\) \(\Rightarrow x = 32^{o}\)Therefore,

\(\angle AOC\) = 3×32^{o}+7=103^{o}

\(\angle COD\) =2×32^{o} – 19 = 45^{o} and

\(\angle BOD\) = 32^{o}

**Question 19: In the adjoining figure, x:y:z = 5:4:6 . If XOY is a straight line, find the value of x,y,z .**

** **

Ans: Let x = 5a; y = 4a and z = 6a

XOY is a straight line. Therefore

\(\angle XOP + \angle POQ + \angle YOQ = 180^{o}\)5a + 4a + 6a = 180^{o}

15a = 180^{o}

a = 12^{o}

Therefore,

x = 5 x 12^{o} = 60^{o}

y = 4 x 12^{o} = 48^{o}

and z = 6 x 12^{o} =72^{o}

**Question 20: In the adjoining figure, what value of x will make AOB, a straight line?**** **

Ans:

AOB will be a straight line, if

3x + 20 + 4x – 36 = 180^{o}

7x = 196^{o}

X = 28^{o}

Hence, x = 28 will make AOB a straight line.

**Question 21: Two line AB and CD intersect at O. If \(\angle AOC=50^{\circ}\), find \(\angle AOC,\angle BOD\; and \; \angle BOC\).**** **

Ans: We know that if two line intersect then the vertically opposite angles are equal. Therefore, \(\angle AOC = \angle BOD = 50^{o}\)

Let \(\angle AOD – \angle BOC = x^{o}\)

Also, we know that the sum of all angles around a point is 360^{o}

Therefore,

\(\angle AOC + \angle AOD + \angle BOD + \angle BOC = 360^{o}\)50 + x + 50 + x = 360^{o}

2x = 260^{o}

X = 130^{o}

Hence, \(\angle AOD = \angle BOC = 130^{o}\)

Therefore,

\(\angle AOD = 130^{o} , \angle BOD = 50^{o} and \angle BOC =130^{o} \)**Question 22: In the adjoining figure, three coplanar lines AB,CD and EF intersect at a point O, forming angles as shown. Find the value of x,y,z and t.**** **

Ans: We know that if two lines intersect, then the vertically opposite angles are equal.

therefore, \(\angle BOD = \angle AOC = 90^{o}\)

Hence, t=90^{o}

Also,

\(\angle DOF = \angle COE = 50^{o}\)Since, AOB is a straight line, we have

\(\angle AOC + \angle COE + \angle BOE = 180^{o}\)90 + 50 + y = 180^{o}

140 + y =180^{o}

Y = 40^{o}

Also,

\(\angle BOE = \angle AOF = 40^{o}\)Hence, x = 40^{o}

Therefore, x = 40^{o}, y = 40^{o}, z = 50^{o} and t = 90^{o}

**Question 23: In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Hence find \(\angle AOD,\angle COE\; and \; \angle AOE\).**** **

** **

Ans: We know that if two line intersect, then the vertically opposite angles are equal.

therefore, \( \angle DOF = \angle COE = 5x^{o}\)

\(\angle AOD = \angle BOC = 2x^{o}\) and

\(\angle AOE = \angle BOF = 3x^{o}\)Since, AOB is a straight line, we have

\(\angle AOE + \angle COE + \angle BOC = 180^{o}\)3x + 5x + 2x =180^{o}

10x = 180^{o}

X = 18^{o}

Therefore,

\(\angle AOD = 2 \times 18^{o} = 36^{o}\) \(\angle COE = 5 \times 18^{o} = 90^{o}\) \(\angle AOE = 3 \times 18^{o} = 54^{o}\)**Question 24: Two adjacent angles on a straight line are in the ration 5:4. Find the measure of each one of these angles.**

Ans: Let the two adjacent angles be 5x and 4x, respectively

Then,

5x + 4x = 180^{o}

9x = 180^{o}

x = 20^{o}

Hence, the two angles are 5 x20^{o} = 100^{o} and 4 x 20^{o} = 80^{o}

**Question 25: If two straight lines intersect each other in such a way that one of the angles formed measures \(90^{\circ}\), show that each of the remaining angles measures \(90^{\circ}\).**

Ans:

We know that if two lines intersect, then the vertically opposite angles are equal.

\(\angle AOC = 90^{o}.\ Then\ \angle AOC = \angle BOD = 90^{o}\)And let \(\angle BOC = \angle AOD = x\)

Also, we know that the sum of all angles around a point is 360^{o}.

90^{o} +90^{o} + x + x = 360^{o}

2x = 180^{o}

x =90^{o}

Hence, \(\angle BOC = \angle AOD = 90^{o}\)

Therefore, angle AOC = angle BOD = angle BOC = angle AOD = 90^{o}

Hence, the measure of each of the remaining angles are 90^{o}

**Question 26: Two lines AB and CD intersect at a point O such that \(angle BOC+angle AOD =280^{\circ}\), as shown in the figure. Find all the four angles.**** **

** **

Ans: We know that if two lines intersect, then the vertically-opposite angles are equal.

Let \(\angle BOC = \angle AOD\)

Then,

x + x = 280

2x = 280

x = 140^{o}

therefore, \(\angle BOC = \angle AOD = 140^{o}\)

Also, let \(\angle AOC = \angle BOD = y^{o}\)

We know that the sum of all angles around a point is 360^{o}.

therefore, \( \angle AOC + \angle BOC + \angle BOD + \angle AOD = 360^{o}\)

y + 140 + y + 140 = 360^{o}

2y = 80^{o}

Y = 40^{o}

Hence, \(\angle AOC = \angle BOD = 40^{o}\)

Therefore, \(\angle BOC = \angle AOD = 140^{o}\ and\ \angle AOC = \angle BOD = 40^{o}\)

**Question 27: In the given figure, ray OC is the bisector of \(\angle AOB\) and OD is the ray opposite to OC. Show that \(\angle AOD=\angle BOD\).**

** **

Ans:

Since DOC is a straight line, we have

\(\angle BOD + \angle COB = 180^{o}\) …(1)

\(\angle AOC + \angle AOD = 180^{o}\) …(2)

From (1) and (2), we get:

\(\angle BOD + \angle COB = \angle AOC + \angle AOD\)Also, \(\angle COB = \angle AOC\) [Since OC is a bisector of \(\angle AOB\) ]

Therefore, \( \angle AOD = \angle BOD\)

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