Lines & Angles

(i) False. A line segment, however small, contains an infinite number of points in it.

(ii) True. A ray has only one end point (its starting point) and a line segment has two end points.

(iii) True. Of all the paths joining two given points, the line segment joining them is the shortest.

(iv) True. Through a single given point an infinite number of straight lines can be drawn.

(v) (a) A line segment AB has 2 end points.

(b) A ray AB has 1 end point.

(c) A line AB has 0 end points.

(vi) A line and a ray extend endlessly, so their lengths are not fixed. Only a line segment has a definite length.

Hence, $\overline{AB}$ has a fixed length.

(vii) Infinite. An infinite number of rays can be drawn through a fixed point O.

(viii) (a) Through three collinear points, only 1 line can be drawn.

(b) Through three non-collinear points, 3 lines can be drawn.

(ix) 40° + 60° = 100° ≠ 90°.

Hence, 40° is not the complement of 60°.

(x) 45° + 45° = 90° ≠ 180°.

Hence, 45° is not the supplement of 45°.

Two angles are adjacent if they have a common vertex, a common arm and their other two arms lie on the opposite sides of the common arm.

In each figure, ∠AOB and ∠AOC have the common vertex O and the common arm OA.

(i) No; since arm OB and arm OC are not on opposite sides of the common arm OA.

(ii) No; since arm OB and arm OC are not on opposite sides of the common arm OA.

(iii) Yes; since arm OB and arm OC are on opposite sides of the common arm OA.

(iv) No; since arm OB and arm OC are not on opposite sides of the common arm OA.

From the figure,

∠AOB = x + 25° and ∠BOC = 3x + 15°

(i) Since AC is a straight line, ∠AOB and ∠BOC form a linear pair.

⇒ ∠AOB + ∠BOC = 180°

⇒ x + 25° + 3x + 15° = 180°

⇒ 4x + 40° = 180°

⇒ 4x = 140°

⇒ x = $\frac{140^{\circ}}{4}$

⇒ x = 35°

Hence, x = 35°.

(ii) ∠AOB = x + 25° = 35° + 25° = 60°

Hence, ∠AOB = 60°.

(iii) ∠BOC = 3x + 15° = 3 × 35° + 15° = 105° + 15° = 120°

Hence, ∠BOC = 120°.

From the figure, AC is a straight line and the rays OB and OD stand on it. Thus, the sum of angles formed = 180°.

⇒ y + 150° − x + x = 180°

⇒ y + 150° = 180°

⇒ y = 30°

Hence, y = 30°.

From the figure, SR is a straight line and the rays QT and QP stand on it.

∠SQT = x + 70° and ∠TQP = 20° − x

⇒ ∠SQP = ∠SQT + ∠TQP

= x + 70° + 20° − x

= 90°

Since SR is a straight line, ∠SQP and ∠PQR form a linear pair.

⇒ ∠SQP + ∠PQR = 180°

⇒ 90° + ∠PQR = 180°

⇒ ∠PQR = 90°

Hence, ∠PQR = 90°.

From the figure, the rays stand on a straight line. Thus, the sum of angles formed = 180°.

⇒ p° + q° + r° = 180°

Given, p° = q° = r°

⇒ p° + p° + p° = 180°

⇒ 3p° = 180°

⇒ p° = $\frac{180^{\circ}}{3}$

⇒ p° = 60°

Hence, p° = q° = r° = 60°.

From the figure, x° and y° are angles on a straight line and so they form a linear pair.

⇒ x° + y° = 180°

Given, x° = 2y°

Substituting values we get :

⇒ 2y° + y° = 180°

⇒ 3y° = 180°

⇒ y° = $\frac{180^{\circ}}{3}$

⇒ y° = 60°

⇒ x° = 2y° = 2 × 60° = 120°

Hence, x° = 120° and y° = 60°.

Given,

b° = a° + c°

From the figure, the rays stand on a straight line.

⇒ a° + b° + c° = 180°

⇒ b° + (a° + c°) = 180°

⇒ b° + b° = 180°

⇒ 2b° = 180°

⇒ b° = $\frac{180^{\circ}}{2}$

⇒ b° = 90°

Hence, b = 90°.

(i) Since AB is perpendicular to BC at B,

⇒ ∠ABC = 90°

From the figure, the rays at B divide ∠ABC into three angles.

⇒ x + 20° + 2x + 1° + 7x − 11° = 90°

⇒ 10x + 10° = 90°

⇒ 10x = 80°

⇒ x = $\frac{80^{\circ}}{10}$

⇒ x = 8°

Hence, x = 8°.

(ii) The complement of an angle = 90° − the angle.

Complement of x = 90° − x = 90° − 8° = 82°

Hence, the complement of angle x is 82°.

The complement of an angle = 90° − the angle.

Complement of 25° = 90° − 25° = 65°

Hence, the complement is 65°.

The complement of an angle = 90° − the angle.

Complement of 90° = 90° − 90° = 0°

Hence, the complement is 0°.

The complement of an angle = 90° − the angle.

Complement of a° = 90° − a

Hence, the complement is 90° − a.

The complement of an angle = 90° − the angle.

Complement of x + 5° = 90° − x − 5°

= 85° − x

Hence, the complement is 85° − x.

The complement of an angle = 90° − the angle.

Complement of 30° − a = 90° − 30° + a

= 60° + a

Hence, the complement is 60° + a.

$\frac{1}{2} \text{ of a right angle} = \frac{1}{2} \times 90^{\circ} = 45^{\circ}$

The complement of an angle = 90° − the angle.

Complement of 45° = 90° − 45° = 45°

Hence, the complement is 45°.

$\frac{1}{3} \text{ of } 180^{\circ} = \frac{1}{3} \times 180^{\circ} = 60^{\circ}$

The complement of an angle = 90° − the angle.

Complement of 60° = 90° − 60° = 30°

Hence, the complement is 30°.

The complement of an angle = 90° − the angle.

Complement of 21° 17' = 90° − 21° 17'

= 89° 60' − 21° 17'     [As 90° = 89° 60']

= 68° 43'

Hence, the complement is 68° 43'.

The supplement of an angle = 180° − the angle.

Supplement of 100° = 180° − 100° = 80°

Hence, the supplement is 80°.

The supplement of an angle = 180° − the angle.

Supplement of 0° = 180° − 0° = 180°

Hence, the supplement is 180°.

The supplement of an angle = 180° − the angle.

Supplement of x° = 180° − x

Hence, the supplement is 180° − x.

The supplement of an angle = 180° − the angle.

Supplement of x + 35° = 180° − x − 35°

= 145° − x

Hence, the supplement is 145° − x.

The supplement of an angle = 180° − the angle.

Supplement of 90° + a + b = 180° − 90° − a − b

= 90° − a − b

Hence, the supplement is 90° − a − b.

The supplement of an angle = 180° − the angle.

Supplement of 110° − x − 2y = 180° − 110° + x + 2y

= 70° + x + 2y

Hence, the supplement is 70° + x + 2y.

$\frac{1}{5}$ of a right angle = $\frac{1}{5} \times 90^{\circ} = 18^{\circ}$

The supplement of an angle = 180° − the angle.

Supplement of 18° = 180° − 18° = 162°

Hence, the supplement is 162°.

The supplement of an angle = 180° − the angle.

Supplement of 80° 49' 25" = 180° − 80° 49' 25"

= 179° 59' 60" − 80° 49' 25"     [As 180° = 179° 59' 60"]

= 99° 10' 35"

Hence, the supplement is 99° 10' 35".

Two angles are complementary if their sum is 90°.

(i) 10° + 80° = 90°

Hence, Yes, the angles are complementary.

(ii) 37° 28' + 52° 33' = 89° 61' = 90° 1'     [As 61' = 1° 1']

Since 90° 1' ≠ 90°,

Hence, No, the angles are not complementary.

(iii) x + 16° + 74° − x = 16° + 74° = 90°

Hence, Yes, the angles are complementary.

(iv) $\frac{2}{5}$ of a right angle = $\frac{2}{5} \times 90^{\circ} = 36^{\circ}$

54° + 36° = 90°

Hence, Yes, the angles are complementary.

Two angles are supplementary if their sum is 180°.

(i) 139° + 39° = 178° ≠ 180°

Hence, No, the angles are not supplementary.

(ii) 26° 59' + 153° 1' = 179° 60' = 180°     [As 60' = 1°]

Hence, Yes, the angles are supplementary.

(iii) $\frac{3}{10}$ of a right angle = $\frac{3}{10} \times 90^{\circ} = 27^{\circ}$

$\frac{4}{15}$ of two right angles = $\frac{4}{15} \times 180^{\circ} = 48^{\circ}$

27° + 48° = 75° ≠ 180°

Hence, No, the angles are not supplementary.

(iv) (2x° + 65°) + (115° − 2x°) = 2x° + 65° + 115° − 2x° = 180°

Hence, Yes, the angles are supplementary.

As the given angles are supplementary, their sum is 180°.

⇒ (3x + 18°) + (2x + 25°) = 180°

⇒ 5x + 43° = 180°

⇒ 5x = 137°

⇒ x = $\frac{137^{\circ}}{5}$

⇒ x = 27.4°

Converting into degrees and minutes,

0.4° = 0.4 × 60' = 24'

⇒ x = 27° 24'

Hence, x = 27.4° = 27° 24'.

As the angles are in the ratio 1 : 5, let the angles be x and 5x.

As the given angles are complementary, their sum is 90°.

⇒ x + 5x = 90°

⇒ 6x = 90°

⇒ x = $\frac{90^{\circ}}{6}$

⇒ x = 15°

So, the angles are x = 15° and 5x = 5 × 15° = 75°.

Hence, the angles are 15° and 75°.

As the angles are in the ratio 2 : 7, let the angles be 2x and 7x.

As the given angles are supplementary, their sum is 180°.

⇒ 2x + 7x = 180°

⇒ 9x = 180°

⇒ x = $\frac{180^{\circ}}{9}$

⇒ x = 20°

So, the angles are 2x = 2 × 20° = 40° and 7x = 7 × 20° = 140°.

Hence, the angles are 40° and 140°.

As the angles are in the ratio 2 : 3 : 7, let the angles be 2x, 3x and 7x.

As the angles add upto 180°,

⇒ 2x + 3x + 7x = 180°

⇒ 12x = 180°

⇒ x = $\frac{180^{\circ}}{12}$

⇒ x = 15°

So, the angles are :

2x = 2 × 15° = 30°

3x = 3 × 15° = 45°

7x = 7 × 15° = 105°

Hence, the angles are 30°, 45° and 105°.

Supplement of 60° = 180° − 60° = 120°

Let the required angle be x.

As 20% of the angle is the supplement of 60°,

⇒ 20% of x = 120°

⇒ $\frac{20}{100} \times x = 120^{\circ}$

⇒ $\frac{x}{5} = 120^{\circ}$

⇒ x = 600°

Hence, the required angle is 600°.

40% of 2x° = $\frac{40}{100} \times 2x^{\circ} = \frac{4x^{\circ}}{5}$

Complement of $\frac{4x^{\circ}}{5}= 90^{\circ} - \frac{4x^{\circ}}{5}$

As 10% of x° is the complement of 40% of 2x°,

$$\begin{aligned}\Rightarrow\frac{10}{100} \times x^{\circ} = 90^{\circ} - \frac{4x^{\circ}}{5}\\[1em] \Rightarrow \frac{x^{\circ}}{10} + \frac{4x^{\circ}}{5} = 90^{\circ}\\[1em] \Rightarrow \frac{x^{\circ} + 8x^{\circ}}{10} = 90^{\circ}\\[1em] \Rightarrow\frac{9x^{\circ}}{10} = 90^{\circ}\\[1em] \Rightarrow 9x^{\circ} = 900^{\circ}\\[1em] \Rightarrow x^{\circ} = \frac{900^{\circ}}{9}\\[1em] \Rightarrow x = 100\end{aligned}$$

Hence, x = 100°.

From the figure,

The angles 4x°, 3x°, 2x° and x° are the angles made on one side of a straight line.

⇒ 4x° + 3x° + 2x° + x° = 180°

⇒ 10x° = 180°

⇒ x° = $\frac{180^{\circ}}{10}$

⇒ x° = 18°

Supplement of x° = 180° − 18° = 162°

Hence, x° = 18° and its supplement = 162°.

From the figure, the angles K − 15°, 30°, 90° and K + 150° are the angles at a point.

As the sum of the angles at a point is 360°,

⇒ K − 15° + 30° + 90° + K + 150° = 360°

⇒ 2K + 255° = 360°

⇒ 2K = 105°

⇒ K = $\frac{105^{\circ}}{2}$

⇒ K = 52.5°

Converting into degrees and minutes,

0.5° = 0.5 × 60' = 30'

⇒ K = 52° 30'

Hence, K = 52.5° = 52° 30'.

From the figure, the angles 42°, 3K, 2K and K are the angles at a point.

As the sum of the angles at a point is 360°,

⇒ 42° + 3K + 2K + K = 360°

⇒ 42° + 6K = 360°

⇒ 6K = 360° - 42°

⇒ 6K = 318°

⇒ K = $\frac{318^{\circ}}{6}$

⇒ K = 53°

Hence, K = 53°.

As the ratio x : y = 1 : 2, let x = a and y = 2a.

From the figure, ∠MOQ and ∠PON (= z°) are vertically opposite angles.

⇒ ∠MOQ = z° = 90°

As RS is a straight line, the angles ∠ROM, ∠MOQ and ∠QOS lie on one side of it.

⇒ ∠ROM + ∠MOQ + ∠QOS = 180°

⇒ y° + 90° + x° = 180°

⇒ 2a° + a° = 90°

⇒ 3a° = 90°

⇒ a° = $\frac{90^{\circ}}{3}$

⇒ a° = 30°

So, x° = a° = 30° and y° = 2a° = 2 × 30° = 60°.

⇒ ∠ROM = y° = 60°

From the figure, ∠POR and ∠QOS are vertically opposite angles.

⇒ ∠POR = ∠QOS = x° = 30°

Hence, ∠ROM = 60° and ∠POR = 30°.

From the figure, the angles 123°, 85°, 5x°, x° and 80° are the angles at the point O.

As the sum of the angles at a point is 360°,

⇒ 123° + 85° + 5x° + x° + 80° = 360°

⇒ 6x° + 288° = 360°

⇒ 6x° = 360° - 288°

⇒ 6x° = 72°

⇒ x° = $\frac{72^{\circ}}{6}$

⇒ x° = 12°

⇒ ∠AOB = 5x° = 5 × 12° = 60°

⇒ ∠BOC = x° = 12°

Hence, ∠AOB = 60° and ∠BOC = 12°.

From the figure,

The angles $2y^{\circ}, 2y^{\circ}, 3\frac{1}{2}y^{\circ} \text{ and }2\frac{1}{2}y^{\circ}$ are the angles at a point.

As the sum of the angles at a point is 360°,

$$\begin{aligned}\Rightarrow 2y^{\circ} + 2y^{\circ} + 3\frac{1}{2}y^{\circ} + 2\frac{1}{2}y^{\circ} = 360^{\circ}\\[1em] \Rightarrow 2y^{\circ} + 2y^{\circ} + \frac{7y^{\circ}}{2} + \frac{5y^{\circ}}{2} = 360^{\circ}\\[1em] \Rightarrow \frac{4y^{\circ} + 4y^{\circ} + 7y^{\circ} + 5y^{\circ}}{2} = 360^{\circ}\\[1em] \Rightarrow \frac{20y^{\circ}}{2} = 360^{\circ}\\[1em] \Rightarrow 10y^{\circ} = 360^{\circ}\\[1em] \Rightarrow y^{\circ} = \frac{360^{\circ}}{10}\\[1em] \Rightarrow y^{\circ} = 36^{\circ}\end{aligned}$$

So, the angles are :

$2y^{\circ} = 2 × 36^{\circ} = 72^{\circ}$

$2y^{\circ} = 2 × 36^{\circ} = 72^{\circ}$

$2\frac{1}{2}y^{\circ} = \frac{5}{2} \times 36^{\circ} = 90^{\circ}$

$3\frac{1}{2}y^{\circ} = \frac{7}{2} \times 36^{\circ} = 126^{\circ}$

Hence, the angles are 72°, 72°, 90° and 126°.

From the figure, EF and AB are intersecting lines,

⇒ ∠AOF = ∠EOB = z°

From figure,

⇒ ∠AOC + ∠AOF + ∠DOF = 180°

⇒ y° + z° + x° = 180°

(i) Given, y = 45° and z = 90°

⇒ 45° + 90° + x° = 180°

⇒ x° + 135° = 180°

⇒ x° = 180° - 135°

⇒ x° = 45°

Hence, x = 45°.

(ii) Given, x = 3a, y = 5x and z = 6x

⇒ y = 5x = 5 × 3a = 15a

⇒ z = 6x = 6 × 3a = 18a

As, x° + y° + z° = 180°

⇒ 3a + 15a + 18a = 180°

⇒ 36a = 180°

⇒ a = $\frac{180^{\circ}}{36}$

⇒ a = 5°

Hence, a = 5°.

(i) ∠3 and ∠6 lie between the two lines and on opposite sides of the transversal.

Hence, they are interior alternate angles.

(ii) ∠2 and ∠4 have a common vertex and a common arm, and they lie on opposite sides of the common arm.

Hence, they are adjacent angles.

(iii) ∠3 and ∠7 lie on the same side of the transversal and in matching positions at the two intersections.

Hence, they are corresponding angles.

(iv) ∠2 and ∠7 lie outside the two lines and on opposite sides of the transversal.

Hence, they are exterior alternate angles.

(v) ∠4 and ∠6 lie between the two lines and on the same side of the transversal.

Hence, they are co-interior angles.

(vi) ∠1 and ∠8 lie outside the two lines and on opposite sides of the transversal.

Hence, they are exterior alternate angles.

(vii) ∠1 and ∠5 lie on the same side of the transversal and in matching positions at the two intersections.

Hence, they are corresponding angles.

(viii) ∠1 and ∠4 are formed by two intersecting lines and lie opposite to each other at the common vertex.

Hence, they are vertically opposite angles.

(ix) ∠5 and ∠7 have a common vertex and a common arm, and lie on opposite sides of the common arm.

Hence, they are adjacent angles.

(i) ∠1 and ∠4 are formed by two intersecting lines and lie opposite to each other at the common vertex.

Hence, they are vertically opposite angles.

(ii) ∠4 and ∠7 lie between the two lines and on opposite sides of the transversal.

Hence, they are interior alternate angles.

(iii) ∠10 and ∠12 are formed by two intersecting lines and lie opposite to each other at the common vertex.

Hence, they are vertically opposite angles.

(iv) ∠7 and ∠13 lie on the same side of the transversal and in matching positions at the two intersections.

Hence, they are corresponding angles.

(v) ∠6 and ∠8 are formed by two intersecting lines and lie opposite to each other at the common vertex.

Hence, they are vertically opposite angles.

(vi) ∠11 and ∠8 lie between the two lines and on the same side of the transversal.

Hence, they are co-interior angles.

(vii) ∠7 and ∠9 are formed by two intersecting lines and lie opposite to each other at the common vertex.

Hence, they are vertically opposite angles.

(viii) ∠4 and ∠5 have a common vertex and a common arm, and lie on opposite sides of the common arm.

Hence, they are adjacent angles.

(ix) ∠4 and ∠6 lie between the two lines and on the same side of the transversal.

Hence, they are co-interior angles.

(x) ∠6 and ∠7 have a common vertex and a common arm, and lie on opposite sides of the common arm.

Hence, they are adjacent angles.

(xi) ∠2 and ∠13 lie between the two lines and on the same side of the transversal.

Hence, they are co-interior angles.

From figure,

∠a and ∠c are interior alternate angles, and the given lines are parallel.

⇒ ∠a = ∠c     [Interior alternate angles are equal]

∠b and ∠c are vertically opposite angles.

⇒ ∠b = ∠c     [Vertically opposite angles are equal]

Hence, ∠a = ∠b = ∠c.

From figure,

∠x and ∠y are vertically opposite angles.

⇒ ∠x = ∠y     [Vertically opposite angles are equal]

∠y and ∠l are interior alternate angles, and the given lines are parallel.

⇒ ∠y = ∠l     [Interior alternate angles are equal]

∠l and ∠n are vertically opposite angles.

⇒ ∠l = ∠n     [Vertically opposite angles are equal]

∠n and ∠r are corresponding angles, and the given lines are parallel.

⇒ ∠n = ∠r     [Corresponding angles are equal]

⇒ ∠x = ∠y = ∠l = ∠n = ∠r

Again, ∠k and ∠m are vertically opposite angles.

⇒ ∠k = ∠m     [Vertically opposite angles are equal]

∠k and ∠q are corresponding angles, and the given lines are parallel.

⇒ ∠k = ∠q     [Corresponding angles are equal]

⇒ ∠k = ∠m = ∠q

Hence, ∠x = ∠y = ∠l = ∠n = ∠r and ∠k = ∠m = ∠q.

From the figure, the two given lines with arrow are parallel and the third line is a transversal.

The 110° angle and ∠e form a linear pair.

⇒ 110° + e = 180°

⇒ e = 70°

∠f is vertically opposite to the 110° angle.

⇒ f = 110°     [Vertically opposite angles are equal]

∠g is vertically opposite to ∠e.

⇒ g = e = 70°     [Vertically opposite angles are equal]

∠a and the 110° angle are corresponding angles.

⇒ a = 110°     [Corresponding angles are equal]

∠a and ∠b form a linear pair.

⇒ b = 180° − 110° = 70°

∠c is vertically opposite to ∠b and ∠d is vertically opposite to ∠a.

⇒ c = b = 70° and d = a = 110°     [Vertically opposite angles are equal]

Hence, a = 110°, b = 70°, c = 70°, d = 110°, e = 70°, f = 110° and g = 70°.

(i) The angles 120° and 50° are co-interior angles.

⇒ 120° + 50° = 170° ≠ 180°

Since the co-interior angles are not supplementary, the dotted line segments are not parallel.

(ii)

From the figure,

∠COD = ∠AOB = 45° (vertically opposite angles are equal)

The angles ∠FDO and ∠COD are co-interior angles.

⇒ 135° + 45° = 180°

Since the co-interior angles are supplementary, the dotted line segments are parallel.

(iii) The angles 120° and 130° are corresponding angles.

But 120° ≠ 130°

Since the corresponding angles are not equal, the dotted line segments are not parallel.

(iv)

From the figure,

∠COD = ∠AOB = 110° (vertically opposite angles are equal)

The angles ∠EDO and ∠COD are co-interior angles.

⇒ 110° + 70° = 180°

Since the co-interior angles are supplementary, the dotted line segments are parallel.

(v)

Taking l₃ as the transversal, the angles 100° and 70° are co-interior angles.

⇒ 100° + 70° = 170° ≠ 180°

As, sum of co-interior angles is not 180°.

⇒ Line l₁ and l₂ are not parallel.

As, ∠AOD and ∠AOB forms a linear pair.

⇒ ∠AOD + ∠AOB = 180°

⇒ 100° + ∠AOB = 180°

⇒ ∠AOB = 180° - 100° = 80°

Taking l₃ as the transversal,

⇒ ∠AOB and ∠OBC are interior alternate angles.

As, ∠AOB = ∠OBC = 80°

⇒ Line l₂ and l₄ are parallel.

Now taking l₄ as a transversal.

70° ≠ 80°

⇒ Corresponding angles of line l₃ and l₅ are not equal.

⇒ Line l₃ and l₅ are not parallel.

Hence, only l₂ and l₄ are parallel.

(vi) The angles 50° and 40° are alternate angles.

But 50° ≠ 40°

Since the alternate angles are not equal, the dotted line segments are not parallel.

From the figure,

⇒ b = 60°     [Vertically opposite angles are equal]

⇒ a = b = 60°     [Corresponding angles are equal]

⇒ c = a = 60°     [Vertically opposite angles are equal]

Hence, a = 60°, b = 60° and c = 60°.

From the figure,

⇒ y = 55°     [Vertically opposite angles are equal]

⇒ 55° + z = 180°     [Linear pair]

⇒ z = 180° - 55°

⇒ z = 125°

⇒ x = z = 125°     [Corresponding angles are equal]

Hence, x = z = 125° and y = 55°.

From the figure,

⇒ a + 120° = 180°     [Co-interior angles are supplementary]

⇒ a = 60°

⇒ b = a = 60°     [Vertically opposite angles are equal]

⇒ b + c = 180°     [Linear pair]

⇒ 60° + c = 180°

⇒ c = 180° − 60° = 120°

Hence, a = 60°, b = 60° and c = 120°.

From the figure,

⇒ x = 50°     [Interior alternate angles are equal]

⇒ y + 120° = 180°     [Co-interior angles are supplementary]

⇒ y = 60°

As z is the reflex angle at the vertex,

⇒ z = 360° − (x + y)

⇒ z = 360° − (50° + 60°) = 360° − 110° = 250°

Hence, x = 50°, y = 60° and z = 250°.

Since the two vertical lines are parallel and the first one is perpendicular to the horizontal line.

As, 90° and x are co-interior angles.

⇒ 90° + x = 180°

⇒ x = 90°

In the triangle formed by the horizontal line, the slant line and the second vertical line,

⇒ 30° + x + z = 180°

⇒ 30° + 90° + z = 180°

⇒ 120° + z = 180°

⇒ z = 180° − 120° = 60°

Now, z and k are alternate interior angles.

⇒ k = z = 60°

As, y and z are on the straight line, they form a linear pair.

⇒ y + z = 180°

⇒ y + 60° = 180°

⇒ y = 180° − 60° = 120°

Hence, k = 60°, x = 90°, y = 120° and z = 60°.

From the figure,

⇒ x = 110°     [Vertically opposite angles are equal]

⇒ s + 110° = 180°     [Linear pair]

⇒ s = 70°

⇒ s = ∠COD = 70° (vertically opposite angles are equal)

As, ∠COD and y are corresponding angles

⇒ y = 70°     [Corresponding angles are equal]

⇒ p = 60°     [Corresponding angles are equal]

⇒ p + q = 180°     [Linear pair]

⇒ q = 180° − 60° = 120°

⇒ 60° + r = 180°     [Linear pair]

⇒ r = 120°

⇒ t = r = 120°     [Vertically opposite angles are equal]

Hence, x = 110°, s = 70°, y = 70°, p = 60°, q = 120°, r = 120° and t = 120°.

From the figure,

⇒ y = 110°     [Vertically opposite angles are equal]

⇒ 110° + ∠COD = 180°     [Co-interior angles are supplementary]

⇒ ∠COD = 70°

⇒ z = 70°     [Vertically opposite angles are equal]

⇒ 120° + x = 180°     [Linear pair]

⇒ x = 60°

⇒ p = x = 60°     [alternate angles are equal]

⇒ p + q = 180°     [Linear pair]

⇒ q = 180° − 60° = 120°

Hence, y = 110°, z = 70°, x = 60°, p = 60° and q = 120°.

From the figure,

∠COD and ∠COA forms a linear pair,

⇒ ∠COD + ∠COA = 180°

⇒ ∠COD + 112° = 180°

⇒ ∠COD = 180° − 112° = 68°

Since, line l and m are parallel to each other,

⇒ x = ∠COD = 68°     [Corresponding angles are equal]

The angles z, 75° and x lie on a straight line.

⇒ z + 75° + x = 180°

⇒ z + 75° + 68° = 180°

⇒ z + 143° = 180°

⇒ z = 180° − 143° = 37°

By the angle sum property of the triangle,

⇒ y + ∠COD + z = 180°

⇒ y + 68° + 37° = 180°

⇒ y + 105° = 180°

⇒ y = 180° − 105° = 75°

Hence, y = 75°, z = 37° and x = 68°.

From the figure,

∠BOC and ∠BOA forms a linear pair,

⇒ ∠BOC + ∠BOA = 180°

⇒ ∠BOC + 115° = 180°

⇒ ∠BOC = 180° - 115° = 65°

∠BCO and ∠BCD forms a linear pair,

⇒ ∠BCO + ∠BCD = 180°

⇒ ∠BCO + 120° = 180°

⇒ ∠BCO = 180° − 120° = 60°

By the angle sum property of the triangle,

⇒ b + ∠BOC + ∠BCO = 180°

⇒ b + 65° + 60° = 180°

⇒ b = 180° − 125° = 55°

Since the two horizontal lines are parallel,

⇒ a = ∠BOC = 65°     [Interior alternate angles are equal]

⇒ c = ∠BCO = 60°     [Interior alternate angles are equal]

Hence, a = 65°, b = 55° and c = 60°.

From the figure, the opposite sides of the figure are parallel.

⇒ 110° + x = 180°     [Co-interior angles are supplementary]

⇒ x = 70°

⇒ y = 110°     [Opposite angles of a parallelogram are equal]

The interior angle at the top right vertex = x = 70°     [Opposite angles of a parallelogram are equal]

⇒ z + 70° = 180°     [Linear pair]

⇒ z = 110°

Hence, z = 110°, x = 70° and y = 110°.

Through the vertex where x is marked, draw a line parallel to the two given parallel lines.

∠ADO and ∠DOB are co-interior angles,

⇒ ∠ADO + ∠DOB = 180°     [Co-interior angles are supplementary]

⇒ 160° + ∠DOB = 180°

⇒ ∠DOB = 180° − 160° = 20°

∠CEO and ∠EOB are co-interior angles,

⇒ ∠CEO + ∠EOB = 180°     [Co-interior angles are supplementary]

⇒ 130° + ∠EOB = 180°

⇒ ∠EOB = 180° − 130° = 50°

⇒ x = 20° + 50° = 70°

As y is the reflex angle at the same vertex,

⇒ y = 360° − x = 360° − 70° = 290°

Hence, x = 70° and y = 290°.

Through the vertex O, draw a line m parallel to the two given parallel lines l and n.

Since l ∥ m and AO is a transversal,

⇒ ∠CAO = ∠FOA     [Alternate interior angles are equal]

⇒ ∠FOA = 50°

Since m ∥ n and BO is a transversal,

⇒ ∠EBO = ∠FOB     [Alternate interior angles are equal]

⇒ ∠FOB = 40°

Now,

⇒ b = ∠FOA + ∠FOB

⇒ b = 50° + 40° = 90°

Since a is the reflex angle at O,

⇒ a = 360° − b

⇒ a = 360° − 90° = 270°

Hence, a = 270° and b = 90°.

From the figure, the reflex angle at the upper vertex is 270°.

Sum of angles at a point = 360°

⇒ 270° + x + 40° = 360°

⇒ x + 310° = 360°

⇒ x = 360° - 310°

⇒ x = 50°

Since the two horizontal rays are parallel and the two slant rays are parallel,

⇒ y = 40°     [Corresponding angles are equal]

⇒ z = x = 50°     [Corresponding angles are equal]

As p and z are the angles on the straight slant line at the lower vertex,

⇒ p + z = 180°     [Linear pair]

⇒ p + 50° = 180°

⇒ p = 180° − 50° = 130°

Hence, x = 50°, y = 40°, z = 50° and p = 130°.

From the figure,

The angles 110°, p and 25° lie on one side of a straight line.

⇒ 110° + p + 25° = 180°

⇒ p + 135° = 180°

⇒ p = 45°

Since the two directed rays are parallel,

⇒ x = p = 45°     [Interior alternate angles are equal]

By the angle sum property of the triangle,

⇒ x + 25° + y = 180°

⇒ 45° + 25° + y = 180°

⇒ 70° + y = 180°

⇒ y = 180° − 70° = 110°

Hence, x = 45°, y = 110° and p = 45°.

The angles 2x and x are co-interior angles.

⇒ 2x + x = 180°     [Co-interior angles are supplementary]

⇒ 3x = 180°

⇒ x = $\frac{180^{\circ}}{3}$

⇒ x = 60°

Hence, x = 60°.

From the figure,

∠DOB = ∠EOF = 5x (vertically opposite angles are equal)

The angles ∠ABO and ∠DOB are co-interior angles.

⇒ ∠ABO + ∠DOB = 180°     [Co-interior angles are supplementary]

⇒ 4x + 5x = 180°

⇒ 9x = 180°

⇒ x = $\frac{180^{\circ}}{9}$

⇒ x = 20°

Hence, x = 20°.

From the figure,

∠EOF = ∠AOB = x (vertically opposite angles are equal)

The angles ∠EOF and ∠DEO are co-interior angles.

⇒ ∠EOF + ∠DEO = 180°     [Co-interior angles are supplementary]

⇒ x + 4x = 180°

⇒ 5x = 180°

⇒ x = $\frac{180^{\circ}}{5}$

⇒ x = 36°

Hence, x = 36°.

The angles 2x + 5° and 3x + 55° are co-interior angles.

⇒ 2x + 5° + 3x + 55° = 180°     [Co-interior angles are supplementary]

⇒ 5x + 60° = 180°

⇒ 5x = 120°

⇒ x = $\frac{120^{\circ}}{5}$

⇒ x = 24°

Hence, x = 24°.

From the figure,

∠COD = ∠AOB = 3x + 25° (vertically opposite angles are equal)

The angles ∠COD and ∠EDO are co-interior angles.

⇒ ∠COD + ∠EDO = 180°     [Co-interior angles are supplementary]

⇒ 3x + 25° + 2x + 20° = 180°

⇒ 5x + 45° = 180°

⇒ 5x = 135°

⇒ x = $\frac{135^{\circ}}{5}$

⇒ x = 27°

Hence, x = 27°.

Through the vertex where 130° is marked, draw a line parallel to the two given parallel lines.

As, line l and m are parallel.

⇒ ∠ACB = ∠COE = 4x (corresponding angles are equal)

Now, line m and n are parallel.

⇒ ∠EOD = ∠FDH = 6x (corresponding angles are equal)

Since, ∠COD = ∠EOD + ∠EOC

⇒ 130° = 4x + 6x

⇒ 10x = 130°

⇒ x = $\frac{130^{\circ}}{10}$

⇒ x = 13°

Hence, x = 13°.

As the angles are in the ratio 1 : 2, let the angles be x and 2x.

As the angles are complementary, their sum is 90°.

⇒ x + 2x = 90°

⇒ 3x = 90°

⇒ x = $\frac{90^{\circ}}{3}$

⇒ x = 30°

So, the angles are 30° and 60°, and the smaller angle is 30°.

Hence, option 2 is the correct option.

As the angles are supplementary, their sum is 180°.

⇒ 3x − 20° + 2x + 30° = 180°

⇒ 5x + 10° = 180°

⇒ 5x = 170°

⇒ x = $\frac{170^{\circ}}{5}$

⇒ x = 34°

Hence, option 1 is the correct option.

Given,

y = 30°

From the figure, AOB is a straight line.

⇒ 2y + 30° + 3x = 180°     [Angles on a straight line]

⇒ 2 × 30° + 30° + 3x = 180°

⇒ 90° + 3x = 180°

⇒ 3x = 90°

⇒ x = $\frac{90^{\circ}}{3}$

⇒ x = 30°

Hence, option 3 is the correct option.

Through the point C, draw a line CF parallel to BA and DE.

Since BA ∥ CF and BC is a transversal,

⇒ ∠BCF + ∠ABC = 180°     [Co-interior angles are supplementary]

⇒ ∠BCF + 100° = 180°

⇒ ∠BCF = 80°

Since, CF ∥ DE and CD is a transversal,

⇒ ∠FCD + ∠CDE = 180°     [Co-interior angles are supplementary]

⇒ ∠FCD + 150° = 180°

⇒ ∠FCD = 30°

⇒ ∠BCD = ∠BCF + ∠FCD = 80° + 30° = 110°

Hence, option 3 is the correct option.

Since AB ∥ CD and PQ is a transversal,

⇒ ∠a = ∠b     [Corresponding angles are equal]

⇒ 3x − 30° = 2x + 10°

⇒ 3x − 2x = 10° + 30°

⇒ x = 40°

Hence, option 2 is the correct option.

Since BA ∥ CE and BD is a transversal,

⇒ ∠ABC = ∠ECD     [Corresponding angles are equal]

⇒ ∠ABC = 60°

Hence, option 2 is the correct option.

Through the point O, draw a line EO parallel to AB and so parallel to CD.

Let the ray from O meet AB at P and CD at Q.

From the figure, the angle marked 130° and ∠OQD form a linear pair.

⇒ ∠OQD + 130° = 180°

⇒ ∠OQD = 180° − 130° = 50°

Since the line through O is parallel to CD and OQ is a transversal,

⇒ ∠EOQ = ∠OQD = 50°     [Interior alternate angles are equal]

Since the line OE is parallel to AB and OP is a transversal,

⇒ ∠EOP = x     [corresponding angles are equal]

Now, the angle at O = 95°

⇒ ∠EOP + ∠EOQ = 95°

⇒ x + 50° = 95°

⇒ x = 45°

Hence, option 4 is the correct option.

Since BA ∥ DC and MN is a transversal,

⇒ ∠CEO = ∠FED = 5x     [Vertically opposite angles are equal]

As, BA ∥ DC and MN is a transversal,

⇒ ∠CEO = ∠a     (Corresponding angles are equal)

⇒ ∠a = 5x

Also, ∠a and ∠AOE form a linear pair.

⇒ ∠a + 3x − 20° = 180°

⇒ 5x + 3x − 20° = 180°

⇒ 8x − 20° = 180°

⇒ 8x = 200°

⇒ x = $\frac{200^{\circ}}{8}$

⇒ x = 25°

⇒ ∠a = 5x = 5 × 25° = 125°.

Hence, option 2 is the correct option.

Since PQ ∥ RO and AB is a transversal,

⇒ ∠AOR = 68°     [Alternate interior angles are equal]

Since AB ∥ CD and RO is a transversal,

⇒ ∠a = ∠AOR     [Corresponding angles are equal]

⇒ ∠a = 68°

Hence, option 1 is the correct option.

Let the angle be x.

Supplement of x = 180° − x and complement of x = 90° − x

As the supplement of the angle is four times its complement,

⇒ 180° − x = 4(90° − x)

⇒ 180° − x = 360° − 4x

⇒ 4x − x = 360° − 180°

⇒ 3x = 180°

⇒ x = $\frac{180^{\circ}}{3}$

⇒ x = 60°

Hence, option 3 is the correct option.

In the given figure, on the line q the four points L, H, G and F are lying, and on the line p the four points L, E, D and C are lying. So, there are two lines on which four points are collinear. So, Statement 1 is false.

Three or more points in a plane which lie on the same straight line are called collinear points. So, Statement 2 is true.

Hence, option 4 is the correct option.

In the given figure, ∠AOB and ∠BOC have the common vertex O, the common arm OB, and their other arms OA and OC lie on the opposite sides of the common arm OB. So, they are adjacent angles and Statement 1 is true.

For two angles to be adjacent, the other two arms must lie on the opposite sides of the common arm only, and not on the same side. So, Statement 2 is false.

Hence, option 3 is the correct option.

Sum of the given angles

= 90° + a − b + b − a + 90°

= 90° + 90°

= 180°

Since the sum is 180°, the angles are supplementary. So, Assertion (A) is true.

Two angles whose sum is 180° are called supplementary angles. So, Reason (R) is true.

Hence, option 3 is the correct option.

From the figure, AB is a straight line and the rays at O make the angles 2x + 10°, 3x − 10° and 40° on one side of it.

⇒ 2x + 10° + 3x − 10° + 40° = 180°

⇒ 5x + 40° = 180°

⇒ 5x = 140°

⇒ x = $\frac{140^{\circ}}{5}$

⇒ x = 28°

Since x = 28° and not 44°, Assertion (A) is false.

Adjacent angles which form a straight line are called a linear pair and their sum is 180°. So, Reason (R) is true.

Hence, option 2 is the correct option.

From the figure, ∠4 and ∠6 lie between AB and CD and on the same side of the transversal PQ, so they are co-interior angles. Similarly, ∠3 and ∠5 are co-interior angles.

Since AB is parallel to CD,

⇒ ∠4 + ∠6 = 180° and ∠3 + ∠5 = 180°     [Co-interior angles are supplementary]

So, Assertion (A) is true.

If a transversal cuts two lines such that the co-interior angles are supplementary, then the two lines are parallel. So, Reason (R) is true.

Hence, option 3 is the correct option.