Lines & Angles — Exercise 16(A)

State, true or false :

(i) A line segment 4 cm long can have only 2000 points in it.

(ii) A ray has one end point and a line segment has two end-points.

(iii) A line segment is the shortest distance between any two given points.

(iv) An infinite number of straight lines can be drawn through a given point.

(v) Write the number of end points in

(a) a line segment AB

(b) a ray AB

(c) a line AB

(vi) Out of $\overleftrightarrow{AB}, \overrightarrow{AB}, \overleftarrow{AB} \text{ and } \overline{AB}$, which one has a fixed length ?

(vii) How many rays can be drawn through a fixed point O ?

(viii) How many lines can be drawn through three

(a) collinear points ?

(b) non-collinear points ?

(ix) Is 40° the complement of 60° ?

(x) Is 45° the supplement of 45° ?

Answer

(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°.

In which of the following figures, are ∠AOB and ∠AOC adjacent angles ? Give, in each case, reason for your answer.

Answer

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.

In the given figure, AC is a straight line. Find :

(i) x,

(ii) ∠AOB,

(iii) ∠BOC.

Answer

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 = $\dfrac{140\degree}{4}$

⇒ 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°.

Find y in the given figure.

Answer

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°.

In the given figure, find ∠PQR.

Answer

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°.

In the given figure, p° = q° = r°, find each.

Answer

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° = $\dfrac{180\degree}{3}$

⇒ p° = 60°

Hence, p° = q° = r° = 60°.

In the given figure, if x° = 2y°, find x° and y°.

Answer

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° = $\dfrac{180\degree}{3}$

⇒ y° = 60°

⇒ x° = 2y° = 2 × 60° = 120°

Hence, x° = 120° and y° = 60°.

In the adjoining figure, if b° = a° + c°, find b.

Answer

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° = $\dfrac{180\degree}{2}$

⇒ b° = 90°

Hence, b = 90°.

In the given figure, AB is perpendicular to BC at B.

Find :

(i) the value of x,

(ii) the complement of angle x.

Answer

(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 = $\dfrac{80\degree}{10}$

⇒ 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°.

Write the complement of :

25°

Answer

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

Complement of 25° = 90° − 25° = 65°

Hence, the complement is 65°.

Write the complement of :

90°

Answer

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

Complement of 90° = 90° − 90° = 0°

Hence, the complement is 0°.

Write the complement of :

a°

Answer

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

Complement of a° = 90° − a

Hence, the complement is 90° − a.

Write the complement of :

x + 5°

Answer

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

Complement of x + 5° = 90° − x − 5°

= 85° − x

Hence, the complement is 85° − x.

Write the complement of :

30° − a

Answer

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

Complement of 30° − a = 90° − 30° + a

= 60° + a

Hence, the complement is 60° + a.

Write the complement of :

$\dfrac{1}{2}$ of a right angle

Answer

$\dfrac{1}{2} \text{ of a right angle} = \dfrac{1}{2} \times 90\degree = 45\degree$

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

Complement of 45° = 90° − 45° = 45°

Hence, the complement is 45°.

Write the complement of :

$\dfrac{1}{3}$ of 180°

Answer

$\dfrac{1}{3} \text{ of } 180\degree = \dfrac{1}{3} \times 180\degree = 60\degree$

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

Complement of 60° = 90° − 60° = 30°

Hence, the complement is 30°.

Write the complement of :

21° 17'

Answer

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'.

Write the supplement of :

100°

Answer

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

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

Hence, the supplement is 80°.

Write the supplement of :

0°

Answer

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

Supplement of 0° = 180° − 0° = 180°

Hence, the supplement is 180°.

Write the supplement of :

x°

Answer

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

Supplement of x° = 180° − x

Hence, the supplement is 180° − x.

Write the supplement of :

x + 35°

Answer

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

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

= 145° − x

Hence, the supplement is 145° − x.

Write the supplement of :

90° + a + b

Answer

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.

Write the supplement of :

110° − x − 2y

Answer

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.

Write the supplement of :

$\dfrac{1}{5}$ of a right angle

Answer

$\dfrac{1}{5}$ of a right angle = $\dfrac{1}{5} \times 90\degree = 18\degree$

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

Supplement of 18° = 180° − 18° = 162°

Hence, the supplement is 162°.

Write the supplement of :

80° 49' 25"

Answer

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".

Are the following pairs of angles complementary ?

(i) 10° and 80°

(ii) 37° 28' and 52° 33'

(iii) x + 16° and 74° − x

(iv) 54° and $\dfrac{2}{5}$ of a right angle.

Answer

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) $\dfrac{2}{5}$ of a right angle = $\dfrac{2}{5} \times 90\degree = 36\degree$

54° + 36° = 90°

Hence, Yes, the angles are complementary.

Are the following pairs of angles supplementary ?

(i) 139° and 39°.

(ii) 26° 59' and 153° 1'.

(iii) $\dfrac{3}{10}$ of a right angle and $\dfrac{4}{15}$ of two right angles.

(iv) 2x° + 65° and 115° - 2x°.

Answer

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) $\dfrac{3}{10}$ of a right angle = $\dfrac{3}{10} \times 90\degree = 27\degree$

$\dfrac{4}{15}$ of two right angles = $\dfrac{4}{15} \times 180\degree = 48\degree$

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.

If 3x + 18° and 2x + 25° are supplementary, find the value of x.

Answer

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

⇒ (3x + 18°) + (2x + 25°) = 180°

⇒ 5x + 43° = 180°

⇒ 5x = 137°

⇒ x = $\dfrac{137\degree}{5}$

⇒ x = 27.4°

Converting into degrees and minutes,

0.4° = 0.4 × 60' = 24'

⇒ x = 27° 24'

Hence, x = 27.4° = 27° 24'.

If two complementary angles are in the ratio 1 : 5, find them.

Answer

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 = $\dfrac{90\degree}{6}$

⇒ x = 15°

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

Hence, the angles are 15° and 75°.

If two supplementary angles are in the ratio 2 : 7, find them.

Answer

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 = $\dfrac{180\degree}{9}$

⇒ x = 20°

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

Hence, the angles are 40° and 140°.

Three angles which add upto 180° are in the ratio 2 : 3 : 7. Find them.

Answer

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 = $\dfrac{180\degree}{12}$

⇒ 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°.

20% of an angle is the supplement of 60°. Find the angle.

Answer

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°

⇒ $\dfrac{20}{100} \times x = 120\degree$

⇒ $\dfrac{x}{5} = 120\degree$

⇒ x = 600°

Hence, the required angle is 600°.

10% of x° is the complement of 40% of 2x°. Find x.

Answer

40% of 2x° = $\dfrac{40}{100} \times 2x\degree = \dfrac{4x\degree}{5}$

Complement of $\dfrac{4x\degree}{5}= 90\degree - \dfrac{4x\degree}{5}$

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

$\Rightarrow\dfrac{10}{100} \times x\degree = 90\degree - \dfrac{4x\degree}{5}\\[1em] \Rightarrow \dfrac{x\degree}{10} + \dfrac{4x\degree}{5} = 90\degree\\[1em] \Rightarrow \dfrac{x\degree + 8x\degree}{10} = 90\degree\\[1em] \Rightarrow\dfrac{9x\degree}{10} = 90\degree\\[1em] \Rightarrow 9x\degree = 900\degree\\[1em] \Rightarrow x\degree = \dfrac{900\degree}{9}\\[1em] \Rightarrow x = 100$

Hence, x = 100°.

Use the adjacent figure, to find angle x° and its supplement.

Answer

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° = $\dfrac{180\degree}{10}$

⇒ x° = 18°

Supplement of x° = 180° − 18° = 162°

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

Find K in the given figure.

Answer

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 = $\dfrac{105\degree}{2}$

⇒ K = 52.5°

Converting into degrees and minutes,

0.5° = 0.5 × 60' = 30'

⇒ K = 52° 30'

Hence, K = 52.5° = 52° 30'.

Find K in the given figure.

Answer

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 = $\dfrac{318\degree}{6}$

⇒ K = 53°

Hence, K = 53°.

In the given figure, lines PQ, MN and RS intersect at O. If x : y = 1 : 2 and z = 90°, find ∠ROM and ∠POR.

Answer

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° = $\dfrac{90\degree}{3}$

⇒ 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°.

In the given figure, find ∠AOB and ∠BOC.

Answer

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° = $\dfrac{72\degree}{6}$

⇒ x° = 12°

⇒ ∠AOB = 5x° = 5 × 12° = 60°

⇒ ∠BOC = x° = 12°

Hence, ∠AOB = 60° and ∠BOC = 12°.

Find each angle shown in the figure.

Answer

From the figure,

The angles $2y\degree, 2y\degree, 3\dfrac{1}{2}y\degree \text{ and }2\dfrac{1}{2}y\degree$ are the angles at a point.

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

$\Rightarrow 2y\degree + 2y\degree + 3\dfrac{1}{2}y\degree + 2\dfrac{1}{2}y\degree = 360\degree\\[1em] \Rightarrow 2y\degree + 2y\degree + \dfrac{7y\degree}{2} + \dfrac{5y\degree}{2} = 360\degree\\[1em] \Rightarrow \dfrac{4y\degree + 4y\degree + 7y\degree + 5y\degree}{2} = 360\degree\\[1em] \Rightarrow \dfrac{20y\degree}{2} = 360\degree\\[1em] \Rightarrow 10y\degree = 360\degree\\[1em] \Rightarrow y\degree = \dfrac{360\degree}{10}\\[1em] \Rightarrow y\degree = 36\degree$

So, the angles are :

$2y\degree = 2 × 36\degree = 72\degree$

$2y\degree = 2 × 36\degree = 72\degree$

$2\dfrac{1}{2}y\degree = \dfrac{5}{2} \times 36\degree = 90\degree$

$3\dfrac{1}{2}y\degree = \dfrac{7}{2} \times 36\degree = 126\degree$

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

AB, CD and EF are three lines intersecting at the same point.

(i) Find x°, if y° = 45° and z° = 90°.

(ii) Find a°, if x° = 3a°, y° = 5x° and z° = 6x°.

Answer

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 = $\dfrac{180\degree}{36}$

⇒ a = 5°

Hence, a = 5°.