Lines and Angles MCQ Quiz - Objective Question with Answer for Lines and Angles - Download Free PDF

Given:

AB and CD are two parallel lines.

PQ is a transversal intersecting AB at R and CD at S.

∠ARP = 2xº

∠ARQ = yº

∠DSP = 4xº

Formula Used:

Using the property of parallel lines, corresponding angles and linear pair property:

1. Corresponding angles are equal.

2. Sum of angles on a straight line is 180º.

Calculation:

Since AB || CD and PQ is a transversal:

∠DSP corresponds to ∠BRP. Hence:

⇒ ∠BRP = 4x° 

Using the linear pair property at point R:

∠ARP + ∠BRP = 180º

⇒ 2x + 4x = 180

⇒ 6x = 180

⇒ x = 30

Using the linear pair property at point R:

∠ARP + ∠ARQ = 180º

⇒ 2x + y = 180

Substitute x = 30:

⇒ 2(30) + y = 180

⇒ 60 + y = 180

⇒ y = 120

The values of x and y are 30 and 120 respectively.

Given:

AB is parallel to CD.

A transversal PQ intersects AB and CD at E and F, respectively.

∠PEB = 49°.

G is a point between AB and CD such that:

∠BEG = 35°

∠GFE = 16°

We need to find the measure of ∠EGF.

Formula Used:

Sum of angles in a triangle = 180°

Calculation:

Use the linear pair on ∠PEB and ∠BEF:

∠PEB + ∠BEF = 180° (on a straight line)

⇒ 49° + ∠BEF = 180°

⇒ ∠BEF = 180° - 49° = 131°

Now, ∠BEF = ∠BEG + ∠GEF

∠GEF = ∠BEF - ∠BEG

∠GEF = 131° - 35° = 96°

Use the angle sum property in triangle EGF:

∠GEF + ∠GFE + ∠EGF = 180°

⇒ 96° + 16° + ∠EGF = 180°

⇒ ∠EGF = 180° − 112° = 68°

∴ The measure of ∠EGF is 68°.

Given:

AB || CD

PQ is a transversal intersecting AB at R and CD at S.

∠BRP = (2x + 13)°

∠DSP = (3x − 22)°

Formula used:

When two parallel lines are intersected by a transversal:

1. Alternate exterior angles are equal.

2. Angles on a straight line (linear pair) sum up to 180°.

Calculations:

∠BRP and ∠DSP are alternate exterior angles.

Since AB || CD, we have:

∠BRP = ∠DSP

⇒ 2x + 13 = 3x - 22

⇒ 13 + 22 = 3x - 2x

⇒ 35 = x

Now, substitute the value of x into ∠DSP:

∠DSP = (3 × 35 - 22)°

⇒ ∠DSP = (105 - 22)°

⇒ ∠DSP = 83°

∠CSP and ∠DSP form a linear pair on the line CD.

∠CSP + ∠DSP = 180°

⇒ ∠CSP + 83° = 180°

⇒ ∠CSP = 180° - 83°

⇒ ∠CSP = 97°

∴ ∠CSP = 97°.

Given:

AB is parallel to CD.

A transversal PQ intersects AB and CD at E and F, respectively.

∠PEB = 59º.

G is a point between AB and CD such that:

∠BEG = 50º

∠GFE = 33º

We need to find the measure of ∠EGF.

Formula Used:

Sum of angles in a triangle = 180º

Calculation:

Use linear pair on ∠PEB and ∠BEF

∠PEB + ∠BEF = 180° (on a straight line)

⇒ 59° + ∠BEF = 180° 

So the angle ∠BEF = 180 - 59 = 121° 

Now, ∠BEF = ∠BEG + ∠GEF

∠GEF = ∠BEF - ∠BEG

∠GEF = 121 - 50 = 71° 

Use angle sum property in triangle EGF

∠GEF + ∠GFE + ∠EGF = 180°

⇒ 71° + 33° + ∠EGF = 180°

⇒ ∠EGF = 180° − 104° = 76°

∴ The measure of ∠EGF = 76° 

Given:

AB is parallel to CD.

Transversal PQ intersects AB and CD at E and F respectively.

∠PEB = 56º

∠BEG = 32º

∠GFE = 47º

Formula used:

In a triangle, the sum of angles is 180º:

∠GEF + ∠GFE + ∠EGF = 180º

Calculation:

∠BEG = 32º

∠GFE = 47º

⇒ ∠GEF = 180º - (∠BEG + ∠PEB)

⇒ ∠GEF = 180º - (32º + 56º)

⇒ ∠GEF = 180º - 88º

⇒ ∠GEF = 92º

Now, ∠GEF + ∠GFE + ∠EGF = 180º

⇒ 92° + 47° + ∠EGF = 180º

⇒ 139° + ∠EGF = 180º

⇒ ∠EGF = 180° - 139° 

⇒ ∠EGF = 41° 

∴ The corect answer is option (3).

Given:

One of the supplement angles is 130°.

Concept used:

For supplementary angle: The sum of two angles is 180°.

For complementary angle: The sum of two angles is 90°.

Calculation:

The supplement angle of 130° = 180° - 130° = 50°

The complement angle of 50° = 90° - 50° = 40°

∴ The complement angle of the supplement angle of 130° is 40°

Mistake PointsPlease note that first, we have to find the supplementary angle of 130° & after that, we will find the complementary angle of the resultant value.

GIVEN :

Sum of the measure of the interior angles of a polygon is 1620°.

FORMULA USED :
Sum of interior angles of a polygon = (n – 2) × 180°

Where n is the number of sides.

CALCULATION :

Applying the formula :

1620° = (n – 2) × 180°

⇒ (n – 2) = 1620°/180°

⇒ (n – 2) = 9

⇒ n = 11

Hence,

Given,

∠A, ∠B and ∠C are three angles of a triangle.

∠A/4 + ∠B/4 + ∠C/5 = 41°

Formula:

Sum of all three angles of a triangle is 180°.

Calculation:

∠A/4 + ∠B/4 + ∠C/5 = 41°

⇒ (5∠A + 5∠B + 4∠C)/20 = 41°

⇒ (∠A + 4∠A + ∠B + 4∠B + 4∠C)/20 = 41°

⇒ (∠A + ∠B + 4∠A + 4∠B + 4∠C)/20 = 41°

⇒ ∠A + ∠B + 4(∠A + ∠B + ∠C) = 41° × 20

⇒ ∠A + ∠B + 4 × 180° = 820°

⇒ ∠A + ∠B = 820° - 720°

⇒ ∠A + ∠B = 100°

∴ The value of ∠A + ∠B is 100°

Given:

A is 26° more than its complementary angle.

B is 30° less than its supplementary angle.

Concept used:

Complementary angles are those angles whose sum is 90°

Supplementary angles are those whose sum is 180° 

Calculation:

A + A - 26 = 90

⇒ 2A = 116

⇒ A = 58

B + B + 30 = 180

⇒ 2B = 150

⇒ B = 75

So,

A - B

⇒ 58 - 75

⇒ - 17

∴ required value is - 17

Given:

The supplement of an angle is 15° more than three times its complement.

Concept Used:

In this type of question, supplement of angle = 180° – x,

Complement of angle = 90° – x

Calculation:

Let angle be x°

According to the question,

180° – x = 3(90° – x) + 15°

⇒ 180° – x° = 270° – 3x° + 15°

⇒ 2x = 105

⇒ x = 52.5

Given:

∠ABD = 55° and ∠ACD = 30°

Calculation:

∠BAD = α and ∠CAD = β   

So, ∠BAC = y = α + β

So Referring to the triangle ΔABD and ΔACD,

∠ADB =  180°-  α - 55°

∠ADC =  180°-   β   - 30°

For point D,

∠ADB +∠ADC + x = 360°

⇒ 180°-  α - 55° +  180°-   β   - 30° + x =  360°

⇒ 360 - α - β - 85° + x = 360

⇒ x - (α + β) - 85° = 0

⇒ x - y - 85° = 0

⇒  x -y = 85°

∴ The correct answer is option (1).

Concept used:

Obtuse angled triangle: A triangle in which one angle is greater than 90° is called an obtuse angle triangle.

Sum of all the angles of the triangle is 180°

Calculation:

As we know,

Sum of all three angles of triangles is 180.

According to the question

⇒ x + 3x + 20 + 6x = 180

⇒ 10x + 20 = 180

⇒ 10x = 180 – 20

⇒ 10x = 160

⇒ x = 160/10

⇒ x = 16

First angle = x = 16°

Second angle = 3x + 20 = 3 × 16 + 20 = 48 + 20 = 68°

Third angle = 6x = 6 × 16 = 96°

GIVEN :

Each of the exterior angles 24°

CONCEPT :

Sum of exterior angles of a regular polygon = 360°

FORMULA USED :
Each of the exterior angles of a regular polygon = 360/n

And

Number of diagonals = n(n – 3)/2

Where n = number of sides

CALCULATION :

Number of sides = 360/24 = 15

So,

Let the angles be ∠A=2x, ∠B=3x, ∠C=4x

The sum of all angles of a triangle is 180°.

⇒ 2x + 3x + 4x = 180°

⇒ 9x = 180°

⇒ x = 20°

So, ∠A = 2 × 20° = 40°

∠B = 3 × 20° = 60°

∠C = 4 × 20° = 80°

Given AB || CD, So, AC acts as a transversal line.

The diagram is,

∠BAC = ∠ACD

i.e. ∠ACD = 40°

Hence, the correct answer is "40°".

Concept:

The radian is the SI unit for measuring angles, and is the standard unit of angular measure used in many areas of mathematics. The length of an arc of a unit circle is numerically equal to the measurement in radians of the angle that it subtends.

Now, π radian = 180°

⇒ 1 radian = 180°/π

⇒ 1 radian = 180°/(22/7)