Lines and Angles: An Overview

Lines and angles are fundamental concepts in geometry, essential for understanding various geometric relationships and properties. In class 9, students explore these concepts through definitions, types, and properties. The lines and angles class 9 worksheet and test paper pdf provide practical problems and examples to enhance learning. Key topics include parallel lines and vertical angles, and parallel lines and interior angles, which are often examined through diagrams and formulas. Additionally, parallel and transversal lines create alternate interiors and corresponding angles, which are vital for solving geometric questions. Tools like the parallel and transversal lines calculator can aid in visualizing and solving problems effectively.

Definition of Lines

A straight one-dimensional figure that extends infinitely in both ways without any thickness or curvature is called a Line. A line does not curve or bend and can be horizontal or vertical.

Definition of Angles

An angle is a fundamental concept in geometry that describes the figure formed by two rays or line segments sharing a common endpoint. The common endpoint is called the vertex of the angle and rays or line segments are called arms or sides of the angle. Angles measure the amount of turn or rotation between the two rays.

Types of Lines

There are several types of lines depending on the characteristics.

In the context of endpoints and length, lines are of two types.

• Line segment

• Ray

Lines can be classified based on their properties and relationships with other lines. Here are the types of lines:

• Straight line

• Horizontal line

• Vertical line

• Parallel line

• Perpendicular line

• Intersecting line

• Transversal line

• Skew line

Line Segment and Ray

Ray is a type of line which has a starting point on one side but no endpoint on the other side. It continues infinitely in one end.

In the picture, A is the starting point and it crosses through point B and goes infinitely. It is denoted by Arror over AB.

A line segment is a type of line which has a starting point and endpoint. It is of finite length and does not extend beyond endpoints.

In the picture, A is the starting point or endpoint on one side and B is the endpoint or starting point on another side. It is denoted by bar AB.

Parallel lines, Perpendicular lines, Transversal lines

Types of Angles

There are many types of angles based on their measurement or characteristics or their position.

Types of angles based on their measurement are:

1. Acute angle

2. Right angle

3. Obtuse Angle

4. Straight Angle

5. Reflex Angle

6. Full Angle/ Complete Angle

Types of angles based on comparison with the other angle are:

1. Supplementary Angles

2. Complementary Angles

3. Corresponding Angles

4. Adjacent Angles

5. Vertically Opposite Angles

6. Alternate interior Angles

7. Alternate Exterior Angles

8. Co-interior Angles

General Classification

Classification based on the relation between two angles

Tips and Tricks

• If one angle is given, then subtract it from 180° to get its supplementary angle.

• If one angle is given, then subtract it from 90° to get its complementary angle.

• In vertical angles, opposite angles formed by two intersecting lines are equal.

• Angles on opposite sides of the transversal but inside the two lines are equal.

• Angles on opposite sides of the transversal but outside the two lines are equal.

• If two lines are perpendicular, they form four right angles at their intersection.

• Co-interior angles lie between the two parallel lines on the same side of the transversal line.

Practice Questions

Q1. If the sum and difference of two angles are 135° and $\frac{\pi}{12}$, respectively, then the value of the angles in degree measure is:

1. 70° and 65°

2. 75° and 60°

3. 45° and 90°

4. 80° and 55°

Hint: Use ${\pi}=180°$ and form linear equations.

Answer:

Let the two angles be x and y.

So, x + y = 135°------------------------------(1)

The difference of the two angles is $\frac{\pi}{12}$.

So, x – y = $\frac{\pi}{12}=\frac{180}{12}$ = 15° ---------------------------(2)

Solving equations (1) and (2), we get,

x = 75°

So, y = 135 – 75 = 60°

Hence, the correct answer is 75° and 60°.

Q2. In an isosceles triangle, if the unequal angle is 4 times the sum of the equal angles, then each equal angle is:

1. 25°

2. 30°

3. 21°

4. 18°

Hint: Use and set up an equation as given as the unequal angle is 4 times the sum of the equal angles and the angle sum property of a triangle.

Answer:

In an isosceles triangle, two angles are equal.

Let the equal angles be $\angle B$ and $\angle C$ ($\angle B=\angle C$). The third angle which is unequal is $\angle A$.

Given that the unequal angle is $4$ times the sum of the equal angles.

$=4(\angle B+\angle C)$

$=4(\angle B+\angle B)$

$=4(2\angle B)$

$=8\angle B$

The sum of the angles in a triangle is always $180^{\circ}$.

$\Rightarrow 2\angle B+ 8\angle B = 180^{\circ}$

$\Rightarrow 10\angle B = 180^{\circ}$

$\Rightarrow \angle B= 18^{\circ}$

Hence, the correct answer is $18^{\circ}$.

Q3. In $\triangle \mathrm{ABC}$, $\angle \mathrm{ABC} = 90^{\circ}$, $\mathrm{BP}$ is drawn perpendicular to $\mathrm{AC}$. If $\angle \mathrm{BAP} = 50^{\circ},$ what is the value of $\angle \mathrm{PBC}?$

1. 30°

2. 45°

3. 50°

4. 60°

Hint: Use the angle sum property and complementary angles to find $\angle \mathrm{PBC}$.

Answer:

In right-angled $\triangle \mathrm{ABC}$, where $\angle \mathrm{ABC} = 90^{\circ}$, $\mathrm{BP}$ is drawn perpendicular to $\mathrm{AC}$. Given that $\angle \mathrm{BAP }= 50^{\circ}$

In $\triangle\mathrm{ ABC}$,

$\mathrm{\angle C = 180^{\circ} - \angle A - \angle B = 180^{\circ} - 50^{\circ} - 90^{\circ} = 40^{\circ}}$

In $\mathrm{\triangle BPC}$,

$\mathrm{\angle PBC +\angle BPC +\angle PCB =180^{\circ}}$

⇒ $\mathrm{\angle PBC = 90^{\circ} - \angle PCB = 90^{\circ} - 40^{\circ} = 50^{\circ}}$

Hence, the correct answer is $50^{\circ}$.

Q4. ABC is an isosceles triangle having AB = AC and $\angle$A = 40$^\circ$. Bisector PO and OQ of the exterior angle $\angle$ ABD and $\angle$ ACE formed by producing BC on both sides, meet at O, then the value of $\angle$BOC is:

1. 70°

2. 110°

3. 80°

4. 55°

Hint: Use the angle sum property and the concept of vertically opposite angles to solve this.

Answer:

Given, an isosceles $\triangle$ ABC with AB = BC

And, $\angle$ A = 40$^\circ$ with bisectors PO and OQ of $\angle$ ABD and $\angle$ ACE.

In $\triangle$ ABC,

$\angle$A + $\angle$ B+ $\angle$ C = 180$^\circ$

Or, 40$^\circ$ + 2$\angle$C = 180$^\circ$

Or, $\angle$ B = $\angle$ C = 70$^\circ$

Now, $\angle$ ABD + $\angle$ ABC = 180$^\circ$

Or, $\angle$ ABD + 70$^\circ$ = 180$^\circ$

Or, $\angle$ ABD = 110$^\circ$

Since PO is an angle bisector, $\angle$ PBD = $\frac{110^\circ}{2}=55^\circ$

Similarly, $\angle$ QCE = 55$^\circ$

In $\triangle$ BCO,

$\angle$BOC + $\angle$ CBO+ $\angle$ BCO = 180$^\circ$

Or, $\angle$ BOC + $\angle$ PBD+ $\angle$ QCE = 180$^\circ$ (vertically opposite angles)

Or, $\angle$ BOC + 55$^\circ$ + 55$^\circ$ = 180$^\circ$

Or, $\angle$ BOC = 70$^\circ$

Hence, the correct answer is 70$^\circ$.

Q5. In the following figure, if l || m, then find the measures of angles marked by a and b.

1. a = 90$^{\circ}$ and b = 90$^{\circ}$

2. a = 55$^{\circ}$ and b = 125$^{\circ}$

3. a = 70$^{\circ}$ and b = 110$^{\circ}$

4. a = 60$^{\circ}$ and b = 120$^{\circ}$

Hint: For two parallel lines, the corresponding angles are equal.

A triangle's exterior angle equals the sum of opposite interior angles.

Answer:

Since $l \parallel m$,

⇒ $\angle$ c = 110$^{\circ}$ (corresponding angles are equal)

Also, $\angle$ a + $\angle$ c = 180$^{\circ}$ (supplementary angles)

⇒ $\angle$ a = 180$^{\circ}$ – 110$^{\circ}$ = 70$^{\circ}$

Now, $\angle$ b = $\angle$ a + 40$^{\circ}$ (exterior angle of a triangle)

= 70$^{\circ}$ + 40$^{\circ}$

= 110$^{\circ}$

Hence, the correct answer is a = 70$^{\circ}$ and b = 110$^{\circ}$.

Q6. An angle is greater than its complementary angle by 30°. Find the measure of the angle.

1. 30°

2. 90°

3. 60°

4. 120°

Hint: Let $x$ be the angle and the complement of the angle be $(90°-x)$.

Answer:

Let $x$ be the angle and the complement of the angle be $(90°-x)$.

According to the question,

$(90°-x)+30°=x$

⇒ $2x=120°$

$\therefore x =60°$

Hence, the correct angle is 60°.

Q7. An angle $\frac{1}{3}$rd of its supplementary angle. Find the measure of the angle.

1. 45°

2. 90°

3. 135°

4. 22.5°

Hint: Let $x$ be the angle.

The supplement of the angle is $(180°-x)$.

Answer:

Let $x$ be the angle.

The supplement of the angle is $(180°-x)$.

According to the question,

$\frac{1}{3}(180°-x)=x$

⇒ $180°-x=3x$

⇒ $4x=180°$

$\therefore x =45°$

Hence, the correct answer is 45°.