NCERT Solutions for Class 9 Maths Chapter 6: Lines and Angles — Free PDF

Chapter 6 is the gateway to Euclidean geometry in the CBSE Class 9 syllabus. It systematically builds from basic angle relationships to the angle sum property of a triangle.

Topics covered:
- Linear pair axiom and vertically opposite angles
- Angles made by a transversal with two parallel lines (corresponding, alternate interior, co-interior)
- Lines parallel to the same line
- Angle sum property of a triangle
- Exterior angle theorem

The chapter has three exercises with a mix of numerical problems and proofs. Exercise 6.1 deals with linear pair and vertically opposite angles. Exercise 6.2 covers parallel lines cut by a transversal. Exercise 6.3 focuses on the angle sum property and exterior angle theorem.

Geometry questions from this chapter appear consistently in CBSE board exams, typically carrying $5$-$8$ marks. The key to scoring full marks is to state the correct reason (axiom, theorem, or property) alongside every numerical step. Examiners look for both the calculation and the justification. This chapter also lays the groundwork for Chapter 7 (Triangles) and Chapter 8 (Quadrilaterals), so a strong understanding here pays dividends throughout the year.

Line: A straight path extending infinitely in both directions. It has no endpoints.

Line segment: A part of a line with two endpoints.

Ray: A part of a line with one endpoint, extending infinitely in one direction.

Angle: Formed by two rays with a common starting point (vertex). Measured in degrees ($^\circ$).

Types of angles:
- Acute angle: $0^\circ < \theta < 90^\circ$
- Right angle: $\theta = 90^\circ$
- Obtuse angle: $90^\circ < \theta < 180^\circ$
- Straight angle: $\theta = 180^\circ$
- Reflex angle: $180^\circ < \theta < 360^\circ$

Complementary angles: Two angles whose sum is $90^\circ$.

Supplementary angles: Two angles whose sum is $180^\circ$.

Adjacent angles: Two angles that share a common vertex and a common arm, with no overlap.

Linear pair: Two adjacent angles formed when a ray stands on a line. They are always supplementary ($\angle 1 + \angle 2 = 180^\circ$).

Vertically opposite angles: When two lines intersect, the angles opposite each other at the vertex are equal.

Transversal: A line that intersects two or more lines at distinct points.

When a transversal cuts two parallel lines, it creates 8 angles grouped into special pairs:
- Corresponding angles (4 pairs): Equal when lines are parallel — same side of transversal, one interior and one exterior
- Alternate interior angles (2 pairs): Equal when lines are parallel — opposite sides of transversal, both between the lines
- Alternate exterior angles (2 pairs): Equal when lines are parallel — opposite sides of transversal, both outside the lines
- Co-interior angles (2 pairs, also called consecutive interior or allied angles): Sum to $180^\circ$ when lines are parallel — same side, both between the lines

Angle sum property of a triangle: The sum of the three interior angles of any triangle is $180^\circ$: $\angle A + \angle B + \angle C = 180^\circ$.

Exterior angle theorem: An exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles.

Problem 1: In the figure, lines $AB$ and $CD$ intersect at $O$. If $\angle AOC + \angle BOE = 70^\circ$ and $\angle BOD = 40^\circ$, find $\angle BOE$ and reflex $\angle COE$.

Solution:
Since $\angle AOC$ and $\angle BOD$ are vertically opposite angles:

Given $\angle AOC + \angle BOE = 70^\circ$:

Now $\angle AOC + \angle COE + \angle BOE = 180^\circ$ (straight line $AB$):

Reflex $\angle COE = 360^\circ - 110^\circ = 250^\circ$

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Problem 2: If a ray stands on a line, prove that the sum of two adjacent angles so formed is $180^\circ$.

Solution:
This is the Linear Pair Axiom. Let ray $OC$ stand on line $AB$ at point $O$.

Angles $\angle AOC$ and $\angle BOC$ are formed. Since $AB$ is a straight line:

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Problem 3: Prove that vertically opposite angles are equal.

Solution:
Let lines $AB$ and $CD$ intersect at $O$.

$\angle AOC + \angle COB = 180^\circ$ (linear pair on line $AB$) ...(i)
$\angle COB + \angle BOD = 180^\circ$ (linear pair on line $CD$) ...(ii)

From (i) and (ii):
$\angle AOC + \angle COB = \angle COB + \angle BOD$
$\angle AOC = \angle BOD \quad \blacksquare$

Similarly, $\angle COB = \angle AOD$.

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Problem 4: Two lines intersect. If one angle is $75^\circ$, find all four angles.

Solution:
Let the four angles be $\angle 1, \angle 2, \angle 3, \angle 4$ in order.

$\angle 1 = 75^\circ$ (given)
$\angle 2 = 180^\circ - 75^\circ = 105^\circ$ (linear pair)
$\angle 3 = 75^\circ$ (vertically opposite to $\angle 1$)
$\angle 4 = 105^\circ$ (vertically opposite to $\angle 2$)

Verification: $75 + 105 + 75 + 105 = 360^\circ$ ✓

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Problem 5: In the figure, ray $OS$ stands on line $PQ$. Ray $OR$ and ray $OT$ are angle bisectors of $\angle POS$ and $\angle SOQ$ respectively. Prove that $\angle ROT = 90^\circ$.

Solution:
$\angle POS + \angle SOQ = 180^\circ$ (linear pair)

Since $OR$ bisects $\angle POS$: $\angle ROS = \dfrac{1}{2}\angle POS$.
Since $OT$ bisects $\angle SOQ$: $\angle SOT = \dfrac{1}{2}\angle SOQ$.

Problem 1: In the figure, $p \parallel q$ and the transversal $t$ intersects them at points $G$ and $H$ respectively. If $\angle AGH = 110^\circ$, find all other angles.

Solution:
$\angle AGH = 110^\circ$ (given)

$\angle BGH = 180^\circ - 110^\circ = 70^\circ$ (linear pair)

Since $p \parallel q$:
- $\angle GHD = \angle AGH = 110^\circ$ (alternate interior angles)
- $\angle GHC = \angle BGH = 70^\circ$ (alternate interior angles)
- $\angle CHF = \angle AGH = 110^\circ$ (corresponding angles)
- $\angle DHF = \angle BGH = 70^\circ$ (corresponding angles)

All eight angles are determined: four of $110^\circ$ and four of $70^\circ$.

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Problem 2: If $AB \parallel CD$, $\angle APQ = 50^\circ$ and $\angle PRD = 127^\circ$, find $x$ and $y$.

Solution:
$\angle APQ = \angle PQR = 50^\circ$ (alternate interior angles, $AB \parallel CD$)
So $x = 50^\circ$.

Using co-interior angles:
$\angle APR + \angle PRD = 180^\circ$ (co-interior angles, $AB \parallel CD$)

$\angle APR = \angle APQ + y = 50^\circ + y$

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Problem 3: In the figure, $AB \parallel CD$ and $CD \parallel EF$. Also $EA \perp AB$. If $\angle BEF = 55^\circ$, find the values of $x$, $y$, and $z$.

Solution:
Since $AB \parallel CD \parallel EF$ (lines parallel to the same line are parallel to each other):

$\angle BEF = 55^\circ$ (given)
$EA \perp AB$ so $\angle EAB = 90^\circ$

Since $AB \parallel EF$: $\angle ABE + \angle BEF = 180^\circ$ (co-interior angles)
$x + 55^\circ = 180^\circ \implies x = 125^\circ$

Since $AB \parallel CD$: $x + y = 180^\circ$ (co-interior angles)
$125^\circ + y = 180^\circ \implies y = 55^\circ$

Since $CD \parallel EF$: $y + z = 180^\circ$ (co-interior angles)
$55^\circ + z = 180^\circ \implies z = 125^\circ$

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Problem 4: Two parallel lines $l$ and $m$ are cut by a transversal. If one pair of co-interior angles are $(2x + 10)^\circ$ and $(3x - 15)^\circ$, find $x$ and both angles.

Solution:
Co-interior angles are supplementary:

The angles are: $2(37) + 10 = 84^\circ$ and $3(37) - 15 = 96^\circ$.
Verification: $84 + 96 = 180^\circ$ ✓

Problem 1: In $\triangle PQR$, $\angle Q = 2\angle P$ and $2\angle R = 3\angle Q$. Find the angles.

Solution:
Let $\angle P = x$. Then $\angle Q = 2x$.
$2\angle R = 3\angle Q = 6x \implies \angle R = 3x$

By angle sum property:

$\angle P = 30^\circ$, $\angle Q = 60^\circ$, $\angle R = 90^\circ$

Verification: $30 + 60 + 90 = 180^\circ$ ✓. This is a $30$-$60$-$90$ triangle.

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Problem 2: The exterior angle of a triangle is $110^\circ$ and one of the interior opposite angles is $35^\circ$. Find the other two angles of the triangle.

Solution:
By exterior angle theorem:

The third angle (adjacent to the exterior angle):

Verification: $35^\circ + 75^\circ + 70^\circ = 180^\circ$ ✓

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Problem 3: The angles of a triangle are in the ratio $2:3:4$. Find all angles.

Solution:
Let the angles be $2x$, $3x$, $4x$.

Angles: $40^\circ$, $60^\circ$, $80^\circ$.

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Problem 4: In $\triangle ABC$, $\angle A + \angle B = 115^\circ$ and $\angle B + \angle C = 130^\circ$. Find all three angles.

Solution:
From $\angle A + \angle B + \angle C = 180^\circ$:

$\angle C = 180^\circ - 115^\circ = 65^\circ$ (using $\angle A + \angle B = 115^\circ$)

$\angle A = 180^\circ - 130^\circ = 50^\circ$ (using $\angle B + \angle C = 130^\circ$)

$\angle B = 115^\circ - 50^\circ = 65^\circ$

Verification: $50 + 65 + 65 = 180^\circ$ ✓

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Problem 5: In the figure, $PQ \parallel ST$, $\angle PQR = 110^\circ$ and $\angle RST = 130^\circ$. Find $\angle QRS$.

Solution:
Draw a line through $R$ parallel to $PQ$ (and $ST$).

Let this line be $XY$. Since $XY \parallel PQ$:
$\angle PQR + \angle QRX = 180^\circ$ (co-interior angles)
$\angle QRX = 180^\circ - 110^\circ = 70^\circ$

Since $XY \parallel ST$:
$\angle SRY + \angle RST = 180^\circ$ (co-interior angles)
$\angle SRY = 180^\circ - 130^\circ = 50^\circ$

$\angle QRS = \angle QRX + \angle SRY = 70^\circ + 50^\circ = 120^\circ$

This technique of drawing an auxiliary parallel line is very powerful for solving angle problems.

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Problem 6: In $\triangle ABC$, $\angle A = 50^\circ$. If the bisectors of $\angle B$ and $\angle C$ meet at point $I$, find $\angle BIC$.

Solution:
Let $\angle B = 2\beta$ and $\angle C = 2\gamma$.

Then $50 + 2\beta + 2\gamma = 180^\circ \implies \beta + \gamma = 65^\circ$.

In $\triangle BIC$:
$\angle BIC = 180^\circ - \beta - \gamma = 180^\circ - 65^\circ = 115^\circ$

General formula: $\angle BIC = 90^\circ + \dfrac{\angle A}{2} = 90^\circ + 25^\circ = 115^\circ$.

Example 1: Finding angles using multiple properties

Two parallel lines are cut by a transversal. One pair of co-interior angles are $(3x + 10)^\circ$ and $(2x + 30)^\circ$. Find $x$ and both angles.

Solution:
Co-interior angles sum to $180^\circ$:

The angles are: $3(28) + 10 = 94^\circ$ and $2(28) + 30 = 86^\circ$.

Verification: $94 + 86 = 180^\circ$ ✓

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Example 2: Proving lines are parallel

A transversal cuts two lines making angles of $65^\circ$ and $115^\circ$ on the same side of the transversal. Are the lines parallel?

Solution:
$65^\circ + 115^\circ = 180^\circ$.

Since co-interior angles are supplementary, the two lines are parallel.

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Example 3: Triangle with an exterior angle

In $\triangle ABC$, side $BC$ is produced to $D$. If $\angle ACD = 128^\circ$ and $\angle ABC = 43^\circ$, find $\angle BAC$ and $\angle ACB$.

Solution:
$\angle ACB = 180^\circ - 128^\circ = 52^\circ$ (linear pair)

By exterior angle theorem: $\angle ACD = \angle BAC + \angle ABC$
$128^\circ = \angle BAC + 43^\circ$
$\angle BAC = 85^\circ$

Verification: $85 + 43 + 52 = 180^\circ$ ✓

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Example 4: Using alternate interior angles to find unknown angles

In the figure, $l \parallel m$ and $t$ is a transversal. If $\angle 1 = (3x + 20)^\circ$ and $\angle 2 = (5x - 40)^\circ$ are alternate interior angles, find both angles.

Solution:
Alternate interior angles are equal (since $l \parallel m$):

Both angles $= 3(30) + 20 = 110^\circ$.

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Example 5: Angle sum property with exterior angle

Two sides of a triangle are produced. The exterior angles formed are $100^\circ$ and $120^\circ$. Find all three interior angles.

Solution:
Interior angle adjacent to $100^\circ$ exterior $= 180^\circ - 100^\circ = 80^\circ$.
Interior angle adjacent to $120^\circ$ exterior $= 180^\circ - 120^\circ = 60^\circ$.
Third angle $= 180^\circ - 80^\circ - 60^\circ = 40^\circ$.

Verification: The exterior angle at the third vertex $= 80^\circ + 60^\circ = 140^\circ$.

Mistake 1: Not stating the reason for each step.
In geometry proofs, every step must have a reason. Writing "$\angle 1 = \angle 2$" without stating "(alternate interior angles, $l \parallel m$)" will lose marks. CBSE examiners award separate marks for reasons.

Mistake 2: Confusing alternate interior angles with co-interior angles.
Alternate interior angles are on opposite sides of the transversal and are equal (when lines are parallel). Co-interior angles are on the same side and are supplementary ($180^\circ$). Mixing these up reverses the equation.

Mistake 3: Using the angle sum property for non-triangles.
The angle sum of $180^\circ$ applies only to triangles. A quadrilateral has an angle sum of $360^\circ$, a pentagon $540^\circ$, and in general an $n$-gon has $(n-2) \times 180^\circ$.

Mistake 4: Assuming angles are equal without checking for parallel lines.
Corresponding angles are equal ONLY when the lines are parallel. If the problem does not state or prove that the lines are parallel, you cannot assume equal corresponding angles.

Mistake 5: Forgetting the reflex angle.
When a problem asks for the reflex angle, remember reflex $\angle = 360^\circ - \text{angle}$. Students often give the non-reflex angle instead.

Q1. Two lines intersect, forming four angles. If one angle is $48^\circ$, find the other three.

Answer: The four angles are $48^\circ$, $132^\circ$, $48^\circ$, $132^\circ$ (using linear pair and vertically opposite angles).

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Q2. If $l \parallel m$ and a transversal makes an angle of $72^\circ$ with $l$, find all eight angles.

Answer: The eight angles are four of $72^\circ$ and four of $108^\circ$ (using corresponding, alternate, and co-interior angle relationships).

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Q3. The angles of a triangle are $(x + 10)^\circ$, $(2x + 20)^\circ$, and $(3x - 30)^\circ$. Find $x$ and all three angles.

Answer: $(x+10) + (2x+20) + (3x-30) = 180 \implies 6x = 180 \implies x = 30$. Angles: $40^\circ$, $80^\circ$, $60^\circ$.

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Q4. An exterior angle of a triangle is $140^\circ$. If the two interior opposite angles are equal, find each.

Answer: Let each interior opposite angle $= a$. Then $a + a = 140^\circ \implies a = 70^\circ$. The third angle $= 180^\circ - 140^\circ = 40^\circ$.

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Q5. In the figure, $AB \parallel CD$. A transversal makes $\angle BAE = 50^\circ$. Another transversal makes $\angle DCE = 60^\circ$. Find $\angle AEC$.

Answer: In $\triangle AEC$: $\angle AEC = 180^\circ - \angle EAC - \angle ECA$. Using alternate interior angles: $\angle EAC = 50^\circ$ and $\angle ECA = 60^\circ$. So $\angle AEC = 180^\circ - 50^\circ - 60^\circ = 70^\circ$.