NCERT Solutions for Class 7 Maths Chapter 5: Lines and Angles — Complete Guide with Step-by-Step Solutions

Two angles are complementary if their sum is $90°$.

Examples:
- $30°$ and $60°$ are complementary because $30° + 60° = 90°$.
- $45°$ and $45°$ are complementary because $45° + 45° = 90°$.
- $15°$ and $75°$ are complementary because $15° + 75° = 90°$.

Finding the complement of an angle:

So the complement of $65°$ is $90° - 65° = 25°$.

Important observations:
- Both complementary angles must be acute (less than $90°$).
- An obtuse angle ($> 90°$) or a right angle ($= 90°$) cannot have a complement.
- Two right angles cannot be complementary (their sum would be $180°$, not $90°$).

Two angles are supplementary if their sum is $180°$.

Examples:
- $110°$ and $70°$ are supplementary because $110° + 70° = 180°$.
- $90°$ and $90°$ are supplementary because $90° + 90° = 180°$.
- $135°$ and $45°$ are supplementary because $135° + 45° = 180°$.

Finding the supplement of an angle:

So the supplement of $105°$ is $180° - 105° = 75°$.

Important observations:
- Two acute angles cannot be supplementary (their sum is less than $180°$).
- Two obtuse angles cannot be supplementary (their sum exceeds $180°$).
- Two right angles are supplementary ($90° + 90° = 180°$).
- A supplementary pair always has one acute angle and one obtuse angle, OR both are $90°$.

Two angles are adjacent if they:
1. Share a common vertex (corner point).
2. Share a common arm (one side is shared).
3. Their non-common arms are on opposite sides of the common arm.
4. They do not overlap.

Think of adjacent angles as two angles sitting side by side, like neighbours sharing a fence. The common arm is the fence.

Example: If two rays $OA$ and $OC$ are drawn from a point $O$ on a line $AB$, then $\angle AOC$ and $\angle COB$ are adjacent angles. They share vertex $O$ and common arm $OC$.

Note: Adjacent angles do NOT have to add up to $90°$ or $180°$ — that depends on whether they also form a complementary pair, supplementary pair, or linear pair.

A linear pair is a special case of adjacent angles where the non-common arms form a straight line (i.e., they are opposite rays).

Key property: Angles in a linear pair are always supplementary (they add up to $180°$).

Example: If a ray $OC$ stands on a line $AB$ at point $O$, then $\angle AOC$ and $\angle BOC$ form a linear pair.

If $\angle AOC = 125°$, then $\angle BOC = 180° - 125° = 55°$.

Important: Every linear pair is supplementary, but not every pair of supplementary angles forms a linear pair. Supplementary angles don't have to be adjacent — for example, $110°$ and $70°$ are supplementary even if they are in different parts of a figure.

When two straight lines intersect, they form two pairs of vertically opposite angles (also called vertical angles).

Key property: Vertically opposite angles are always equal.

If two lines intersect at a point and one angle is $\angle 1$, then:
- The angle vertically opposite to $\angle 1$ is equal to $\angle 1$.
- The two angles adjacent to $\angle 1$ are each equal to $180° - \angle 1$.

Example: If two lines intersect and one of the angles is $75°$, then:

So the four angles are $75°, 105°, 75°, 105°$.

Proof (why vertically opposite angles are equal):
$\angle 1 + \angle 2 = 180°$ (linear pair) $\Rightarrow \angle 2 = 180° - \angle 1$
$\angle 2 + \angle 3 = 180°$ (linear pair) $\Rightarrow \angle 3 = 180° - \angle 2 = 180° - (180° - \angle 1) = \angle 1$

Therefore $\angle 1 = \angle 3$.

A transversal is a line that intersects two or more lines at distinct points.

When a transversal crosses two lines, it creates 8 angles at the two points of intersection. These 8 angles are divided into specific pairs with special names and properties.

Let the two lines be $l$ and $m$, and the transversal be $t$. At the intersection with line $l$, we get angles $\angle 1, \angle 2, \angle 3, \angle 4$. At the intersection with line $m$, we get angles $\angle 5, \angle 6, \angle 7, \angle 8$.

The region between the two lines is called the interior. The region outside the two lines is called the exterior.

Interior angles: $\angle 3, \angle 4, \angle 5, \angle 6$
Exterior angles: $\angle 1, \angle 2, \angle 7, \angle 8$

Corresponding angles are in the same position at each intersection point — both on the same side of the transversal, and both either above or below their respective line.

The four pairs of corresponding angles are:
- $\angle 1$ and $\angle 5$
- $\angle 2$ and $\angle 6$
- $\angle 3$ and $\angle 7$
- $\angle 4$ and $\angle 8$

**When $l \parallel m$: All pairs of corresponding angles are equal**.

Memory trick: Corresponding angles are like the same person standing at two different bus stops — same position, same angle.

Alternate interior angles are between the two lines (interior) and on opposite sides of the transversal.

The two pairs of alternate interior angles are:
- $\angle 3$ and $\angle 6$
- $\angle 4$ and $\angle 5$

**When $l \parallel m$: Alternate interior angles are equal**.

Memory trick: Alternate interior angles form a Z-shape (or reverse Z-shape) with the transversal.

Alternate exterior angles are outside the two lines (exterior) and on opposite sides of the transversal.

The two pairs are:
- $\angle 1$ and $\angle 8$
- $\angle 2$ and $\angle 7$

**When $l \parallel m$: Alternate exterior angles are equal**.

Co-interior angles (also called same-side interior angles or consecutive interior angles) are between the two lines and on the same side of the transversal.

The two pairs are:
- $\angle 3$ and $\angle 5$
- $\angle 4$ and $\angle 6$

**When $l \parallel m$: Co-interior angles are supplementary** (they add up to $180°$).

This is different from the others! Corresponding and alternate angles are EQUAL; co-interior angles ADD UP TO $180°$. This is the most common source of confusion.

Memory trick: Co-interior angles form a U-shape (or C-shape). They are on the same side and add to $180°$.

The converse of the above properties is also true. If a transversal cuts two lines and ANY of the following conditions hold, then the lines must be parallel:

1. A pair of corresponding angles is equal.
2. A pair of alternate interior angles is equal.
3. A pair of co-interior angles is supplementary (adds to $180°$).

This is extremely useful for proving that two lines are parallel in geometry problems.

Problem: Find the complement of each angle: (i) $20°$ (ii) $63°$ (iii) $57°$

Solution:
The complement of angle $x$ is $90° - x$.

(i) Complement of $20° = 90° - 20° = 70°$

(ii) Complement of $63° = 90° - 63° = 27°$

(iii) Complement of $57° = 90° - 57° = 33°$

Answer: $70°, 27°, 33°$.

Problem: Find the supplement of each angle: (i) $105°$ (ii) $87°$ (iii) $154°$

Solution:
The supplement of angle $x$ is $180° - x$.

(i) Supplement of $105° = 180° - 105° = 75°$

(ii) Supplement of $87° = 180° - 87° = 93°$

(iii) Supplement of $154° = 180° - 154° = 26°$

Answer: $75°, 93°, 26°$.

Problem: Two supplementary angles are in the ratio $4:5$. Find each angle.

Solution:
Let the angles be $4x$ and $5x$.

Since they are supplementary:

The angles are:

Verification: $80° + 100° = 180°$. Correct.

Answer: The angles are $80°$ and $100°$.

Problem: Two complementary angles are in the ratio $2:3$. Find them.

Solution:
Let the angles be $2x$ and $3x$.

Since they are complementary:

The angles are:

Verification: $36° + 54° = 90°$. Correct.

Answer: The angles are $36°$ and $54°$.

Problem: Can two obtuse angles be supplementary? Can two acute angles be supplementary?

Solution:

Two obtuse angles: Each obtuse angle is greater than $90°$. So their sum is greater than $90° + 90° = 180°$. Since supplementary angles must sum to exactly $180°$, two obtuse angles cannot be supplementary.

Two acute angles: Each acute angle is less than $90°$. So their sum is less than $90° + 90° = 180°$. Since supplementary angles must sum to exactly $180°$, two acute angles cannot be supplementary.

Conclusion: A supplementary pair must consist of one acute angle and one obtuse angle, or both angles must be exactly $90°$.

Problem: In the figure, a ray $OC$ stands on line $AB$. If $\angle AOC = (2x + 30)°$ and $\angle BOC = (3x - 10)°$, find both angles.

Solution:
Since $\angle AOC$ and $\angle BOC$ form a linear pair:

Therefore:

Verification: $94° + 86° = 180°$. Correct.

Answer: $\angle AOC = 94°$ and $\angle BOC = 86°$.

Problem: Two lines intersect at a point. If one of the angles is $75°$, find all four angles.

Solution:
Let the four angles be $\angle 1, \angle 2, \angle 3, \angle 4$ (going clockwise).

Given: $\angle 1 = 75°$.

$\angle 3 = 75°$ (vertically opposite to $\angle 1$).

$\angle 1$ and $\angle 2$ form a linear pair:

$\angle 4 = 105°$ (vertically opposite to $\angle 2$).

Answer: The four angles are $75°, 105°, 75°, 105°$.

Problem: Two lines intersect. One pair of vertically opposite angles is $(3x + 5)°$ and $(5x - 25)°$. Find $x$ and all four angles.

Solution:
Vertically opposite angles are equal:

First pair of vertically opposite angles:

Second pair (linear pair with $50°$):

Answer: $x = 15$. The four angles are $50°, 130°, 50°, 130°$.

Problem: Two rays $OC$ and $OD$ stand on line $AB$ such that $\angle AOC = 50°$, $\angle COD = 70°$. Find $\angle DOB$.

Solution:
Since $A$, $O$, $B$ are on a straight line:

Answer: $\angle DOB = 60°$.

Problem: The supplement of an angle is three times its complement. Find the angle.

Solution:
Let the angle be $x$.

Supplement $= 180° - x$.
Complement $= 90° - x$.

Given: Supplement $= 3 \times$ Complement

Verification:
Supplement of $45° = 135°$.
Complement of $45° = 45°$.
$135° = 3 \times 45°$. Correct.

Answer: The angle is $45°$.

Problem: In the figure, $l \parallel m$ and a transversal $t$ cuts them. If $\angle 1 = 110°$, find all other angles.

Solution:
$\angle 1 = 110°$ (given).

Using linear pair:

Using vertically opposite angles:

Using corresponding angles ($l \parallel m$):

Answer: $\angle 1 = \angle 3 = \angle 5 = \angle 7 = 110°$ and $\angle 2 = \angle 4 = \angle 6 = \angle 8 = 70°$.

Key insight: When a transversal cuts two parallel lines and you know any one of the 8 angles, you can find all 8.

Problem: If $l \parallel m$ and a transversal makes $\angle 3 = 55°$, find $\angle 6$.

Solution:
$\angle 3$ and $\angle 6$ are alternate interior angles.

Since $l \parallel m$, alternate interior angles are equal.

Answer: $\angle 6 = 55°$.

Problem: If $l \parallel m$ and co-interior angles are $(2x + 10)°$ and $(3x - 5)°$, find $x$ and both angles.

Solution:
Co-interior angles are supplementary when lines are parallel:

The angles are:

Verification: $80° + 100° = 180°$. Correct.

Answer: $x = 35$. The angles are $80°$ and $100°$.

Problem: Lines $PQ$ and $RS$ are parallel. A transversal cuts them making $\angle PXY = 125°$ (at the intersection with $PQ$). Find $\angle XYS$ (the angle on the same side at the intersection with $RS$).

Solution:
$\angle PXY$ and $\angle XYS$ are co-interior angles (same side of the transversal, between the parallel lines).

Since $PQ \parallel RS$:

Answer: $\angle XYS = 55°$.

Problem: In the figure, $AB \parallel CD$. A transversal intersects $AB$ at $E$ and $CD$ at $F$. If $\angle AEF = (3x + 15)°$ and $\angle EFD = (5x - 35)°$, find $x$ and both angles.

Solution:
$\angle AEF$ and $\angle EFD$ are alternate interior angles (they are between the parallel lines and on opposite sides of the transversal).

Since $AB \parallel CD$, alternate interior angles are equal:

Both angles $= 3(25) + 15 = 90°$.

Verification: $5(25) - 35 = 125 - 35 = 90°$. Both equal $90°$. Correct.

Answer: $x = 25$. Both angles are $90°$ (the transversal is perpendicular to both parallel lines).

Problem: A transversal $t$ crosses two lines $l$ and $m$. The pair of alternate interior angles are $\angle 3 = 70°$ and $\angle 6 = 70°$. Are the lines parallel?

Solution:
Alternate interior angles: $\angle 3 = 70°$ and $\angle 6 = 70°$.

Since $\angle 3 = \angle 6$ (alternate interior angles are equal), by the converse of the alternate interior angles theorem, **$l \parallel m$**.

Answer: Yes, the lines are parallel.

When faced with any angle problem, follow this systematic approach:

Step 1: Identify the given angles and relationships (are lines parallel? Is it a linear pair? Are lines intersecting?).

Step 2: Label all angles clearly. If two lines intersect, you get 4 angles. If a transversal cuts two lines, you get 8 angles.

Step 3: Apply the correct property:
- Lines intersecting? $\to$ Vertically opposite angles are equal.
- Ray on a line? $\to$ Linear pair (sum $= 180°$).
- Transversal + parallel lines? $\to$ Identify which pair (corresponding, alternate, or co-interior) and apply the correct rule.

Step 4: Set up an equation if there are unknowns, and solve.

Step 5: Verify your answer by checking that all angle sums are consistent.

Answer 1:
Complement of $37° = 90° - 37° = 53°$.
Supplement of $37° = 180° - 37° = 143°$.

Answer 2:
Let the smaller angle be $x$. The larger is $x + 20$.

Answer 3:
Vertically opposite angles are equal:

Answer 4:
$\angle DCE = \angle BAE = 72°$ (alternate interior angles, $AB \parallel CD$).
$\angle BCE = 180° - 72° = 108°$ (co-interior angles are supplementary).

Answer 5:
Co-interior angles are supplementary:

Answer 6:
Equal to its complement: $x = 90° - x \Rightarrow 2x = 90° \Rightarrow x = 45°$. Yes, $45°$.
Equal to its supplement: $x = 180° - x \Rightarrow 2x = 180° \Rightarrow x = 90°$. Yes, $90°$.

Problem: In the figure, two lines $AB$ and $CD$ intersect at $O$. Ray $OE$ bisects $\angle AOC$. If $\angle AOC = 130°$, find $\angle BOE$.

Solution:
Since $OE$ bisects $\angle AOC$:

$\angle AOE$ and $\angle BOE$ form a linear pair (since $A$, $O$, $B$ are on a straight line):

Answer: $\angle BOE = 115°$.

Problem: In the figure, $l \parallel m \parallel n$. A transversal makes $\angle 1 = 60°$ at line $l$. Find $\angle 2$ at line $m$ (co-interior with $\angle 1$) and $\angle 3$ at line $n$ (corresponding to $\angle 2$).

Solution:
$\angle 1 = 60°$ (at line $l$).

Since $l \parallel m$, and $\angle 1$ and $\angle 2$ are co-interior:

Since $m \parallel n$, and $\angle 2$ and $\angle 3$ are corresponding:

Answer: $\angle 2 = 120°$, $\angle 3 = 120°$.

Problem: $AB \parallel CD$. A transversal intersects $AB$ at $P$ and $CD$ at $Q$. The bisectors of $\angle APQ$ and $\angle PQD$ meet at point $R$. If $\angle APQ = 70°$, find $\angle PRQ$.

Solution:
Since $AB \parallel CD$:
$\angle APQ$ and $\angle PQD$ are co-interior angles.

The bisector of $\angle APQ$ makes $\angle RPQ = \frac{70°}{2} = 35°$.

The bisector of $\angle PQD$ makes $\angle RQP = \frac{110°}{2} = 55°$.

In triangle $PQR$:

Answer: $\angle PRQ = 90°$.

Key insight: When the bisectors of co-interior angles meet, they always form a right angle ($90°$). This is a powerful result to remember!