NCERT Solutions for Class 6 Maths Chapter 2: Lines and Angles — Complete Guide

Point: A point is an exact location in space. It has no length, width, or height — it is simply a position. We represent a point with a dot and label it with a capital letter, such as $A$, $B$, or $P$.

Line: A line is a straight path that extends infinitely in both directions. It has no endpoints and no fixed length. We denote a line passing through points $A$ and $B$ as $\overleftrightarrow{AB}$. An important property: through any two distinct points, exactly one line can be drawn.

Ray: A ray has one fixed endpoint (called the initial point or origin) and extends infinitely in one direction. We write a ray starting at $A$ and passing through $B$ as $\overrightarrow{AB}$. Note that $\overrightarrow{AB}$ and $\overrightarrow{BA}$ are different rays — they have different starting points!

Line Segment: A line segment is the part of a line between two endpoints. It has a definite, measurable length. We write the segment with endpoints $A$ and $B$ as $\overline{AB}$. The length of $\overline{AB}$ is denoted $AB$ (without the bar).

Collinear points are points that lie on the same straight line. For example, if $P$, $Q$, $R$ all lie on line $\ell$, they are collinear.

Non-collinear points are points that do NOT all lie on the same straight line. Any three non-collinear points determine a unique triangle.

Key fact: Two points are always collinear (you can always draw a line through two points). It takes at least three points to test for collinearity.

If $Q$ lies between $P$ and $R$ on a line, then:

This is called the segment addition postulate and is fundamental to solving many problems.

An angle is formed when two rays share a common initial point. The common point is called the vertex of the angle, and the two rays are called the arms (or sides) of the angle.

We denote an angle in several ways:
- $\angle ABC$ — the angle at vertex $B$ formed by rays $\overrightarrow{BA}$ and $\overrightarrow{BC}$
- $\angle B$ — if there is only one angle at $B$
- $\angle 1$ or $\angle x$ — using numbers or variables

Interior and Exterior: Every angle divides the plane into two regions. The region between the arms is the interior of the angle. The region outside is the exterior.

Measuring Angles: Angles are measured in degrees (symbol: $^\circ$). A full turn is $360^\circ$. Half a turn is $180^\circ$. A quarter turn is $90^\circ$.

Angles are classified based on their measure:

Zero angle: Exactly $0^\circ$. The two rays overlap completely.

Acute angle: Greater than $0^\circ$ but less than $90^\circ$. Examples: $30^\circ$, $45^\circ$, $60^\circ$, $89^\circ$.

Right angle: Exactly $90^\circ$. The two rays are perpendicular. We mark right angles with a small square at the vertex.

Obtuse angle: Greater than $90^\circ$ but less than $180^\circ$. Examples: $91^\circ$, $120^\circ$, $150^\circ$, $179^\circ$.

Straight angle: Exactly $180^\circ$. The two rays form a straight line.

Reflex angle: Greater than $180^\circ$ but less than $360^\circ$. Examples: $200^\circ$, $270^\circ$, $350^\circ$.

Complete angle (Full turn): Exactly $360^\circ$. The ray makes a complete rotation back to its starting position.

Memory aid: Think of a clock. At $3{:}00$, the hands form a $90^\circ$ angle (right). At $6{:}00$, they form $180^\circ$ (straight). At $12{:}00$, they form $0^\circ$ or $360^\circ$.

Complementary Angles: Two angles whose measures add up to $90^\circ$. Example: $30^\circ$ and $60^\circ$ are complementary because $30 + 60 = 90$.

Supplementary Angles: Two angles whose measures add up to $180^\circ$. Example: $110^\circ$ and $70^\circ$ are supplementary because $110 + 70 = 180$.

Adjacent Angles: Two angles that share a common vertex and a common arm, but do not overlap. The non-common arms are on opposite sides of the common arm.

Linear Pair: Two adjacent angles whose non-common arms form a straight line. A linear pair always adds up to $180^\circ$.

Vertically Opposite Angles: When two lines intersect, they form two pairs of vertically opposite angles. Vertically opposite angles are always equal.

If two lines intersect forming angles $a$, $b$, $c$, $d$ (going clockwise), then:

Problem: How is a line different from a line segment and a ray?

Solution:

A line extends infinitely in both directions. It has no endpoints and cannot be measured. Notation: $\overleftrightarrow{AB}$.

A ray has one fixed endpoint (initial point) and extends infinitely in one direction. It is half of a line. Notation: $\overrightarrow{AB}$ (starts at $A$, passes through $B$).

A line segment has two endpoints and a definite, finite length that can be measured with a ruler. Notation: $\overline{AB}$.

Key difference: A line segment is the only one of the three that has a measurable length. A line and ray both extend to infinity (in two directions and one direction, respectively).

Answer: A line has no endpoints and infinite length; a ray has one endpoint and extends infinitely in one direction; a line segment has two endpoints and a finite measurable length.

Problem: How many line segments can be drawn using $4$ points (no three of which are collinear)?

Solution:

Let the points be $A$, $B$, $C$, $D$.

Step 1: List all possible line segments by choosing every pair of points:

Step 2: Count them. We choose $2$ points out of $4$:

General formula: With $n$ points (no three collinear), the number of line segments is:

Verification:
- $3$ points $\Rightarrow \frac{3 \times 2}{2} = 3$ segments $\checkmark$
- $5$ points $\Rightarrow \frac{5 \times 4}{2} = 10$ segments
- $6$ points $\Rightarrow \frac{6 \times 5}{2} = 15$ segments

Answer: $6$ line segments can be drawn.

Problem: Three points $P$, $Q$, $R$ lie on a straight line with $Q$ between $P$ and $R$. If $PQ = 4$ cm and $QR = 6$ cm, find $PR$.

Solution:

Since $Q$ lies between $P$ and $R$ on the same line, by the segment addition postulate:

Answer: $PR = 10$ cm.

Important note: This only works because $P$, $Q$, $R$ are collinear and $Q$ is between $P$ and $R$. If $Q$ were not between them, the relationship would be different.

Problem: $M$ is the midpoint of $\overline{AB}$. If $AB = 14$ cm, find $AM$ and $MB$.

Solution:

The midpoint of a line segment divides it into two equal parts.

Since $M$ is the midpoint of $\overline{AB}$:

Verification: $AM + MB = 7 + 7 = 14 = AB$ $\checkmark$

Answer: $AM = MB = 7$ cm.

Problem: How many rays can be drawn from a single point? How many rays can be drawn with a given initial point passing through $5$ given points?

Solution:

Part 1: From a single point, infinitely many rays can be drawn — one in every direction. Think of a point as the center of a circle; you can draw a ray toward any point on the circle.

Part 2: If we have a fixed initial point $O$ and $5$ other points $A, B, C, D, E$ (none coinciding with $O$), we can draw exactly $5$ rays:

However, if two of these points are collinear with $O$ and on the same side, those rays coincide and count as one.

Answer: Infinitely many rays can be drawn from a single point; exactly $5$ rays if passing through $5$ distinct non-collinear points.

Problem: How many lines can be drawn through (a) one point? (b) two points? (c) three non-collinear points?

Solution:

(a) Through one point: Infinitely many lines can pass through a single point, each in a different direction.

(b) Through two points: Exactly one line. This is a fundamental axiom of geometry: through any two distinct points, there is exactly one line.

(c) Through three non-collinear points: No single line can pass through all three (that is what "non-collinear" means). But we can draw $3$ lines, each passing through a pair of points: $\overleftrightarrow{AB}$, $\overleftrightarrow{BC}$, $\overleftrightarrow{AC}$.

Answer: (a) Infinitely many (b) Exactly one (c) No line through all three, but $3$ lines through pairs.

Problem: Give real-life examples of (a) parallel lines, (b) intersecting lines, and (c) perpendicular lines.

Solution:

(a) Parallel lines — lines in the same plane that never meet:
- Railway tracks
- Opposite edges of a ruler
- Horizontal lines on ruled paper
- Opposite sides of a rectangular door

(b) Intersecting lines — lines that cross at exactly one point:
- The letter $X$
- Scissors (the two blades)
- Crossroads where two roads meet
- Hands of a clock at most times

(c) Perpendicular lines — lines that intersect at a $90^\circ$ angle:
- The corner of a room (wall meets floor)
- The letter $T$ or $L$
- The horizontal and vertical edges of a window
- A flagpole standing on flat ground

Answer: Examples listed above for each category.

Problem: Points $A$, $B$, $C$, $D$ lie on a line in that order. If $AB = 3$ cm, $BC = 5$ cm, and $CD = 4$ cm, find $AD$, $AC$, and $BD$.

Solution:

Since the points lie in order $A$-$B$-$C$-$D$:

Verification: $AD = AC + CD = 8 + 4 = 12$ $\checkmark$
Also: $AD = AB + BD = 3 + 9 = 12$ $\checkmark$

Answer: $AD = 12$ cm, $AC = 8$ cm, $BD = 9$ cm.

Problem: Name the angle formed by rays $\overrightarrow{BA}$ and $\overrightarrow{BC}$. Identify the vertex and arms.

Solution:

The two rays are $\overrightarrow{BA}$ and $\overrightarrow{BC}$. Both start from point $B$.

Vertex: $B$ (the common initial point)
Arms: $\overrightarrow{BA}$ and $\overrightarrow{BC}$
Angle name: $\angle ABC$ or $\angle CBA$ or simply $\angle B$

Important rule: When naming an angle with three letters, the vertex is ALWAYS the middle letter. So $\angle ABC$ has vertex $B$, not $A$ or $C$.

Answer: The angle is $\angle ABC$ with vertex $B$ and arms $\overrightarrow{BA}$, $\overrightarrow{BC}$.

Problem: Classify each angle as acute, right, obtuse, straight, or reflex:
(a) $45^\circ$ (b) $90^\circ$ (c) $135^\circ$ (d) $180^\circ$ (e) $250^\circ$ (f) $15^\circ$

Solution:

(a) $45^\circ$: Since $0^\circ < 45^\circ < 90^\circ$, this is acute.

(b) $90^\circ$: This is exactly $90^\circ$, so it is a right angle.

(c) $135^\circ$: Since $90^\circ < 135^\circ < 180^\circ$, this is obtuse.

(d) $180^\circ$: This is exactly $180^\circ$, so it is a straight angle.

(e) $250^\circ$: Since $180^\circ < 250^\circ < 360^\circ$, this is a reflex angle.

(f) $15^\circ$: Since $0^\circ < 15^\circ < 90^\circ$, this is acute.

Answer: (a) Acute (b) Right (c) Obtuse (d) Straight (e) Reflex (f) Acute.

Problem: What angle is formed by the hands of a clock at (a) $3{:}00$ (b) $6{:}00$ (c) $12{:}00$ (d) $9{:}00$ (e) $1{:}00$? Classify each.

Solution:

The clock face is a full circle of $360^\circ$. There are $12$ hours, so each hour mark is $\frac{360^\circ}{12} = 30^\circ$ apart.

(a) At $3{:}00$: The hands are $3$ hour-marks apart.

(b) At $6{:}00$: The hands are $6$ hour-marks apart.

(c) At $12{:}00$: Both hands overlap.

(d) At $9{:}00$: The hands are $3$ hour-marks apart (same as $3{:}00$).

(e) At $1{:}00$: The hands are $1$ hour-mark apart.

Answer: (a) $90^\circ$ — right (b) $180^\circ$ — straight (c) $0^\circ$ — zero (d) $90^\circ$ — right (e) $30^\circ$ — acute.

Problem: Four rays $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$, $\overrightarrow{OD}$ emanate from point $O$. How many angles are formed?

Solution:

We need to count angles formed by choosing any two rays from the four.

Number of pairs $= \frac{4 \times 3}{2} = 6$.

The six angles are:

However, each of these can be measured as the smaller angle or the larger (reflex) angle, giving up to $12$ different angle measures. Typically, we count only the $6$ non-reflex angles.

General formula: $n$ rays from a common point form $\frac{n(n-1)}{2}$ angles.

Answer: $6$ angles are formed.

Problem: How many angles can you spot in each of these capital letters: $L$, $T$, $X$, $V$, $Z$?

Solution:

• **$L$:** $1$ angle (a right angle at the corner)
  - **$T$:** $2$ angles (both right angles where the vertical meets the horizontal)
  - **$X$:** $4$ angles (two pairs of vertically opposite angles)
  - **$V$:** $1$ angle (an acute angle at the bottom)
  - **$Z$:** $2$ angles (both acute, at the top-right and bottom-left bends)

Answer: $L$: $1$, $T$: $2$, $X$: $4$, $V$: $1$, $Z$: $2$.

Problem: For $\angle PQR = 60^\circ$, a point $S$ lies inside the angle and a point $T$ lies outside. If $\angle PQS = 25^\circ$, find $\angle SQR$.

Solution:

Since $S$ lies in the interior of $\angle PQR$, ray $\overrightarrow{QS}$ divides $\angle PQR$ into two parts:

Answer: $\angle SQR = 35^\circ$.

Problem: Without using a protractor, arrange these angles from smallest to largest: a right angle, a straight angle, an acute angle of $45^\circ$, an obtuse angle of $120^\circ$.

Solution:

Converting all to degrees:
- Acute angle: $45^\circ$
- Right angle: $90^\circ$
- Obtuse angle: $120^\circ$
- Straight angle: $180^\circ$

From smallest to largest:

Answer: $45^\circ$ (acute) $<$ $90^\circ$ (right) $<$ $120^\circ$ (obtuse) $<$ $180^\circ$ (straight).

Problem: What angle do the hands of a clock make at $4{:}30$?

Solution:

At $4{:}30$:
- The minute hand points at $6$ (i.e., at $180^\circ$ from $12$).
- The hour hand has moved past $4$. At $4{:}00$ it was at $4 \times 30^\circ = 120^\circ$. In $30$ minutes, it moves an additional $\frac{30}{60} \times 30^\circ = 15^\circ$. So the hour hand is at $120^\circ + 15^\circ = 135^\circ$.

Angle between the hands:

Answer: The angle at $4{:}30$ is $45^\circ$ (acute).

Problem: Describe the step-by-step process of measuring an angle using a protractor.

Solution:

Step 1: Place the protractor so that its centre point (the small hole or mark at the base) is exactly on the vertex of the angle.

Step 2: Align the baseline (the $0^\circ$-$180^\circ$ line) of the protractor along one arm of the angle.

Step 3: Read the scale where the other arm of the angle crosses the protractor. Use the scale that starts from $0^\circ$ on the arm you aligned.

Choosing the correct scale: If the arm you aligned is on the right side, use the inner scale (which reads $0^\circ$ on the right). If aligned on the left side, use the outer scale (which reads $0^\circ$ on the left).

Quick check: If the angle looks acute, the reading should be less than $90^\circ$. If it looks obtuse, the reading should be more than $90^\circ$. If your reading contradicts what the angle looks like, you are using the wrong scale.

Answer: Place centre on vertex, align baseline with one arm, read the correct scale where the other arm crosses.

Problem: An angle measures $72^\circ$ on the protractor. Classify it and find its complement and supplement.

Solution:

Classification: Since $0^\circ < 72^\circ < 90^\circ$, this is an acute angle.

Complement: The complement of an angle $\theta$ is $90^\circ - \theta$.

Supplement: The supplement of an angle $\theta$ is $180^\circ - \theta$.

Verification: $72^\circ + 18^\circ = 90^\circ$ $\checkmark$ and $72^\circ + 108^\circ = 180^\circ$ $\checkmark$.

Answer: Acute angle; complement $= 18^\circ$; supplement $= 108^\circ$.

Problem: Draw an angle of $55^\circ$ using a protractor.

Solution:

Step 1: Draw a ray $\overrightarrow{OA}$ (this will be one arm of the angle).

Step 2: Place the protractor with its centre on $O$ and baseline along $\overrightarrow{OA}$.

Step 3: Starting from $0^\circ$ on the scale aligned with $\overrightarrow{OA}$, count to $55^\circ$ and mark a point $B$.

Step 4: Remove the protractor and draw ray $\overrightarrow{OB}$.

Step 5: Label the angle: $\angle AOB = 55^\circ$.

Verification: Re-measure the angle by placing the protractor on it again. It should read $55^\circ$.

Answer: Constructed $\angle AOB = 55^\circ$.

Problem: Two angles on a straight line are $\angle AOB = 130^\circ$ and $\angle BOC$. Find $\angle BOC$.

Solution:

Angles on a straight line add up to $180^\circ$ (they form a linear pair):

Answer: $\angle BOC = 50^\circ$.

Problem: Three rays $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$ are on the same side of line $\overleftrightarrow{PQ}$, passing through $O$. If $\angle POA = 40^\circ$, $\angle AOB = 55^\circ$, find $\angle BOQ$.

Solution:

Since $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ form a straight line:

Answer: $\angle BOQ = 85^\circ$.

Problem: How would you measure a reflex angle using a protractor that only goes up to $180^\circ$?

Solution:

Method: Measure the non-reflex angle (the smaller angle between the two arms) and subtract from $360^\circ$.

For example, if the non-reflex angle measures $110^\circ$, then the reflex angle is:

Why this works: The reflex angle and the non-reflex angle together make a complete turn of $360^\circ$.

Example: If the non-reflex angle between clock hands at $10{:}00$ is:

Then the reflex angle is:

Answer: Measure the non-reflex angle and subtract from $360^\circ$.

Problem: Draw a right angle and bisect it. What type of angles do you get?

Solution:

Step 1: Draw $\angle AOB = 90^\circ$ using a protractor.

Step 2: To bisect means to divide into two equal parts. The bisector ray $\overrightarrow{OC}$ splits $\angle AOB$ into:

Step 3: Classify: $45^\circ$ is an acute angle.

Answer: Bisecting a right angle gives two $45^\circ$ acute angles.

Problem: Lines $AB$ and $CD$ intersect at $O$. If $\angle AOC = 90^\circ$, find all four angles and state whether the lines are perpendicular.

Solution:

When two lines intersect, they form two pairs of vertically opposite angles.

$\angle AOC = 90^\circ$ (given)

$\angle BOD = 90^\circ$ (vertically opposite to $\angle AOC$)

$\angle AOD = 180^\circ - 90^\circ = 90^\circ$ (linear pair with $\angle AOC$)

$\angle BOC = 180^\circ - 90^\circ = 90^\circ$ (linear pair with $\angle AOC$)

All four angles are $90^\circ$!

Since the lines intersect at right angles, $AB \perp CD$ (the lines are perpendicular).

Answer: All four angles are $90^\circ$. Yes, $AB$ and $CD$ are perpendicular.

Problem: Find the complement of each angle: (a) $35^\circ$ (b) $60^\circ$ (c) $48^\circ$ (d) $0^\circ$ (e) $90^\circ$

Solution:

The complement of $\theta$ is $90^\circ - \theta$.

(a) $90^\circ - 35^\circ = 55^\circ$

(b) $90^\circ - 60^\circ = 30^\circ$

(c) $90^\circ - 48^\circ = 42^\circ$

(d) $90^\circ - 0^\circ = 90^\circ$

(e) $90^\circ - 90^\circ = 0^\circ$

Note: Only acute angles (less than $90^\circ$) have complements that are positive. An obtuse angle cannot have a complement (you would get a negative value, which does not make sense for angles).

Answer: (a) $55^\circ$ (b) $30^\circ$ (c) $42^\circ$ (d) $90^\circ$ (e) $0^\circ$.

Problem: Find the supplement of each angle: (a) $70^\circ$ (b) $120^\circ$ (c) $90^\circ$ (d) $180^\circ$ (e) $5^\circ$

Solution:

The supplement of $\theta$ is $180^\circ - \theta$.

(a) $180^\circ - 70^\circ = 110^\circ$

(b) $180^\circ - 120^\circ = 60^\circ$

(c) $180^\circ - 90^\circ = 90^\circ$

(d) $180^\circ - 180^\circ = 0^\circ$

(e) $180^\circ - 5^\circ = 175^\circ$

Note: A right angle ($90^\circ$) is its own supplement! Any angle from $0^\circ$ to $180^\circ$ has a valid supplement.

Answer: (a) $110^\circ$ (b) $60^\circ$ (c) $90^\circ$ (d) $0^\circ$ (e) $175^\circ$.

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

Solution:

Let the two angles be $2x$ and $3x$.

Since they are complementary:

The angles are:

Verification: $36^\circ + 54^\circ = 90^\circ$ $\checkmark$

Answer: The two angles are $36^\circ$ and $54^\circ$.

Problem: Two supplementary angles differ by $40^\circ$. Find both angles.

Solution:

Let the smaller angle be $x$. The larger angle is $x + 40^\circ$.

Since they are supplementary:

The angles are $70^\circ$ and $70^\circ + 40^\circ = 110^\circ$.

Verification: $70^\circ + 110^\circ = 180^\circ$ $\checkmark$ and $110^\circ - 70^\circ = 40^\circ$ $\checkmark$.

Answer: The two angles are $70^\circ$ and $110^\circ$.

Problem: Find the complement of the complement of $35^\circ$.

Solution:

Step 1: Complement of $35^\circ$ is $90^\circ - 35^\circ = 55^\circ$.

Step 2: Complement of $55^\circ$ is $90^\circ - 55^\circ = 35^\circ$.

So the complement of the complement of an angle is the angle itself!

Proof (general): Let the angle be $\theta$.
Complement $= 90^\circ - \theta$.
Complement of complement $= 90^\circ - (90^\circ - \theta) = \theta$.

Answer: $35^\circ$. In general, the complement of the complement of any angle equals the angle itself.

Problem: Find the supplement of the complement of $40^\circ$.

Solution:

Step 1: Complement of $40^\circ = 90^\circ - 40^\circ = 50^\circ$.

Step 2: Supplement of $50^\circ = 180^\circ - 50^\circ = 130^\circ$.

General formula: Supplement of complement of $\theta = 180^\circ - (90^\circ - \theta) = 90^\circ + \theta$.

Check: $90^\circ + 40^\circ = 130^\circ$ $\checkmark$.

Answer: $130^\circ$.

Problem: (a) What angle is equal to its complement? (b) What angle is equal to its supplement?

Solution:

(a) Let the angle be $x$. Its complement is $90^\circ - x$.

A $45^\circ$ angle is equal to its own complement.

(b) Let the angle be $x$. Its supplement is $180^\circ - x$.

A $90^\circ$ angle is equal to its own supplement.

Answer: (a) $45^\circ$ (b) $90^\circ$.

Problem: Two lines intersect at a point $O$. If one of the angles formed is $65^\circ$, find all four angles.

Solution:

Let the four angles be $a$, $b$, $c$, $d$ going clockwise, where $a = 65^\circ$.

Vertically opposite angles are equal:

Adjacent angles form a linear pair ($180^\circ$):

So the four angles are: $65^\circ, 115^\circ, 65^\circ, 115^\circ$.

Verification: $65 + 115 + 65 + 115 = 360^\circ$ $\checkmark$ (angles at a point)

Answer: The four angles are $65^\circ$, $115^\circ$, $65^\circ$, $115^\circ$.

Problem: Three angles at a point are $x$, $2x$, and $3x$. Find the value of $x$ and each angle.

Solution:

Angles at a point add up to $360^\circ$:

The three angles are:

Verification: $60 + 120 + 180 = 360^\circ$ $\checkmark$.

Answer: $x = 60^\circ$; the angles are $60^\circ$, $120^\circ$, and $180^\circ$.

Problem: Two lines $AB$ and $CD$ intersect at $O$ such that $\angle AOC = 3x + 10^\circ$ and $\angle BOD = 5x - 20^\circ$. Find $x$ and all four angles.

Solution:

$\angle AOC$ and $\angle BOD$ are vertically opposite angles, so they are equal:

$\angle AOC = 3(15) + 10 = 55^\circ$
$\angle BOD = 5(15) - 20 = 55^\circ$ $\checkmark$

$\angle AOD = 180^\circ - 55^\circ = 125^\circ$ (linear pair)
$\angle BOC = 125^\circ$ (vertically opposite to $\angle AOD$)

Verification: $55 + 125 + 55 + 125 = 360^\circ$ $\checkmark$.

Answer: $x = 15$; angles are $55^\circ$, $125^\circ$, $55^\circ$, $125^\circ$.

Problem: Find the value of $x$: angles of $x$, $2x$, and $3x$ lie on a straight line.

Solution:

Angles on a straight line sum to $180^\circ$:

The three angles are: $30^\circ$, $60^\circ$, $90^\circ$.

Verification: $30 + 60 + 90 = 180^\circ$ $\checkmark$.

Answer: $x = 30^\circ$.

Problem: Five angles around a point are $80^\circ$, $55^\circ$, $95^\circ$, $x$, and $60^\circ$. Find $x$.

Solution:

Angles around a point sum to $360^\circ$:

Verification: $80 + 55 + 95 + 70 + 60 = 360^\circ$ $\checkmark$.

Answer: $x = 70^\circ$.

Problem: Ray $\overrightarrow{OC}$ is between rays $\overrightarrow{OA}$ and $\overrightarrow{OB}$, where $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are opposite rays. If $\angle AOC = (3x + 7)^\circ$ and $\angle COB = (2x - 2)^\circ$, find $x$ and both angles.

Solution:

Since $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are opposite rays, $AOB$ is a straight line:

$\angle AOC = 3(35) + 7 = 112^\circ$
$\angle COB = 2(35) - 2 = 68^\circ$

Verification: $112 + 68 = 180^\circ$ $\checkmark$.

Answer: $x = 35$; $\angle AOC = 112^\circ$ and $\angle COB = 68^\circ$.

Problem: The angles of a triangle are $50^\circ$, $60^\circ$, and $x$. Find $x$.

Solution:

The sum of angles in a triangle is $180^\circ$:

The triangle has angles $50^\circ$, $60^\circ$, $70^\circ$. Since all angles are acute (less than $90^\circ$), this is an acute triangle.

Answer: $x = 70^\circ$.

Problem: The angles of a triangle are in the ratio $1 : 2 : 3$. Find all three angles. What type of triangle is this?

Solution:

Let the angles be $x$, $2x$, $3x$.