CBSE Class 9 Mathematics Lines And Angles Worksheet Set 06

CBSE Class 9 Mathematics Worksheet Lines and Angles (6)

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Lines and Angles

<1M>

1.If two supplementary angles are in the ratio 2:7, then the angles are

(A) 40°, 140°

(B) 85°, 95°

(C) 40°, 50°

(D) 60°, 120°.

2.Supplementary angle of 103.5° is

(A) 70.5°

(B) 76.5°

(C) 70°

(D) 72.5°

3.Measure of an obtuse angle is

(A) > 0°, < 90°

(B) > 90º < 180°

(C) > 0°, < 270°

(D) > 0°, < 180°

4.In the figure, AD is the bisector of Ain ABC then

(A) AB > BD

(B) AC < AB

(C) BC = AD

(D) None.

5.If two lines are parallel to same line then these line will be _______ to each other.

6.

If two parallel lines are intersected by a transversal, then bisectors of any two corresponding angles

___________

7.If a ray stands on a line, and then the sum of the adjacent angles so formed is ______

8.Two adjacent angles are said to form a linear pair of angles, if their non-common arms are

____________

9.The measure of an angle is twice the measure of its supplimentary angle. Find its measure.

10.Two supplementary angles are in the ratio 4:5. Find the angles.

11.Find out the two pairs of adjacent angles

Please click the below link to access CBSE Class 9 Mathematics Worksheet Lines and Angles (6)

Lines and Angles

<1M>

Question 1. If two supplementary angles are in the ratio 2:7, then the angles are
(a) 40°, 140°
(b) 85°, 95°
(c) 40°, 50°
(d) 60°, 120°
Answer: (a) 40°, 140°
Let the two supplementary angles be represented by \( 2x \) and \( 7x \). Because supplementary angles sum to \( 180^\circ \):
\( 2x + 7x = 180^\circ \)
\( \implies 9x = 180^\circ \)
\( \implies x = 20^\circ \)
The two angles are:
\( 2x = 2 \times 20^\circ = 40^\circ \)
\( 7x = 7 \times 20^\circ = 140^\circ \)
In simple words: Add the ratio parts 2 and 7 to get 9 parts. Dividing 180 by 9 gives 20 degrees per part, so the angles are 40 degrees and 140 degrees.

Exam Tip: Supplementary angles always add up to \( 180^\circ \), whereas complementary angles add up to \( 90^\circ \); never mix up the two sums.

Question 2. Supplementary angle of 103.5° is
(a) 70.5°
(b) 76.5°
(c) 70°
(d) 72.5°
Answer: (b) 76.5°
The supplement of any angle is obtained by subtracting its value from \( 180^\circ \):
\( \text{Supplement} = 180^\circ - 103.5^\circ = 76.5^\circ \)
In simple words: Subtract 103.5 from 180 to find the missing angle, which is 76.5 degrees.

Exam Tip: Align the decimal point carefully when performing subtraction from 180.0 to prevent arithmetic slips.

Question 3. Measure of an obtuse angle is
(a) > 0°, < 90°
(b) > 90º < 180°
(c) > 0°, < 270°
(d) > 0°, < 180°
Answer: (b) > 90º < 180°
By definition, an obtuse angle has a measure strictly greater than one right angle (\( 90^\circ \)) and strictly less than two right angles (\( 180^\circ \)).
In simple words: An obtuse angle is wider than a right angle of 90 degrees but narrower than a straight line of 180 degrees.

Exam Tip: Keep the angle classifications clear: acute is between \( 0^\circ \) and \( 90^\circ \), right is exactly \( 90^\circ \), obtuse is between \( 90^\circ \) and \( 180^\circ \), and reflex is between \( 180^\circ \) and \( 360^\circ \).

Question 4. In the figure, AD is the bisector of ∠A in ΔABC then
(a) AB > BD
(b) AC < AB
(c) BC = AD
(d) None of the options
Answer: (a) AB > BD
In \( \Delta ADC \), side \( CD \) is extended to \( B \). The exterior angle \( \angle ADB \) equals the sum of the two interior opposite angles:
\( \angle ADB = \angle CAD + \angle C \)
Since \( AD \) bisects \( \angle BAC \), we know \( \angle CAD = \angle BAD \):
\( \angle ADB = \angle BAD + \angle C \)
\( \implies \angle ADB > \angle BAD \)
In \( \Delta ABD \), the side opposite to the larger angle is strictly greater than the side opposite to the smaller angle. Therefore, \( AB > BD \).
In simple words: The exterior angle at D is bigger than half of angle A. In a triangle, a bigger angle always opens up opposite a longer side, so AB is longer than BD.

Exam Tip: Quote the triangle inequality theorem: in any triangle, the side opposite the greater angle is always longer.

Question 5. If two lines are parallel to same line then these line will be _______ to each other.
Answer: parallel
According to the parallel lines theorem (transitivity of parallel lines), two or more lines in a plane that are each parallel to a common third line are parallel to each other.
In simple words: If two straight paths never touch a third path, they also point the same way and will never touch each other.

Exam Tip: State the geometric theorem: lines which are parallel to the same line are parallel to each other.

Question 6. If two parallel lines are intersected by a transversal, then bisectors of any two corresponding angles ___________
Answer: are parallel
When a transversal cuts two parallel lines, the pairs of corresponding angles are equal. The rays bisecting these equal angles make identical angles with the transversal line, so by the converse of the corresponding angles axiom, the two bisector rays are parallel.
In simple words: Halves of equal corresponding angles are still equal, so the bisector lines run in the same direction and stay parallel.

Exam Tip: Full marks require stating that halves of equal angles are equal, which proves the bisectors form equal corresponding angles.

Question 7. If a ray stands on a line, and then the sum of the adjacent angles so formed is ______
Answer: 180°
By the linear pair axiom, whenever a ray originates from and stands upon a straight line, the measures of the two resulting adjacent angles add up to \( 180^\circ \).
In simple words: A ray standing on a flat line cuts the straight angle of 180 degrees into two parts that together still total 180 degrees.

Exam Tip: Always cite the "Linear Pair Axiom" by name when using the property that angles on a straight line sum to \( 180^\circ \).

Question 8. Two adjacent angles are said to form a linear pair of angles, if their non-common arms are ____________
Answer: two opposite rays
Adjacent angles create a linear pair if and only if their non-common sides point in directly opposite directions, forming a single unbroken straight line.
In simple words: A linear pair happens when the outside arms of two touching angles form one straight line.

Exam Tip: The exact terminology expected by examiners is "two opposite rays".

Question 9. The measure of an angle is twice the measure of its supplimentary angle. Find its measure.
Answer: Let the angle be \( x \). Its supplementary angle is \( 180^\circ - x \).
According to the condition:
\( x = 2(180^\circ - x) \)
\( x = 360^\circ - 2x \)
\( \implies 3x = 360^\circ \)
\( \implies x = 120^\circ \)
Hence, the required angle measures \( 120^\circ \).
In simple words: The angle is double its supplement, so dividing 180 degrees into 3 equal shares of 60 gives the angle two shares, which is 120 degrees.

Exam Tip: Write down the relation \( x + \text{supplement} = 180^\circ \) clearly before setting up the equation.

Question 10. Two supplementary angles are in the ratio 4:5. Find the angles.
Answer: Let the two supplementary angles be \( 4x \) and \( 5x \).
Since supplementary angles sum to \( 180^\circ \):
\( 4x + 5x = 180^\circ \)
\( \implies 9x = 180^\circ \)
\( \implies x = 20^\circ \)
Therefore, the two angles are:
\( 4x = 4 \times 20^\circ = 80^\circ \)
\( 5x = 5 \times 20^\circ = 100^\circ \)
In simple words: The total ratio parts equal 9. Each part is 180 divided by 9, or 20 degrees, making the two angles 80 degrees and 100 degrees.

Exam Tip: Verify your final answer by checking that the sum of the calculated values equals \( 180^\circ \) (\( 80^\circ + 100^\circ = 180^\circ \)).

Question 11. Find out the two pairs of adjacent angles
Answer: Two angles are adjacent if they share a common vertex, share a common arm, and have their non-common arms situated on opposite sides of the common arm.
From the given figure with intersecting lines \( AB \) and \( CD \) meeting at point \( O \), two pairs of adjacent angles are:
1. \( \angle AOD \) and \( \angle DOB \) (sharing vertex \( O \) and common arm \( OD \))
2. \( \angle DOB \) and \( \angle BOC \) (sharing vertex \( O \) and common arm \( OB \))
In simple words: Any two side-by-side angles that share the center point and the ray between them form a pair of adjacent angles.

Exam Tip: Specify the common vertex and the common arm for each pair to get full marks on definition-based questions.

Question 12. In Fig, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 2 : 3 , find angle POR and angle ROQ (1 Marks)
Answer: Because \( POQ \) is a straight line, ray \( OR \) stands upon it and forms a linear pair:
\( \angle POR + \angle ROQ = 180^\circ \)
Let \( \angle POR = 2x \) and \( \angle ROQ = 3x \):
\( 2x + 3x = 180^\circ \)
\( \implies 5x = 180^\circ \)
\( \implies x = 36^\circ \)
Now calculate both angles:
\( \angle POR = 2 \times 36^\circ = 72^\circ \)
\( \angle ROQ = 3 \times 36^\circ = 108^\circ \)
In simple words: The two angles lie on the straight line PQ and total 180 degrees. Dividing 180 into 5 parts gives 36 degrees each, so the angles are 72 degrees and 108 degrees.

Exam Tip: State the linear pair relation explicitly before introducing the common ratio variable \( x \).

Question 13. From the adjoining figure, the value of y is
(a) 35°
(b) 37°
(c) 39°
(d) 10°
Answer: (d) 10°
In the figure, three straight lines \( AD \), \( BE \), and \( CF \) intersect at point \( O \). The vertically opposite angle to \( \angle AOB = 3y^\circ \) is \( \angle DOE = 3y^\circ \).
Because line \( CD \) is a straight line, the adjacent angles situated on one side of line \( CD \) form a straight angle summing to \( 180^\circ \):
\( \angle FOD + \angle FOE + \angle EOB + \angle BOD = 180^\circ \)
Summing the angles on the straight line \( BOE \):
\( \angle AOB + \angle AOF + \angle FOE = 180^\circ \)
Using the given angles \( 3y^\circ \), \( 7y^\circ \), and \( 8y^\circ \) which lie on one half-plane of the line:
\( 3y^\circ + 7y^\circ + 8y^\circ = 180^\circ \)
\( \implies 18y^\circ = 180^\circ \)
\( \implies y = 10^\circ \)
In simple words: The three angles add together to cover a flat straight line of 180 degrees. 18y equals 180, which means y is 10 degrees.

Exam Tip: Use vertically opposite angles to transfer angles onto one continuous straight line before setting their sum to \( 180^\circ \).

Question 14. The interior and boundary of a triangle is called
(a) exterior.
(b) interior.
(c) triangular region.
(d) plane.
Answer: (c) triangular region.
The geometric combination of the planar area bounded inside a triangle together with the perimeter line of the triangle itself is formally called a triangular region.
In simple words: The inside space plus the border lines of a triangle make up what is known as the triangular region.

Exam Tip: Remember the distinction: the "interior" excludes the boundary, whereas the "triangular region" includes both the interior and the boundary.

Question 15. Which of the following pair is complementary?
(a) 37°, 45°
(b) 38°,54°
(c) 55°, 35°
(d) 74°, 25°
Answer: (c) 55°, 35°
Two angles are complementary if their measures add up to precisely \( 90^\circ \). Checking the choices:
(a) \( 37^\circ + 45^\circ = 82^\circ \neq 90^\circ \)
(b) \( 38^\circ + 54^\circ = 92^\circ \neq 90^\circ \)
(c) \( 55^\circ + 35^\circ = 90^\circ \)
(d) \( 74^\circ + 25^\circ = 99^\circ \neq 90^\circ \)
In simple words: Add each pair of angles. Only 55 and 35 add up to exactly 90 degrees, making them complementary.

Exam Tip: Quickly sum the units digits first: \( 5 + 5 = 10 \) immediately highlights option (c) as the candidate for 90°.

Question 16. The value of internal and the external bisectors of linear pair angle is
(a) 90°
(b) 45°
(c) 360°
(d) 270°
Answer: (a) 90°
Let two angles forming a linear pair be \( \angle 1 \) and \( \angle 2 \), such that \( \angle 1 + \angle 2 = 180^\circ \).
The angle between their angle bisectors is:
\( \frac{1}{2}\angle 1 + \frac{1}{2}\angle 2 = \frac{1}{2}(\angle 1 + \angle 2) = \frac{1}{2} \times 180^\circ = 90^\circ \)
In simple words: Since the two big angles make 180 degrees, taking half of each angle gives half of 180, which is 90 degrees.

Exam Tip: The angle formed by the bisectors of a linear pair is always a right angle (\( 90^\circ \)); this is a standard textbook property.

Question 17. Measure of ∠y is
(a) 70°
(b) 65°
(c) 90°
(d) 55°
Answer: (d) 55°
By the exterior angle theorem, an exterior angle of a triangle equals the sum of its two interior opposite angles:
\( \angle A + \angle B = \text{Exterior } \angle C \)
\( 60^\circ + y^\circ = 115^\circ \)
\( \implies y = 115^\circ - 60^\circ = 55^\circ \)
In simple words: The outer angle of 115 degrees equals the sum of the two opposite inside angles. Subtracting 60 from 115 gives 55 degrees for y.

Exam Tip: State the Exterior Angle Theorem explicitly: \( \text{Exterior angle} = \text{Sum of opposite interior angles} \).

Question 18. From the adjoining figure, if ∠1 = 55° and ∠6 = 60° ,then the lines m and n are
(a) parallel
(b) not parallel
(c) can't say
(d) Perpendicular.
Answer: (b) not parallel
In the figure, angles \( \angle 1 \) and \( \angle 2 \) form a linear pair on line \( m \):
\( \angle 1 + \angle 2 = 180^\circ \)
\( \implies \angle 2 = 180^\circ - 55^\circ = 125^\circ \)
If lines \( m \) and \( n \) were parallel, the corresponding angles \( \angle 2 \) and \( \angle 6 \) would have to be equal.
Since \( \angle 2 = 125^\circ \) and \( \angle 6 = 60^\circ \), \( \angle 2 \neq \angle 6 \).
Therefore, lines \( m \) and \( n \) are not parallel.
In simple words: If the lines were parallel, matching angles would be identical. Here they do not match, so the lines are not parallel.

Exam Tip: To show lines are not parallel, calculate either a pair of corresponding angles or alternate angles and show that they are unequal.

Question 19. Sum of all angles round a point is equal to ______
Answer: 360°
The total angular measure formed by all consecutive adjacent angles surrounding a single point in a plane completes a full rotation, which equals \( 360^\circ \) (four right angles).
In simple words: Going all the way around a point makes a full circle of 360 degrees.

Exam Tip: You may write either \( 360^\circ \) or "four right angles"; both forms are accepted in marking schemes.

Question 20. In the adjoining figure AB || CD, ∠1 : ∠2 = 3 : 2 Then ∠6 is
(a) 72º
(b) 35º
(c) 45º
(d) 190º
Answer: (a) 72º
Angles \( \angle 1 \) and \( \angle 2 \) form a linear pair on line \( AB \):
\( \angle 1 + \angle 2 = 180^\circ \)
Let \( \angle 1 = 3x \) and \( \angle 2 = 2x \):
\( 3x + 2x = 180^\circ \)
\( \implies 5x = 180^\circ \)
\( \implies x = 36^\circ \)
Thus, \( \angle 2 = 2 \times 36^\circ = 72^\circ \).
Because \( AB \parallel CD \), corresponding angles are equal:
\( \implies \angle 6 = \angle 2 = 72^\circ \)
In simple words: The ratio parts add to 5, so each part is 36 degrees. Angle 2 is 72 degrees, and angle 6 matches angle 2 because the lines are parallel.

Exam Tip: First find the interior or corresponding angle on the top parallel line using the linear pair, then transfer it directly to the bottom line.

Question 21. All linear pairs are
(a) supplementary
(b) complementary
(c) right angles
(d) None of the options
Answer: (a) supplementary
By the linear pair axiom, the two angles of every linear pair lie on a straight line and their sum is always \( 180^\circ \), which means they are invariably supplementary.
In simple words: Every linear pair forms a straight line of 180 degrees, which is the definition of supplementary angles.

Exam Tip: Remember: all linear pairs are supplementary, but not all supplementary angles are linear pairs (they only form a linear pair if they are adjacent).

Question 22. If one of the linear pair is acute, then the measure of the other angle is
(a) supplementary
(b) obtuse
(c) complementary
(d) None of the options
Answer: (b) obtuse
Let the two angles of the linear pair be \( \angle A \) and \( \angle B \), where \( \angle A + \angle B = 180^\circ \).
If \( \angle A \) is acute (\( \angle A < 90^\circ \)), then:
\( \angle B = 180^\circ - \angle A > 180^\circ - 90^\circ = 90^\circ \)
Since \( \angle B > 90^\circ \) and \( \angle B < 180^\circ \), \( \angle B \) must be an obtuse angle.
In simple words: If one angle is less than 90 degrees, the other must be greater than 90 degrees to add up to 180, making it obtuse.

Exam Tip: If one angle of a linear pair is acute, the other is obtuse; if one is a right angle, the other is also a right angle.

Question 23. The common end point of an angle is called
(a) vertex.
(b) zero.
(c) end point.
(d) All of the options
Answer: (a) vertex.
An angle is formed by two rays that share the same origin point. This shared initial point is known as the vertex of the angle.
In simple words: The corner where the two arms of an angle meet is called the vertex.

Exam Tip: "Vertex" is the specific geometric term for the common origin point of the rays forming an angle.

Question 24. An exterior angle of a triangle is 110° and one of the interior of opposite angles is 60°. Then the other two angles of a triangle are
(a) 70°, 50°
(b) 70°, 40°
(c) 110°, 40°
(d) 110°, 75°
Answer: (a) 70°, 50°
By the exterior angle theorem, the exterior angle equals the sum of its two interior opposite angles:
\( \text{Exterior angle} = \text{Interior opposite}_1 + \text{Interior opposite}_2 \)
\( 110^\circ = 60^\circ + \text{Interior opposite}_2 \)
\( \implies \text{Interior opposite}_2 = 110^\circ - 60^\circ = 50^\circ \)
The adjacent interior angle forms a linear pair with the exterior angle:
\( 180^\circ - 110^\circ = 70^\circ \)
Therefore, the remaining two angles of the triangle are \( 70^\circ \) and \( 50^\circ \).
In simple words: Subtract 60 from 110 to get the other opposite inside angle (50 degrees), and subtract 110 from 180 to find the third angle (70 degrees).

Exam Tip: Verify that the three interior angles of the triangle add up to \( 180^\circ \): \( 60^\circ + 50^\circ + 70^\circ = 180^\circ \).

Question 25. From the adjoining figure the value of x is
(a) 75°
(b) 90°
(c) 85°
(d) None of the options
Answer: (a) 75°
In the figure, parallel lines cut by a transversal form consecutive interior angles that are supplementary:
\( 105^\circ + x^\circ = 180^\circ \)
\( \implies x = 180^\circ - 105^\circ = 75^\circ \)
In simple words: The two inside angles on the same side of the parallel lines total 180 degrees. 180 minus 105 leaves 75 degrees.

Exam Tip: Identify co-interior angles quickly: when lines are parallel, interior angles on the same side of the transversal always sum to \( 180^\circ \).

Question 26. In the adjoining figure, it is given that AB||CD. ∠AOC = _____
(a) 120°
(b) 72°
(c) 112°
(d) 150°
Answer: (c) 112°
Construct a line through point \( O \) parallel to both \( AB \) and \( CD \), extending towards the left as ray \( OE \).
Since \( OE \parallel AB \), the consecutive interior angles on the same side of transversal \( AO \) sum to \( 180^\circ \):
\( \angle AOE = 180^\circ - 118^\circ = 62^\circ \)
Similarly, since \( OE \parallel CD \), the consecutive interior angles with transversal \( CO \) sum to \( 180^\circ \):
\( \angle COE = 180^\circ - 130^\circ = 50^\circ \)
Combining these adjacent angles gives:
\( \angle AOC = \angle AOE + \angle COE = 62^\circ + 50^\circ = 112^\circ \)
In simple words: Draw a parallel line through the bend point O. The top half is 180 minus 118 (62 degrees), and the bottom half is 180 minus 130 (50 degrees). Adding them gives 112 degrees.

Exam Tip: Drawing an auxiliary parallel line through the reflex or interior vertex point is the standard method for solving zigzag parallel line questions.

Question 27. From the adjoining figure the value of x is
(a) 106°
(b) 180°
(c) 120°
(d) None of the options
Answer: (c) 120°
By the exterior angle property of the non-convex quadrilateral (arrowhead) or by extending the line segment to create two triangles:
The reflex/interior angle at vertex \( E \) equals the sum of the three interior angles:
\( x = 35^\circ + 25^\circ + 60^\circ = 120^\circ \)
In simple words: In an arrowhead shape, the reflex angle at the inner point equals the sum of the three pointed tip angles: 35 plus 25 plus 60 gives 120 degrees.

Exam Tip: Remember the boomerang/arrowhead theorem: the interior angle at the dent equals the sum of the three pointed vertex angles.

Question 28. If B lies between A and C where AC = 17 cm and BC = 9 cm, then AB² is
(a) 306
(b) 144
(c) 64
(d) 24
Answer: (c) 64
Since point \( B \) lies collinear between points \( A \) and \( C \):
\( AB + BC = AC \)
\( AB + 9 = 17 \)
\( \implies AB = 17 - 9 = 8\text{ cm} \)
Now compute the square of segment \( AB \):
\( AB^2 = 8^2 = 64 \)
In simple words: Subtract 9 from 17 to find the length of AB, which is 8 cm. Squaring 8 gives 64.

Exam Tip: Carefully read whether the question asks for the segment length \( AB \) or its square \( AB^2 \).

Question 29. The difference of two complimentory angles is 40°. Then the angles are
(a) 65°, 25°
(b) 70°, 30°
(c) 70°, 45°
(d) 60°, 30°
Answer: (a) 65°, 25°
Let the two complementary angles be denoted by \( x \) and \( y \).
Since they are complementary:
\( x + y = 90^\circ \) ...(1)
Their given difference is:
\( x - y = 40^\circ \) ...(2)
Adding equations (1) and (2):
\( 2x = 130^\circ \implies x = 65^\circ \)
Substituting \( x = 65^\circ \) into equation (1):
\( 65^\circ + y = 90^\circ \implies y = 25^\circ \)
Thus, the two angles are \( 65^\circ \) and \( 25^\circ \).
In simple words: The two angles add to 90 and differ by 40. Half of 130 is 65 degrees, and 90 minus 65 is 25 degrees.

Exam Tip: Always double check that the two angles satisfy both conditions: \( 65^\circ + 25^\circ = 90^\circ \) and \( 65^\circ - 25^\circ = 40^\circ \).

Question 30. In the given figure the corresponding angles are
(a) ∠1 & ∠5
(b) ∠2 & ∠6
(c) ∠3 & ∠7
(d) All of the options
Answer: (d) All of the options
When two lines are intersected by a transversal, corresponding angles occupy matching relative positions at each intersection:
- \( \angle 1 \) and \( \angle 5 \) are both top-left angles.
- \( \angle 2 \) and \( \angle 6 \) are both top-right angles.
- \( \angle 3 \) and \( \angle 7 \) are both bottom-left angles.
Since all of the listed pairs are pairs of corresponding angles, option (d) is correct.
In simple words: Each of the given pairs sits in the exact same matching corner at the two intersections, so all of them are corresponding angles.

Exam Tip: Corresponding angles sit in identical relative positions: top-left with top-left, top-right with top-right, bottom-left with bottom-left, and bottom-right with bottom-right.

Question 31. In the adjoining figure BO, CO are angle bisectors of external angles of ΔABC. Then ∠BOC is
(a) \( 90^\circ - \frac{1}{2}\angle A \)
(b) \( 90^\circ + \frac{1}{2}\angle A \)
(c) \( 180^\circ - \frac{1}{2}\angle A \)
(d) None of the options
Answer: (a) \( 90^\circ - \frac{1}{2}\angle A \)
The exterior angles at vertices \( B \) and \( C \) are \( 180^\circ - \angle B \) and \( 180^\circ - \angle C \).
Their bisectors give:
\( \angle OBC = \frac{1}{2}(180^\circ - \angle B) = 90^\circ - \frac{1}{2}\angle B \)
\( \angle OCB = \frac{1}{2}(180^\circ - \angle C) = 90^\circ - \frac{1}{2}\angle C \)
In \( \Delta BOC \):
\( \angle BOC = 180^\circ - (\angle OBC + \angle OCB) \)
\( = 180^\circ - \left(90^\circ - \frac{1}{2}\angle B + 90^\circ - \frac{1}{2}\angle C\right) \)
\( = 180^\circ - 180^\circ + \frac{1}{2}(\angle B + \angle C) \)
\( = \frac{1}{2}(180^\circ - \angle A) = 90^\circ - \frac{1}{2}\angle A \)
In simple words: When the outer angle bisectors meet, the angle formed is always 90 degrees minus half of the opposite vertex angle A.

Exam Tip: Remember the key sign difference: external angle bisectors give \( 90^\circ - \frac{1}{2}\angle A \), whereas internal angle bisectors give \( 90^\circ + \frac{1}{2}\angle A \).

Question 32. If the arms of one angle are respectively parallel to the arms of another angle, then the two angles are:
(a) Neither equal nor supplementary
(b) Not equal but supplementary
(c) Equal but not supplementary
(d) Either equal or supplementary
Answer: (d) Either equal or supplementary
If two angles have their corresponding arms parallel, they are either both acute, both obtuse (making them equal), or one is acute and the other is obtuse (making their sum \( 180^\circ \), so they are supplementary).
In simple words: When the sides of two angles run in parallel lines, the angles are either exactly equal or they add up to 180 degrees.

Exam Tip: If the parallel arms point in the same direction or both in opposite directions, the angles are equal; if one points in the same direction and the other in the opposite direction, they are supplementary.

Question 33. A, B, C and D are four non-coplanar points. The number of planes that can be drawn passing through any three of these points, is:
(a) 3
(b) 4
(c) 5
(d) 8
Answer: (b) 4
Any three non-collinear points uniquely establish a single plane. The number of unique planes that can be formed from four non-coplanar points taken three at a time is given by the combination formula:
\( \binom{4}{3} = \frac{4!}{3! \times 1!} = 4 \)
In simple words: Choosing groups of 3 points out of 4 gives exactly 4 different combinations, creating 4 distinct planes like the faces of a triangular pyramid.

Exam Tip: Three non-collinear points determine a unique plane; use combinations \( \binom{n}{3} \) to find the number of planes formed by \( n \) points.

Question 34. How many degrees are there in an angle which equals one-fifth of its supplement?
(a) 15°
(b) 30°
(c) 75°
(d) 150°
Answer: (b) 30°
Let the measure of the angle be \( x \). Its supplement is \( 180^\circ - x \).
According to the given condition:
\( x = \frac{1}{5}(180^\circ - x) \)
\( \implies 5x = 180^\circ - x \)
\( \implies 6x = 180^\circ \)
\( \implies x = 30^\circ \)
In simple words: The angle is 1 part and its supplement is 5 parts, giving 6 equal parts. 180 divided by 6 is 30 degrees.

Exam Tip: Clear the fraction immediately by multiplying both sides by 5 to avoid algebraic mistakes.

Question 35. If lines AB, AC, AD and AE are parallel to a line l, then
(a) A, B, C, D, E are collinear points.
(b) A, B, C, D, E are non collinear points.
(c) AB & AC are parallel and AD & AE are perpendicular.
(d) None of the options
Answer: (a) A, B, C, D, E are collinear points.
By Playfair's axiom (Euclid's fifth postulate equivalent), through a given point \( A \) not lying on line \( l \), only one unique line can be drawn parallel to line \( l \).
Because lines \( AB \), \( AC \), \( AD \), and \( AE \) all pass through the identical point \( A \) and are parallel to \( l \), they must all be the exact same line. Thus, points \( A \), \( B \), \( C \), \( D \), and \( E \) all lie on this single line and are collinear.
In simple words: Only one parallel line can pass through point A. Since all these lines go through A and are parallel to line l, they are the same line, making all the points lie on one straight line.

Exam Tip: State Playfair's axiom: through a given point outside a line, only one line can be drawn parallel to the given line.

Question 36. If two lines are parallel ,then the perpendicular distance between them is
(a) decreasing.
(b) increasing.
(c) constant.
(d) None of the options
Answer: (c) constant.
By definition, parallel lines never intersect and remain equidistant from each other throughout their length, which means the perpendicular distance between them remains constant.
In simple words: Parallel lines never get closer and never drift apart; the gap between them stays the same everywhere.

Exam Tip: The constant perpendicular distance between two lines is also known as the height or altitude between parallel lines.

Question 37. Supplementary and complementary angles need not be
(a) equal to 180°, 90°
(b) adjacent
(c) angles
(d) None of the options
Answer: (b) adjacent
Two angles only need to have measures summing to \( 180^\circ \) (for supplementary) or \( 90^\circ \) (for complementary). They do not need to share a common vertex or common arm, and can be located anywhere in the plane.
In simple words: Two angles can be supplementary or complementary even if they are completely separate and do not touch each other.

Exam Tip: Adjacent supplementary angles form a linear pair, but non-adjacent angles whose sum is \( 180^\circ \) are still supplementary.

<2M>

Question 38. It is given that ∠XYZ = 64° and XY is produced to a point P. Draw a figure from the given information. If ray YQ bisect ∠ZYP, find ∠XYQ and reflex ∠QYP
Answer: Since line segment \( XY \) is produced to point \( P \), \( XYP \) forms a straight line.
Ray \( YZ \) stands on line \( XYP \), forming a linear pair:
\( \angle XYZ + \angle ZYP = 180^\circ \)
\( 64^\circ + \angle ZYP = 180^\circ \)
\( \implies \angle ZYP = 180^\circ - 64^\circ = 116^\circ \)
Ray \( YQ \) bisects \( \angle ZYP \):
\( \angle ZYQ = \angle QYP = \frac{116^\circ}{2} = 58^\circ \)
Now calculate \( \angle XYQ \):
\( \angle XYQ = \angle XYZ + \angle ZYQ = 64^\circ + 58^\circ = 122^\circ \)
Now calculate reflex \( \angle QYP \):
\( \text{Reflex } \angle QYP = 360^\circ - \angle QYP = 360^\circ - 58^\circ = 302^\circ \)
In simple words: The straight line leaves 116 degrees for angle ZYP, which ray YQ cuts into two 58-degree angles. Adding 64 and 58 gives 122 degrees for angle XYQ, and subtracting 58 from 360 gives 302 degrees for the reflex angle.

Exam Tip: Remember to calculate the reflex angle by subtracting the interior angle from \( 360^\circ \); finding only the interior angle is a frequent omission.

Question 39. Match the following:
(a) Adjacent angles
(b) Vertically opposite angles
(c) Linear pair of angles
Fig-1 Fig-2 Fig-3
Answer:
(a) Adjacent angles matches Fig-2 (two angles sharing a common vertex \( B \) and common arm \( BD \) where non-common arms do not form a straight line).
(b) Vertically opposite angles matches Fig-1 (two lines intersecting at point \( O \) forming non-adjacent opposite angles).
(c) Linear pair of angles matches Fig-3 (two adjacent angles lying on a single straight line whose non-common arms form opposite rays).
In simple words: Fig-1 shows crossed lines making vertically opposite angles, Fig-2 shows side-by-side adjacent angles, and Fig-3 shows a linear pair sitting on a flat line.

Exam Tip: Clearly write the matching pairs in your answer sheet, such as: (a) - Fig-2, (b) - Fig-1, (c) - Fig-3.

Question 40. In Fig, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7, find all the angles.
Answer: Line \( PQ \) and line \( RS \) cross at point \( O \).
Because ray \( OR \) stands on the straight line \( POQ \), the adjacent angles form a linear pair:
\( \angle POR + \angle ROQ = 180^\circ \)
Let \( \angle POR = 5x \) and \( \angle ROQ = 7x \):
\( 5x + 7x = 180^\circ \)
\( \implies 12x = 180^\circ \)
\( \implies x = 15^\circ \)
This gives:
\( \angle POR = 5 \times 15^\circ = 75^\circ \)
\( \angle ROQ = 7 \times 15^\circ = 105^\circ \)
Using the vertically opposite angles property:
\( \angle QOS = \angle POR = 75^\circ \)
\( \angle POS = \angle ROQ = 105^\circ \)
Hence, the four angles are \( 75^\circ \), \( 105^\circ \), \( 75^\circ \), and \( 105^\circ \).
In simple words: The ratio gives 12 parts totaling 180 degrees, making each part 15 degrees. Angles POR and ROQ are 75 and 105 degrees, and the opposite angles match them.

Exam Tip: Do not stop after finding \( \angle POR \) and \( \angle ROQ \); the question says "find all the angles", so list all four angles.

Question 41. In figure if x + y = w + z, then prove that AOB is a line.
Answer: The sum of all angles around the common point \( O \) equals a complete rotation of \( 360^\circ \):
\( (x + y) + (w + z) = 360^\circ \)
We are given that \( w + z = x + y \). Substituting this into the equation:
\( (x + y) + (x + y) = 360^\circ \)
\( \implies 2(x + y) = 360^\circ \)
\( \implies x + y = 180^\circ \)
Since the sum of two adjacent angles \( x \) and \( y \) is \( 180^\circ \), by the converse of the linear pair axiom, their non-common arms \( OA \) and \( OB \) form two opposite rays.
Therefore, \( AOB \) is a straight line.
In simple words: All four angles around point O add up to 360 degrees. Since x plus y equals w plus z, x plus y must be half of 360, which is 180 degrees, proving AOB is a straight line.

Exam Tip: State the converse of the linear pair axiom: if the sum of two adjacent angles is \( 180^\circ \), then their non-common arms form a straight line.

Question 42. In the figure, find the value of y.
Answer: In the figure, the rays meet at a point on a straight line, forming adjacent angles that add up to a straight angle of \( 180^\circ \):
\( 5y^\circ + 3y^\circ + 2y^\circ = 180^\circ \)
\( \implies 10y^\circ = 180^\circ \)
\( \implies y = 18^\circ \)
Therefore, the value of \( y \) is \( 18^\circ \).
In simple words: The three angles sit on a straight line and add up to 180 degrees. 10y equals 180, so dividing by 10 gives y equal to 18 degrees.

Exam Tip: Combine the like algebraic terms first: \( 5y + 3y + 2y = 10y \), then equate to \( 180^\circ \) using the angles-on-a-straight-line property.

<3M>

Question 43. In Figure two straight lines PQ and RS intersect each other at O. If ∠POT = 75°, find the values of a, b and c
Answer: Lines \( PQ \) and \( RS \) intersect at point \( O \).
Since \( POQ \) is a straight line, ray \( OT \) and ray \( OS \) form angles on it that sum to \( 180^\circ \):
\( \angle POR + \angle POT + \angle TOS = 180^\circ \) (with \( \angle POR = 4b \) and \( \angle TOS = b \)):
\( 4b + 75^\circ + b = 180^\circ \)
\( \implies 5b + 75^\circ = 180^\circ \)
\( \implies 5b = 105^\circ \)
\( \implies b = 21^\circ \)
Now find \( a \):
Angles \( \angle QOS \) and \( \angle POR \) are vertically opposite angles:
\( a = 4b = 4 \times 21^\circ = 84^\circ \)
Now find \( c \):
Angles \( \angle ROQ \) and \( \angle POS \) are vertically opposite angles:
\( \angle POS = \angle POT + \angle TOS = 75^\circ + b = 75^\circ + 21^\circ = 96^\circ \)
\( \implies 2c = 96^\circ \)
\( \implies c = 48^\circ \)
Therefore, \( a = 84^\circ \), \( b = 21^\circ \), and \( c = 48^\circ \).
In simple words: The straight line gives 5b plus 75 equals 180, which makes b equal to 21 degrees. Matching vertically opposite angles gives a as 84 degrees and 2c as 96 degrees, meaning c is 48 degrees.

Exam Tip: Clearly write out which angles form a linear pair and which pairs are vertically opposite before solving for each variable.

Question 44. In figure, lines l₁ and l₂ intersect at O forming angles as shown in the figure. If a = 35° Find the value of b, c and d.
Answer: Lines \( l_1 \) and \( l_2 \) intersect at point \( O \).
1. Angles \( a \) and \( c \) are vertically opposite angles:
\( \implies c = a = 35^\circ \)
2. Angles \( a \) and \( b \) form a linear pair on line \( l_1 \):
\( a + b = 180^\circ \)
\( 35^\circ + b = 180^\circ \)
\( \implies b = 180^\circ - 35^\circ = 145^\circ \)
3. Angles \( b \) and \( d \) are vertically opposite angles:
\( \implies d = b = 145^\circ \)
Therefore, \( b = 145^\circ \), \( c = 35^\circ \), and \( d = 145^\circ \).
In simple words: The angle opposite to a is c, so c is 35 degrees. The side angle b makes a straight line with a, so b is 145 degrees, and d matches b.

Exam Tip: State the reasons for each step clearly: "vertically opposite angles are equal" and "angles in a linear pair sum to \( 180^\circ \)".

Question 45. If the angles of a triangle are in the ration2:3:4, find the three angles.
Answer: Let the measures of the three interior angles of the triangle be \( 2x \), \( 3x \), and \( 4x \).
By the angle sum property of a triangle, the sum of all interior angles equals \( 180^\circ \):
\( 2x + 3x + 4x = 180^\circ \)
\( \implies 9x = 180^\circ \)
\( \implies x = 20^\circ \)
Now compute each individual angle:
\( 2x = 2 \times 20^\circ = 40^\circ \)
\( 3x = 3 \times 20^\circ = 60^\circ \)
\( 4x = 4 \times 20^\circ = 80^\circ \)
Therefore, the three angles of the triangle are \( 40^\circ \), \( 60^\circ \), and \( 80^\circ \).
In simple words: The three angles add up to 9 parts totaling 180 degrees. Each part is 20 degrees, giving angles of 40, 60, and 80 degrees.

Exam Tip: Explicitly mention the "Angle Sum Property of a Triangle" to ensure full method marks.

Question 46. In the figure, side QR of ΔPQR has been produced S, if ∠P : ∠Q : ∠R = 3 : 2 : 1 and RT ⊥ PR, Find ∠TRS
Answer: In \( \Delta PQR \), let the interior angles be \( \angle P = 3x \), \( \angle Q = 2x \), and \( \angle PRQ = 1x \).
By the angle sum property of a triangle:
\( 3x + 2x + 1x = 180^\circ \)
\( \implies 6x = 180^\circ \)
\( \implies x = 30^\circ \)
Thus, the interior angle at \( R \) is:
\( \angle PRQ = 1 \times 30^\circ = 30^\circ \)
Because side \( QR \) is produced to point \( S \), \( QRS \) is a straight line:
\( \angle PRS = 180^\circ - \angle PRQ = 180^\circ - 30^\circ = 150^\circ \)
We are given that \( RT \perp PR \), meaning \( \angle PRT = 90^\circ \).
Ray \( RT \) lies inside \( \angle PRS \), so:
\( \angle TRS = \angle PRS - \angle PRT = 150^\circ - 90^\circ = 60^\circ \)
Therefore, the measure of \( \angle TRS \) is \( 60^\circ \).
In simple words: The ratio gives 30 degrees for inside angle R. The straight line leaves 150 degrees outside, and subtracting the 90-degree right angle leaves 60 degrees for angle TRS.

Exam Tip: Carefully distinguish between the interior angle \( \angle PRQ \) and the exterior angle \( \angle PRS \) on the straight line \( QRS \).

Question 47. In Figure determine the value of y
Answer: In the figure, three straight lines intersect at a common point, forming vertically opposite angles around that point.
Summing the angles situated on one half-plane of the straight line:
\( 2y^\circ + 5y^\circ + 2y^\circ = 180^\circ \)
\( \implies 9y^\circ = 180^\circ \)
\( \implies y = 20^\circ \)
Therefore, the value of \( y \) is \( 20^\circ \).
In simple words: Using vertically opposite angles brings all angles onto one side of a straight line. 9y equals 180 degrees, which means y is 20 degrees.

Exam Tip: Write down the vertically opposite angle equality first so examiners can follow which line forms your \( 180^\circ \) linear sum.

<5M>

Question 48. In the given figure ∠Q > ∠R and M is a point QR such that PM is the bisector of angle P . If the perpendicular from P on QR meets QR at N, then prove that ∠MPN = 1/2 (∠Q - ∠R)
Answer: In \( \Delta PQR \), \( PM \) is the angle bisector of \( \angle P \):
\( \angle QPM = \angle RPM = \frac{1}{2}\angle P \) ...(1)
In right-angled triangle \( \Delta PNQ \), \( PN \perp QR \), so \( \angle PNQ = 90^\circ \):
\( \angle QPN = 90^\circ - \angle Q \) ...(2)
From the figure, ray \( PN \) lies inside \( \angle QPM \):
\( \angle MPN = \angle QPM - \angle QPN \)
Substituting equations (1) and (2):
\( \angle MPN = \frac{1}{2}\angle P - (90^\circ - \angle Q) \) ...(3)
In \( \Delta PQR \), by the angle sum property:
\( \angle P + \angle Q + \angle R = 180^\circ \)
Dividing by 2 gives:
\( 90^\circ = \frac{1}{2}\angle P + \frac{1}{2}\angle Q + \frac{1}{2}\angle R \)
Substitute this expression for \( 90^\circ \) into equation (3):
\( \angle MPN = \frac{1}{2}\angle P - \left(\frac{1}{2}\angle P + \frac{1}{2}\angle Q + \frac{1}{2}\angle R - \angle Q\right) \)
\( = \frac{1}{2}\angle P - \frac{1}{2}\angle P + \angle Q - \frac{1}{2}\angle Q - \frac{1}{2}\angle R \)
\( = \frac{1}{2}\angle Q - \frac{1}{2}\angle R = \frac{1}{2}(\angle Q - \angle R) \)
Hence proved.
In simple words: Write angle MPN as half of angle P minus (90 minus angle Q). Replacing 90 with half of all three angles cancels half of angle P and leaves half of (angle Q minus angle R).

Exam Tip: Substituting \( 90^\circ = \frac{1}{2}(\angle P + \angle Q + \angle R) \) directly is the most elegant and mark-safe way to complete this classic 5-mark proof.

Question 49. In figure the sides AB and AC of are produced to points E and D respectively. If bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O, then prove that.
Answer: We are required to prove that \( \angle BOC = 90^\circ - \frac{1}{2}\angle BAC \).
Ray \( BO \) bisects \( \angle CBE \):
\( \angle CBO = \frac{1}{2}\angle CBE = \frac{1}{2}(180^\circ - \angle B) = 90^\circ - \frac{1}{2}\angle B \) ...(1)
Ray \( CO \) bisects \( \angle BCD \):
\( \angle BCO = \frac{1}{2}\angle BCD = \frac{1}{2}(180^\circ - \angle C) = 90^\circ - \frac{1}{2}\angle C \) ...(2)
In \( \Delta BOC \), the sum of angles equals \( 180^\circ \):
\( \angle BOC + \angle CBO + \angle BCO = 180^\circ \)
\( \angle BOC = 180^\circ - (\angle CBO + \angle BCO) \)
Substitute relations (1) and (2):
\( \angle BOC = 180^\circ - \left(90^\circ - \frac{1}{2}\angle B + 90^\circ - \frac{1}{2}\angle C\right) \)
\( = 180^\circ - \left(180^\circ - \frac{1}{2}(\angle B + \angle C)\right) \)
\( = \frac{1}{2}(\angle B + \angle C) \)
In \( \Delta ABC \), \( \angle A + \angle B + \angle C = 180^\circ \implies \angle B + \angle C = 180^\circ - \angle A \):
\( \angle BOC = \frac{1}{2}(180^\circ - \angle A) = 90^\circ - \frac{1}{2}\angle A \)
\( \implies \angle BOC = 90^\circ - \frac{1}{2}\angle BAC \)
Hence proved.
In simple words: Halving the two exterior angles gives angles inside the bottom triangle. Their sum leaves half of (angle B plus angle C) for angle BOC, which equals 90 minus half of angle A.

Exam Tip: Be sure to substitute \( \angle B + \angle C = 180^\circ - \angle BAC \) at the very end to arrive precisely at the required form.

Question 50. In Figure OP||RS. Determine ∠PQR
Answer: Through vertex \( Q \), draw an auxiliary line \( XQY \) parallel to both ray \( OP \) and ray \( RS \), extending left and right.
1. For parallel lines \( OP \parallel XQ \) cut by transversal \( PQ \):
Consecutive interior angles on the same side of the transversal sum to \( 180^\circ \):
\( \angle OPQ + \angle PQX = 180^\circ \)
\( 110^\circ + \angle PQX = 180^\circ \)
\( \implies \angle PQX = 180^\circ - 110^\circ = 70^\circ \)
2. For parallel lines \( RS \parallel QY \) cut by transversal \( RQ \):
Consecutive interior angles on the same side of transversal \( RQ \) sum to \( 180^\circ \):
\( \angle SRQ + \angle RQY = 180^\circ \)
\( 130^\circ + \angle RQY = 180^\circ \)
\( \implies \angle RQY = 180^\circ - 130^\circ = 50^\circ \)
3. Because \( XQY \) is a straight line, the angles at point \( Q \) form a straight line:
\( \angle PQX + \angle PQR + \angle RQY = 180^\circ \)
\( 70^\circ + \angle PQR + 50^\circ = 180^\circ \)
\( \implies \angle PQR + 120^\circ = 180^\circ \)
\( \implies \angle PQR = 60^\circ \)
In simple words: Draw a parallel line right through Q. The left side takes 70 degrees from 180, and the right side takes 50 degrees from 180. That leaves 60 degrees in the middle for angle PQR.

Exam Tip: Drawing the line through Q parallel to both OP and RS is a graded construction step worth 1 mark in 4-mark and 5-mark questions.

Question 51. ABCDE is a regular pentagon and bisector of ∠BAE meets CD in M. IF bisector of ∠BCD meets AM at P find ∠CPM
Answer: Each interior angle of a regular pentagon with \( n = 5 \) sides has measure:
\( \text{Each interior angle} = \frac{(5 - 2) \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ \)
Thus, \( \angle BAE = 108^\circ \) and \( \angle BCD = 108^\circ \).
Ray \( AM \) bisects \( \angle BAE \):
\( \angle BAM = \frac{108^\circ}{2} = 54^\circ \)
In quadrilateral \( ABCM \), the sum of interior angles is \( 360^\circ \):
\( \angle BAM + \angle B + \angle BCM + \angle AMC = 360^\circ \)
\( 54^\circ + 108^\circ + 108^\circ + \angle AMC = 360^\circ \)
\( 270^\circ + \angle AMC = 360^\circ \)
\( \implies \angle AMC = 90^\circ \)
Next, ray \( CP \) bisects \( \angle BCD = 108^\circ \):
\( \angle PCM = \frac{108^\circ}{2} = 54^\circ \)
Now examine right-angled triangle \( \Delta PMC \):
The angles in \( \Delta PMC \) sum to \( 180^\circ \):
\( \angle CPM + \angle PCM + \angle PMC = 180^\circ \)
\( \angle CPM + 54^\circ + 90^\circ = 180^\circ \)
\( \angle CPM + 144^\circ = 180^\circ \)
\( \implies \angle CPM = 180^\circ - 144^\circ = 36^\circ \)
Therefore, the measure of \( \angle CPM \) is \( 36^\circ \).
In simple words: Each corner of the pentagon is 108 degrees. Line AM cuts straight down at a 90-degree right angle to the bottom side. Half of corner C is 54 degrees, leaving 36 degrees for angle CPM in the small triangle.

Exam Tip: First state the formula for the interior angle of a regular polygon: \( \frac{(n-2) \times 180^\circ}{n} \) to earn the initial method mark.