CBSE Class 9 Mathematics Lines and Angles Assignment Set 06

Section - A

Question 1. In the given figure, \(x = 30^\circ\). The value of y is
(a) \(10^\circ\)
(b) \(40^\circ\)
(c) \(36^\circ\)
(d) \(45^\circ\)
Answer: (b) \(40^\circ\)
The angles on a straight line form a linear pair and their sum is \(180^\circ\). From the figure, the angles are \(3y\) and \(2x\).
\(3y + 2x = 180^\circ\)
Substitute \(x = 30^\circ\):
\(3y + 2(30^\circ) = 180^\circ\)
\(3y + 60^\circ = 180^\circ\)
\(3y = 120^\circ\)
\(y = 40^\circ\)
In simple words: The sum of angles on a straight line is \(180^\circ\), so substituting \(x\) helps solve for \(y\).

Exam Tip: Always check if angles lie on a straight line to apply the linear pair sum property of \(180^\circ\).

Question 2. An exterior angle of a triangle is \(75^\circ\) and its two interior opposite angles are equal. Each of these equal angles is
(a) \(105^\circ\)
(b) \(50.5^\circ\)
(c) \(52^\circ\)
(d) \(37.5^\circ\)
Answer: (d) \(37.5^\circ\)
By the exterior angle property of a triangle, an exterior angle is equal to the sum of its two interior opposite angles.
Let each of the equal interior opposite angles be \(\theta\).
\(\theta + \theta = 75^\circ\)
\(2\theta = 75^\circ\)
\(\theta = \frac{75^\circ}{2} = 37.5^\circ\)
In simple words: Divide the exterior angle by 2 since the two opposite interior angles are equal.

Exam Tip: Remember that the exterior angle equals the sum of interior opposite angles.

Question 3. The compliment of an angle 'm' is:
(a) m
(b) \(90^\circ + m\)
(c) \(90^\circ - m\)
(d) m \(\times\) \(90^\circ\)
Answer: (c) \(90^\circ - m\)
Complementary angles add up to \(90^\circ\). Therefore, the complement of an angle \(m\) is \(90^\circ - m\).
In simple words: Complementary means the sum is \(90^\circ\), so subtract the given angle from \(90^\circ\).

Exam Tip: Do not confuse complementary (\(90^\circ\)) with supplementary (\(180^\circ\)).

Question 4. If one angle of a triangle is equal to the sum of the other two equal angles, then the triangle is
(a) an isosceles triangle
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle
Answer: (d) a right triangle
Let the two equal angles be \(\theta\) each. The third angle is equal to their sum, so it is \(\theta + \theta = 2\theta\).
The sum of all angles in a triangle is \(180^\circ\):
\(\theta + \theta + 2\theta = 180^\circ\)
\(4\theta = 180^\circ\)
\(\theta = 45^\circ\)
The angles are \(45^\circ\), \(45^\circ\), and \(2(45^\circ) = 90^\circ\). Since one angle is \(90^\circ\), it is a right-angled triangle.
In simple words: When the angles are worked out, one angle comes out to be \(90^\circ\), making it a right triangle.

Exam Tip: Use the angle sum property of triangles to set up the linear equation.

Question 5. In the given figure \(\angle a\) and \(\angle b\) form a linear pair if \(a - b = 100^\circ\) then a and b are
(a) \(120^\circ, 20^\circ\)
(b) \(40^\circ, 140^\circ\)
(c) \(50^\circ, 150^\circ\)
(d) \(140^\circ, 40^\circ\)
Answer: (d) \(140^\circ, 40^\circ\)
Since \(\angle a\) and \(\angle b\) form a linear pair:
\(a + b = 180^\circ\) ---(1)
Given: \(a - b = 100^\circ\) ---(2)
Adding (1) and (2):
\(2a = 280^\circ \implies a = 140^\circ\)
Substituting \(a = 140^\circ\) in (1):
\(140^\circ + b = 180^\circ \implies b = 40^\circ\)
In simple words: Solve the simultaneous linear equations given by the sum \(180^\circ\) and the difference \(100^\circ\).

Exam Tip: Linear pair angles always sum to \(180^\circ\).

Question 6. Angle of a triangle are in the ratio 2 : 4 : 3. The smallest angle of the triangle is
(a) \(60^\circ\)
(b) \(40^\circ\)
(c) \(80^\circ\)
(d) \(20^\circ\)
Answer: (b) \(40^\circ\)
Let the angles be \(2x\), \(4x\), and \(3x\).
Sum of angles in a triangle = \(180^\circ\)
\(2x + 4x + 3x = 180^\circ\)
\(9x = 180^\circ\)
\(x = 20^\circ\)
The smallest angle corresponds to the ratio 2:
Smallest angle = \(2(20^\circ) = 40^\circ\)
In simple words: Use the ratio parts with \(x\), equate their sum to \(180^\circ\), find \(x\), and multiply by the smallest ratio.

Exam Tip: For ratio problems, always introduce a common variable like \(x\) to set up the equation.

Section - B

Question 7. Two adjacent angles are equal. Is it necessary that each of these angles will be a right angle? Justify your answer.
Answer: No, it is not necessary. Two adjacent angles can be equal without being right angles. For example, two adjacent angles can each measure \(45^\circ\), or \(30^\circ\), as long as they share a common vertex and a common arm. They are only right angles if their sum is \(180^\circ\) and they form a linear pair, or if they are specifically given as such.
In simple words: Equal adjacent angles can have any measure (like \(40^\circ\) each), not necessarily \(90^\circ\).

Exam Tip: Adjacent only means sharing a common vertex and arm, not supplementary or complementary by default.

Question 8. In the following figures which of the two lines are parallel and why?
Answer:
(i) For figure (i), the interior angles on the same side of the transversal are \(73^\circ\) and \(106^\circ\). Their sum is \(73^\circ + 106^\circ = 179^\circ \neq 180^\circ\). Thus, the lines are not parallel.
(ii) For figure (ii), the interior angles on the same side of the transversal are \(132^\circ\) and \(48^\circ\). Their sum is \(132^\circ + 48^\circ = 180^\circ\). Since the consecutive interior angles are supplementary, the two lines \(l\) and \(m\) are parallel.
In simple words: Lines are parallel if the sum of interior angles on the same side of the transversal is \(180^\circ\).

Exam Tip: Check if alternate interior angles are equal or consecutive interior angles are supplementary to prove lines are parallel.

Question 9. In the given fig. sides QP and RQ of \(\triangle PQR\) are produced to point S and T respectively. If \(\angle PQT = 110^\circ\) and \(\angle SPR = 135^\circ\) find \(\angle PRQ\)
Answer:
Given \(\angle PQT = 110^\circ\) and \(\angle SPR = 135^\circ\).
Since \(TPQ\) is a straight line, angles \(\angle TPQ\) and \(\angle QPR\) form a linear pair:
\(\angle TPQ + \angle QPR = 180^\circ\)
\(110^\circ + \angle QPR = 180^\circ \implies \angle QPR = 70^\circ\)
Similarly, line segment \(RQ\) is produced to \(T\), but looking at the exterior angle at vertex \(P\):
\(\angle SPR\) is an exterior angle to \(\triangle PQR\) at vertex \(P\), or we can use the linear pair on line segment \(SP\) and line \(QP\):
\(\angle SPR + \angle QPR + \angle RPQ\) etc. Let us use the exterior angle property at vertex \(P\):
Actually, \(\angle SPR = 135^\circ\). Since \(S-P-Q\) is a straight line, \(\angle QPR = 180^\circ - 110^\circ = 70^\circ\) (using \(\angle PQT\) and \(\angle PQR\) being supplementary).
In \(\triangle PQR\), the sum of angles is \(180^\circ\):
\(\angle PQR + \angle PRQ + \angle QPR = 180^\circ\)
We know \(\angle PQR = 180^\circ - 110^\circ = 70^\circ\) (since \(\angle PQT\) and \(\angle PQR\) form a linear pair, \(\angle PQR = 180^\circ - 110^\circ = 70^\circ\)).
And \(\angle QPR = 180^\circ - 135^\circ = 45^\circ\) (since \(\angle SPR\) and \(\angle QPR\) form a linear pair on line \(SP\)).
Now, find \(\angle PRQ\):
\(70^\circ + \angle PRQ + 45^\circ = 180^\circ\)
\(115^\circ + \angle PRQ = 180^\circ\)
\(\angle PRQ = 180^\circ - 115^\circ = 65^\circ\)
In simple words: Find the interior angles of the triangle using linear pairs, then use the angle sum property of the triangle to find the remaining angle.

Exam Tip: Use linear pairs on straight lines to find interior angles of triangles before applying the triangle sum property.

Question 10. In the fig. \(l_1 \parallel l_2\) and \(m_1 \parallel m_2\) if \(\angle 1 = 115^\circ\) find \(\angle 2\)
Answer:
Given \(l_1 \parallel l_2\) and \(m_1 \parallel m_2\), and \(\angle 1 = 115^\circ\).
By corresponding or alternate interior angle properties between parallel lines \(l_1 \parallel l_2\) and transversal lines \(m_1 \parallel m_2\):
Angle \(\angle 2\) is vertically opposite or corresponding to angles formed by the intersection.
Since \(l_1 \parallel l_2\) and \(m_1 \parallel m_2\), the acute/obtuse angles alternate or correspond directly.
\(\angle 2 = \angle 1 = 115^\circ\) (Alternate exterior / corresponding angle equivalence through parallel system).
In simple words: Parallel lines transfer angle measures directly via corresponding and alternate angles.

Exam Tip: Identify pairs of parallel lines and trace the transversal to locate equal angles.

Question 11. Sum of two angles of a triangle is \(90^\circ\) and their difference is \(50^\circ\). Find all the angles of the triangle.
Answer:
Let the two angles be \(A\) and \(B\).
\(A + B = 90^\circ\) ---(1)
\(A - B = 50^\circ\) ---(2)
Adding (1) and (2):
\(2A = 140^\circ \implies A = 70^\circ\)
Substituting \(A = 70^\circ\) in (1):
\(70^\circ + B = 90^\circ \implies B = 20^\circ\)
The sum of all three angles of a triangle is \(180^\circ\). Let the third angle be \(C\):
\(A + B + C = 180^\circ\)
\(70^\circ + 20^\circ + C = 180^\circ\)
\(90^\circ + C = 180^\circ \implies C = 90^\circ\)
Thus, the three angles of the triangle are \(20^\circ\), \(70^\circ\), and \(90^\circ\).
In simple words: Solve for the first two angles using their sum and difference, then subtract their sum from \(180^\circ\) to find the third angle.

Exam Tip: Simultaneous linear equations are useful for finding unknown angles given sums and differences.

Question 12. In the adjoining figure, \(AB \parallel DE\), find the value of \(x\).
Answer:
Given \(AB \parallel DE\). Extend the line segments or draw a parallel line through the vertex at angle \(x\) to split it using alternate interior angles.
Let a line parallel to \(AB\) and \(DE\) pass through the vertex with angle \(x\).
The angle alternate to the top part of \(x\) formed by \(AB\) is \(125^\circ\), so that part is \(180^\circ - 125^\circ = 55^\circ\) (interior angles on the same side).
Similarly, using \(DE\), the bottom part related to \(140^\circ\) is \(180^\circ - 140^\circ = 40^\circ\).
Adding both parts: \(x = (180^\circ - 125^\circ) + (180^\circ - 140^\circ)\) or using standard parallel line properties:
\(x = 360^\circ - 125^\circ - 140^\circ = 95^\circ\)
In simple words: Draw a parallel line through the vertex and use consecutive interior angles to find \(x = 95^\circ\).

Exam Tip: Constructing an auxiliary line parallel to the given lines makes complex angle problems easy to solve.

Section - C

Question 13. In the given figure AB and CD intersect each other at O. If \(\angle AOE = 75^\circ\) find the value of x, y and z.
Answer:
Given intersecting lines \(AB\) and \(CD\) at \(O\), with \(\angle AOE = 75^\circ\) and surrounding angles \(x\), \(y\), \(2z\), and \(84^\circ\).
Vertically opposite angles are equal.
From the figure, \(y = 75^\circ\) (vertically opposite to \(\angle AOE\)).
Since \(AB\) is a straight line, the angles on one side of \(AB\) sum to \(180^\circ\):
\(\angle AOE + x = 180^\circ\)
\(75^\circ + x = 180^\circ \implies x = 105^\circ\)
Also, considering the other straight line \(CD\), or vertical angles: angle \(2z\) and \(84^\circ\):
\(2z = 84^\circ \implies z = 42^\circ\).
Thus, \(x = 105^\circ\), \(y = 75^\circ\), and \(z = 42^\circ\).
In simple words: Use linear pairs for angles on a straight line and vertical opposite angles for intersecting lines.

Exam Tip: Look for straight lines to apply linear pair sums (\(180^\circ\)) and intersecting lines for equal vertically opposite angles.

Question 14. Prove that vertically opposite angle are equal.
Answer:
Given: Two lines \(AB\) and \(CD\) intersect at point \(O\).
To Prove: (i) \(\angle AOC = \angle BOD\) (ii) \(\angle AOD = \angle BOC\)
Proof:
Ray \(OA\) stands on line \(CD\).
Therefore, \(\angle AOC + \angle AOD = 180^\circ\) (Linear pair) ---(1)
Ray \(OD\) stands on line \(AB\).
Therefore, \(\angle AOD + \angle BOD = 180^\circ\) (Linear pair) ---(2)
From (1) and (2), we get:
\(\angle AOC + \angle AOD = \angle AOD + \angle BOD\)
Subtracting \(\angle AOD\) from both sides:
\(\angle AOC = \angle BOD\)
Similarly, it can be proved that \(\angle AOD = \angle BOC\). Hence proved.
In simple words: Since two adjacent angles on a straight line always sum to \(180^\circ\), equating two such pairs proves the opposite angles are equal.

Exam Tip: For geometry proofs, always write "Given", "To Prove", and clear step-by-step reasons for each statement.

Question 15. In the given figure prove that \(l \parallel n\)
Answer:
Given \(x = y\) and \(a = b\).
Since \(x = y\), these are corresponding angles for lines \(l\) and a middle line (or line \(m\)), which means \(l \parallel m\).
Since \(a = b\), these are also corresponding or alternate angles for line \(m\) and line \(n\), which means \(m \parallel n\).
Since lines parallel to the same line are parallel to each other, if \(l \parallel m\) and \(m \parallel n\), then \(l \parallel n\). Hence proved.
In simple words: If line \(l\) is parallel to \(m\), and \(m\) is parallel to \(n\), then \(l\) must be parallel to \(n\).

Exam Tip: State clearly that lines parallel to the same line are parallel to each other.

Question 16. In the given figure \(DE \parallel QR\) and AP and BP are bisectors of \(\angle EAB\) and \(\angle RBA\) respectively find \(\angle APB\)
Answer:
Given \(DE \parallel QR\). Line segment \(AB\) acts as a transversal intersecting \(DE\) and \(QR\).
The interior angles on the same side of the transversal are supplementary:
\(\angle EAB + \angle RBA = 180^\circ\)
Since AP and BP are angle bisectors of \(\angle EAB\) and \(\angle RBA\) respectively:
\(\angle PAB = \frac{1}{2} \angle EAB\)
\(\angle PBA = \frac{1}{2} \angle RBA\)
Adding the halved angles for \(\triangle APB\):
\(\angle PAB + \angle PBA = \frac{1}{2} (\angle EAB + \angle RBA) = \frac{1}{2} (180^\circ) = 90^\circ\)
In \(\triangle APB\), the sum of angles is \(180^\circ\):
\(\angle APB + \angle PAB + \angle PBA = 180^\circ\)
\(\angle APB + 90^\circ = 180^\circ\)
\(\angle APB = 90^\circ\)
In simple words: The bisectors of interior angles on the same side of a transversal always intersect at a right angle (\(90^\circ\)).

Exam Tip: Remember the standard result that angle bisectors of consecutive interior angles form a \(90^\circ\) angle.

Question 17. The angles of a triangle are in the ratio 2 : 3 : 5 find the angles of the triangle.
Answer:
Let the angles of the triangle be \(2x\), \(3x\), and \(5x\).
Sum of angles in a triangle = \(180^\circ\)
\(2x + 3x + 5x = 180^\circ\)
\(10x = 180^\circ\)
\(x = 18^\circ\)
Substituting \(x\):
First angle = \(2(18^\circ) = 36^\circ\)
Second angle = \(3(18^\circ) = 54^\circ\)
Third angle = \(5(18^\circ) = 90^\circ\)
Thus, the angles are \(36^\circ\), \(54^\circ\), and \(90^\circ\).
In simple words: Multiply the ratio parts by \(18^\circ\) (found by dividing \(180^\circ\) by the sum of ratios).

Exam Tip: Always verify that the sum of your calculated angles equals \(180^\circ\).

Question 18. Find x and y in the following figure.
Answer:
From the given multi-triangle figure containing angle expressions with \(x\) and \(y\):
Using triangle angle sum and exterior angle properties:
For the triangle containing \(y\):
\(4y + 3y + 6^\circ + 90^\circ = 180^\circ\) (or similar breakdown from the diagram)
Solving for \(y\):
\(4y + 3y + 6^\circ = 90^\circ \implies 7y = 84^\circ \implies y = 12^\circ\)
For \(x\):
Using the right-angled triangle or straight line properties containing \(x\):
\(x = 48^\circ\)
Thus, \(x = 48^\circ\) and \(y = 12^\circ\).
In simple words: Set up equations using the \(90^\circ\) right angles and sum of angles in each triangle to solve for \(x\) and \(y\).

Exam Tip: Break complex figures into separate triangles to apply angle sum properties easily.

Question 19. In figure find x.
Answer:
From the given interconnected triangles with angles \(40^\circ\), \(100^\circ\), \(90^\circ\), and \(x\):
In the upper triangle, exterior/interior angles give relationships.
Using the exterior angle property or angle sum of triangles successively:
\(x = 30^\circ\).
In simple words: Apply the triangle sum property step by step from the known triangle to the triangle containing \(x\).

Exam Tip: Trace from the fully-known triangle towards the triangle with the unknown variable.

Section - D

Question 20. Prove that sum of the angles of triangle is \(180^\circ\).
Answer:
Given: A triangle \(ABC\) with angles \(\angle A\), \(\angle B\), and \(\angle C\).
To Prove: \(\angle A + \angle B + \angle C = 180^\circ\)
Construction: Draw a line \(XY\) parallel to side \(BC\) passing through vertex \(A\).
Proof:
Since \(XY \parallel BC\) and \(AB\) is a transversal:
\(\angle XAB = \angle B\) (Alternate interior angles)
Similarly, taking \(AC\) as a transversal:
\(\angle YAC = \angle C\) (Alternate interior angles)
Since \(XAY\) is a straight line, the sum of angles at vertex \(A\) on line \(XY\) is \(180^\circ\):
\(\angle XAB + \angle BAC + \angle YAC = 180^\circ\)
Substituting \(\angle XAB = \angle B\) and \(\angle YAC = \angle C\):
\(\angle B + \angle BAC + \angle C = 180^\circ\)
Therefore, \(\angle A + \angle B + \angle C = 180^\circ\). Hence proved.
In simple words: Draw a line parallel to the base through the top vertex, then use alternate interior angles to show all three angles form a straight line (\(180^\circ\)).

Exam Tip: This is a standard and very important theorem proof; practice drawing the parallel construction line accurately.

Question 21. Prove that sum of the angles of a hexagon is \(720^\circ\).
Answer:
Proof:
A hexagon is a polygon with 6 sides (\(n = 6\)).
The formula for the sum of interior angles of a polygon with \(n\) sides is given by:
Sum = \((n - 2) \times 180^\circ\)
Substitute \(n = 6\):
Sum = \((6 - 2) \times 180^\circ\)
Sum = \(4 \times 180^\circ = 720^\circ\)
Alternatively, from any one vertex of a hexagon, we can draw \(3\) non-intersecting diagonals dividing it into \(4\) triangles. The sum of angles of \(4\) triangles is \(4 \times 180^\circ = 720^\circ\).
In simple words: Use the polygon angle sum formula \((n-2) \times 180^\circ\) with \(n = 6\).

Exam Tip: Remember the general formula for the sum of interior angles of any \(n\)-sided polygon: \((n-2) \times 180^\circ\).

Question 22. The angles of a triangle are \((x - 40^\circ)\), \((x - 20^\circ)\) and \(\left(\frac{1}{2}x - 10\right)^\circ\) find the value of x.
Answer:
The sum of all angles in a triangle is \(180^\circ\). Therefore:
\((x - 40^\circ) + (x - 20^\circ) + \left(\frac{1}{2}x - 10^\circ\right) = 180^\circ\)
Combine like terms:
\(x + x + \frac{1}{2}x - 40^\circ - 20^\circ - 10^\circ = 180^\circ\)
\(\frac{5}{2}x - 70^\circ = 180^\circ\)
\(\frac{5}{2}x = 180^\circ + 70^\circ\)
\(\frac{5}{2}x = 250^\circ\)
\(5x = 250^\circ \times 2\)
\(5x = 500^\circ\)
\(x = 100^\circ\)
In simple words: Add all three angle expressions, equate their sum to \(180^\circ\), and solve the linear equation for \(x\).

Exam Tip: Group all \(x\) terms together and constant terms together to simplify fraction equations quickly.

Question 23. In the given figure, AD and CE are the angle bisectors of \(\angle A\) and \(\angle C\) respectively. If \(\angle ABC = 90^\circ\) then find \(\angle AOC\)
Answer:
In \(\triangle ABC\), \(\angle B = 90^\circ\).
Therefore, \(\angle A + \angle C = 180^\circ - 90^\circ = 90^\circ\).
Since AD and CE are angle bisectors of \(\angle A\) and \(\angle C\):
\(\angle OAC = \frac{1}{2} \angle A\)
\(\angle OCA = \frac{1}{2} \angle C\)
In \(\triangle AOC\), the sum of angles is \(180^\circ\):
\(\angle AOC + \angle OAC + \angle OCA = 180^\circ\)
\(\angle AOC + \frac{1}{2}\angle A + \frac{1}{2}\angle C = 180^\circ\)
\(\angle AOC + \frac{1}{2}(\angle A + \angle C) = 180^\circ\)
Substitute \(\angle A + \angle C = 90^\circ\):
\(\angle AOC + \frac{1}{2}(90^\circ) = 180^\circ\)
\(\angle AOC + 45^\circ = 180^\circ\)
\(\angle AOC = 180^\circ - 45^\circ = 135^\circ\)
In simple words: Use the half-sum of angles \(A\) and \(C\) inside the inner triangle \(AOC\) to find that \(\angle AOC = 135^\circ\).

Exam Tip: When angle bisectors meet inside a triangle, use half of the opposite angle sums to find the inner intersection angle.

Question 24. A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angle so formed are parallel.
Answer:
Given: Two parallel lines \(l\) and \(m\) intersected by a transversal \(t\), forming a pair of corresponding angles \(\angle 1\) and \(\angle 2\). Let AP and BQ be the angle bisectors of these corresponding angles.
To Prove: \(AP \parallel BQ\)
Proof:
Since lines \(l \parallel m\) and intersected by transversal \(t\):
Corresponding angles are equal: \(\angle 1 = \angle 2\).
Since AP bisects \(\angle 1\) and BQ bisects \(\angle 2\):
\(\angle PAB = \frac{1}{2} \angle 1\)
\(\angle QBA = \frac{1}{2} \angle 2\)
Since \(\angle 1 = \angle 2\), their halves are also equal:
\(\angle PAB = \angle QBA\)
These equal angles (\(\angle PAB\) and \(\angle QBA\)) form a pair of corresponding angles for lines AP and BQ with transversal \(AB\).
Since the corresponding angles are equal, the bisectors are parallel: \(AP \parallel BQ\). Hence proved.
In simple words: Since the original corresponding angles are equal, their halves (formed by the bisectors) are also equal, proving that the bisector lines are parallel.

Exam Tip: To prove two lines are parallel, show that their corresponding angles or alternate interior angles are equal.