CBSE Class 9 Mathematics Lines And Angles Worksheet Set 04

Lines and Angles

Question 1. The complement of an angle is 1/4 of its supplement. Find the angle.
Answer: Let the required angle be denoted by \( x \).
The complementary angle of \( x \) is \( 90^\circ - x \), while its supplementary angle is \( 180^\circ - x \).
From the given condition:
\( 90^\circ - x = \frac{1}{4}(180^\circ - x) \)
Multiplying both sides by 4:
\( 4(90^\circ - x) = 180^\circ - x \)
\( 360^\circ - 4x = 180^\circ - x \)
\( \implies 4x - x = 360^\circ - 180^\circ \)
\( \implies 3x = 180^\circ \)
\( \implies x = 60^\circ \)
Therefore, the measure of the angle is \( 60^\circ \).
In simple words: The complement is 30 degrees and the supplement is 120 degrees. Since 30 is one-fourth of 120, the original angle is 60 degrees.

Exam Tip: Always state the expressions for both the complement (\( 90^\circ - x \)) and the supplement (\( 180^\circ - x \)) clearly before forming the linear equation.

Question 2. The angles of a supplementary pair are in the ratio 3:2. Find the angles.
Answer: Let the two supplementary angles be \( 3x \) and \( 2x \).
Because the sum of supplementary angles is always \( 180^\circ \):
\( 3x + 2x = 180^\circ \)
\( \implies 5x = 180^\circ \)
\( \implies x = 36^\circ \)
Evaluating each angle:
First angle \( = 3 \times 36^\circ = 108^\circ \)
Second angle \( = 2 \times 36^\circ = 72^\circ \)
Thus, the two required angles are \( 108^\circ \) and \( 72^\circ \).
In simple words: Divide 180 degrees into 5 equal parts of 36 degrees each. Three parts give 108 degrees and two parts give 72 degrees.

Exam Tip: Check your final values by adding them together (\( 108^\circ + 72^\circ = 180^\circ \)) to ensure they form a valid supplementary pair.

Question 3. 4 times the complement of an angle is equal to its supplement. Find the angle.
Answer: Let the measure of the unknown angle be \( \theta \).
The complement of this angle is \( 90^\circ - \theta \) and its supplement is \( 180^\circ - \theta \).
According to the problem statement:
\( 4(90^\circ - \theta) = 180^\circ - \theta \)
\( 360^\circ - 4\theta = 180^\circ - \theta \)
\( \implies 4\theta - \theta = 360^\circ - 180^\circ \)
\( \implies 3\theta = 180^\circ \)
\( \implies \theta = 60^\circ \)
Hence, the required angle measures \( 60^\circ \).
In simple words: Four times the complement (30 degrees) gives 120 degrees, which matches the supplement, so the angle must be 60 degrees.

Exam Tip: Be careful when expanding brackets: ensure the factor 4 multiplies both \( 90^\circ \) and the variable \( \theta \).

Question 4. Angles of a Linear pair are in the ratio 3:2. Find the angles.
Answer: Let the two adjacent angles forming the linear pair be \( 3k \) and \( 2k \).
According to the linear pair axiom, the sum of these adjacent angles equals \( 180^\circ \):
\( 3k + 2k = 180^\circ \)
\( \implies 5k = 180^\circ \)
\( \implies k = 36^\circ \)
The angles are therefore:
\( 3k = 3 \times 36^\circ = 108^\circ \)
\( 2k = 2 \times 36^\circ = 72^\circ \)
Hence, the measures of the angles are \( 108^\circ \) and \( 72^\circ \).
In simple words: A linear pair forms a straight line of 180 degrees. Splitting that into a 3 to 2 ratio yields 108 degrees and 72 degrees.

Exam Tip: Always mention the "Linear Pair Axiom" when equating the sum of adjacent angles on a straight line to \( 180^\circ \).

Question 5. An Exterior angle of a triangle is 108° and the interior opposite angles are in the ratio 4:5. Find the angles of the triangle.
Answer: Let the two opposite interior angles be represented by \( 4x \) and \( 5x \).
By the exterior angle theorem, the exterior angle equals the sum of the two opposite interior angles:
\( 4x + 5x = 108^\circ \)
\( \implies 9x = 108^\circ \)
\( \implies x = 12^\circ \)
This gives the two opposite interior angles as:
\( 4x = 4 \times 12^\circ = 48^\circ \)
\( 5x = 5 \times 12^\circ = 60^\circ \)
The third interior angle forms a linear pair with the exterior angle:
\( 180^\circ - 108^\circ = 72^\circ \)
Consequently, the three interior angles of the triangle are \( 48^\circ \), \( 60^\circ \), and \( 72^\circ \).
In simple words: The two inside angles add up to the outside angle of 108 degrees, giving 48 and 60 degrees. The remaining inside angle on the straight line is 72 degrees.

Exam Tip: Remember to calculate all three interior angles of the triangle; finding only the two ratio-based angles leaves the question incomplete.

Question 6. AB||CD. The bisectors of ∠BPQ and ∠DQP meet at O. Prove that ∠POQ = 90°.
Answer: We are given parallel lines \( AB \parallel CD \) cut by the transversal line segment \( PQ \).
Consecutive interior angles on the same side of the transversal are supplementary:
\( \angle BPQ + \angle DQP = 180^\circ \)
Dividing the entire relation by 2:
\( \frac{1}{2}\angle BPQ + \frac{1}{2}\angle DQP = 90^\circ \)
Since ray \( PO \) bisects \( \angle BPQ \) and ray \( QO \) bisects \( \angle DQP \):
\( \angle OPQ = \frac{1}{2}\angle BPQ \quad \text{and} \quad \angle OQP = \frac{1}{2}\angle DQP \)
Substituting these values into the halved equation gives:
\( \angle OPQ + \angle OQP = 90^\circ \)
Now consider \( \Delta POQ \). By the angle sum property of a triangle:
\( \angle POQ + \angle OPQ + \angle OQP = 180^\circ \)
\( \angle POQ + 90^\circ = 180^\circ \)
\( \implies \angle POQ = 180^\circ - 90^\circ = 90^\circ \)
Hence proved.
In simple words: The two consecutive inside angles add up to 180 degrees. Taking half of each gives 90 degrees inside the small triangle, leaving exactly 90 degrees for angle POQ.

Exam Tip: State the property clearly: "the bisectors of consecutive interior angles of parallel lines intersect at right angles (\( 90^\circ \))."

Question 7. The sides of ΔABC are produced as shown in the figure. Prove that ∠CBX + ∠ACY + ∠BAZ = 360°.
Answer: Let the interior angles of \( \Delta ABC \) at vertices \( A \), \( B \), and \( C \) be \( \angle 1 \), \( \angle 2 \), and \( \angle 3 \) respectively.
By the linear pair axiom at each of the three vertices:
\( \angle CBX + \angle 2 = 180^\circ \implies \angle CBX = 180^\circ - \angle 2 \)
\( \angle ACY + \angle 3 = 180^\circ \implies \angle ACY = 180^\circ - \angle 3 \)
\( \angle BAZ + \angle 1 = 180^\circ \implies \angle BAZ = 180^\circ - \angle 1 \)
Adding these three equations together:
\( \angle CBX + \angle ACY + \angle BAZ = (180^\circ - \angle 2) + (180^\circ - \angle 3) + (180^\circ - \angle 1) \)
\( = 540^\circ - (\angle 1 + \angle 2 + \angle 3) \)
According to the angle sum property of a triangle, \( \angle 1 + \angle 2 + \angle 3 = 180^\circ \).
Substituting this sum into the equation:
\( \angle CBX + \angle ACY + \angle BAZ = 540^\circ - 180^\circ = 360^\circ \)
Hence proved.
In simple words: Each outer angle is 180 minus its inside partner. Adding all three gives 540 minus 180, which proves the outer angles total 360 degrees.

Exam Tip: Expressing each exterior angle as \( 180^\circ - \text{interior angle} \) is the most reliable way to prove this standard polygon property.

Question 8. Two lines are intersected by transversal such that the bisectors of a pair of corresponding angles are parallel. Prove that the lines are parallel.
Answer: Let lines \( AB \) and \( CD \) be cut by transversal line \( EF \) at points \( G \) and \( H \) respectively.
Let ray \( GM \) bisect corresponding angle \( \angle EGB \), and ray \( HN \) bisect corresponding angle \( \angle GHD \).
We are given that the bisector rays are parallel:
\( GM \parallel HN \)
Transversal line \( EF \) cuts these two parallel rays at points \( G \) and \( H \).
Since corresponding angles formed by parallel lines are equal:
\( \angle EGM = \angle GHN \)
Because ray \( GM \) bisects \( \angle EGB \):
\( \angle EGM = \frac{1}{2}\angle EGB \implies \angle EGB = 2\angle EGM \)
Similarly, because ray \( HN \) bisects \( \angle GHD \):
\( \angle GHN = \frac{1}{2}\angle GHD \implies \angle GHD = 2\angle GHN \)
Multiplying the equality \( \angle EGM = \angle GHN \) by 2 gives:
\( 2\angle EGM = 2\angle GHN \)
\( \implies \angle EGB = \angle GHD \)
These two angles constitute a pair of corresponding angles formed when transversal \( EF \) intersects lines \( AB \) and \( CD \).
By the converse of the corresponding angles axiom, if a pair of corresponding angles is equal, then the lines are parallel.
Therefore, \( AB \parallel CD \).
In simple words: Because the two bisector rays are parallel, their small angles match. Doubling those small angles makes the full corresponding angles match, which proves the main lines are parallel.

Exam Tip: State the converse theorem explicitly: if a pair of corresponding angles is equal, then the two lines intersected by the transversal are parallel.

Question 9. Lines AB and CD are intersected by the transversal PQ such that the bisectors of ∠APQ and ∠DQP are parallel. Prove that AB||CD.
Answer: Let transversal line \( PQ \) intersect line \( AB \) at point \( P \) and line \( CD \) at point \( Q \).
Let ray \( PR \) bisect alternate interior angle \( \angle APQ \), and ray \( QS \) bisect alternate interior angle \( \angle DQP \).
We are given that:
\( PR \parallel QS \)
With line \( PQ \) acting as the transversal intersecting parallel rays \( PR \) and \( QS \), alternate interior angles are equal:
\( \angle RPQ = \angle SQP \)
Since ray \( PR \) is the angle bisector of \( \angle APQ \):
\( \angle RPQ = \frac{1}{2}\angle APQ \implies \angle APQ = 2\angle RPQ \)
Similarly, since ray \( QS \) is the angle bisector of \( \angle DQP \):
\( \angle SQP = \frac{1}{2}\angle DQP \implies \angle DQP = 2\angle SQP \)
Multiplying the relation \( \angle RPQ = \angle SQP \) by 2:
\( 2\angle RPQ = 2\angle SQP \)
\( \implies \angle APQ = \angle DQP \)
Notice that \( \angle APQ \) and \( \angle DQP \) form a pair of alternate interior angles for lines \( AB \) and \( CD \) cut by transversal \( PQ \).
By the converse of the alternate interior angles theorem, lines having equal alternate interior angles are parallel.
Therefore, \( AB \parallel CD \).
In simple words: The parallel bisector lines create matching half-angles. Doubling them shows that the alternate interior angles match, proving that the two horizontal lines are parallel.

Exam Tip: Clearly show the doubling step \( 2\angle RPQ = 2\angle SQP \) to establish that the complete alternate interior angles are equal.

Question 10. The bisectors of ∠B and ∠C meet at O. Prove that ∠BOC = 90° + ½ ∠A.
Answer: In \( \Delta ABC \), the sum of the interior angles is:
\( \angle A + \angle B + \angle C = 180^\circ \)
\( \implies \angle B + \angle C = 180^\circ - \angle A \)
Dividing by 2:
\( \frac{1}{2}\angle B + \frac{1}{2}\angle C = 90^\circ - \frac{1}{2}\angle A \) ...(1)
Now consider \( \Delta OBC \). The sum of its interior angles is:
\( \angle BOC + \angle OBC + \angle OCB = 180^\circ \)
Since ray \( BO \) bisects \( \angle B \) and ray \( CO \) bisects \( \angle C \):
\( \angle OBC = \frac{1}{2}\angle B \quad \text{and} \quad \angle OCB = \frac{1}{2}\angle C \)
Substituting these into the triangle equation:
\( \angle BOC + \left(\frac{1}{2}\angle B + \frac{1}{2}\angle C\right) = 180^\circ \)
Using relation (1):
\( \angle BOC + \left(90^\circ - \frac{1}{2}\angle A\right) = 180^\circ \)
\( \implies \angle BOC = 180^\circ - \left(90^\circ - \frac{1}{2}\angle A\right) \)
\( \implies \angle BOC = 90^\circ + \frac{1}{2}\angle A \)
Hence proved.
In simple words: The two inside half-angles leave 90 degrees minus half of angle A. Subtracting that from 180 gives 90 plus half of angle A for the angle at point O.

Exam Tip: Remember that interior angle bisectors meet at \( 90^\circ + \frac{1}{2}\angle A \), whereas exterior angle bisectors meet at \( 90^\circ - \frac{1}{2}\angle A \).

Question 11. The bisectors of the exterior angles ∠CBX and ∠BCY meet at O. Prove that ∠BOC = 90° - ½ ∠A.
Answer: Side \( AB \) of \( \Delta ABC \) is produced to point \( X \), and side \( AC \) is produced to point \( Y \).
By the linear pair axiom:
\( \angle CBX = 180^\circ - \angle B \)
\( \angle BCY = 180^\circ - \angle C \)
Ray \( BO \) bisects \( \angle CBX \), so:
\( \angle CBO = \frac{1}{2}(180^\circ - \angle B) = 90^\circ - \frac{1}{2}\angle B \)
Ray \( CO \) bisects \( \angle BCY \), so:
\( \angle BCO = \frac{1}{2}(180^\circ - \angle C) = 90^\circ - \frac{1}{2}\angle C \)
In \( \Delta BOC \), the interior angles sum to \( 180^\circ \):
\( \angle BOC = 180^\circ - (\angle CBO + \angle BCO) \)
\( = 180^\circ - \left(90^\circ - \frac{1}{2}\angle B + 90^\circ - \frac{1}{2}\angle C\right) \)
\( = 180^\circ - \left(180^\circ - \frac{1}{2}(\angle B + \angle C)\right) \)
\( = \frac{1}{2}(\angle B + \angle C) \)
Since \( \angle A + \angle B + \angle C = 180^\circ \), we have \( \angle B + \angle C = 180^\circ - \angle A \):
\( \implies \angle BOC = \frac{1}{2}(180^\circ - \angle A) = 90^\circ - \frac{1}{2}\angle A \)
Hence proved.
In simple words: The exterior bisectors create angles inside the bottom triangle that leave half of (B plus C) for angle BOC. That simplifies to 90 minus half of angle A.

Exam Tip: Keep the parentheses intact when subtracting: \( 180^\circ - (180^\circ - \frac{1}{2}(\angle B + \angle C)) = \frac{1}{2}(\angle B + \angle C) \); dropping parentheses causes sign errors.

Question 12. PQRS is a parallelogram. The bisectors of ∠P and ∠Q meet at O. Prove that ∠POQ = 90°.
Answer: In parallelogram \( PQRS \), consecutive interior angles are supplementary:
\( \angle P + \angle Q = 180^\circ \)
Dividing by 2 gives:
\( \frac{1}{2}\angle P + \frac{1}{2}\angle Q = 90^\circ \)
Because ray \( PO \) bisects \( \angle P \) and ray \( QO \) bisects \( \angle Q \):
\( \angle OPQ = \frac{1}{2}\angle P \quad \text{and} \quad \angle OQP = \frac{1}{2}\angle Q \)
Therefore:
\( \angle OPQ + \angle OQP = 90^\circ \)
In \( \Delta POQ \), applying the angle sum property:
\( \angle POQ + \angle OPQ + \angle OQP = 180^\circ \)
\( \angle POQ + 90^\circ = 180^\circ \)
\( \implies \angle POQ = 90^\circ \)
Hence proved.
In simple words: Neighboring corners of a parallelogram add up to 180 degrees. Taking half of each corner gives 90 degrees, leaving exactly 90 degrees for angle POQ.

Exam Tip: Cite the property: "adjacent angles of a parallelogram are supplementary" before taking half of their sum.

Question 13. The bisectors of ∠PQR and ∠PRS meet at O. Prove that 1} ∠QTR = ½ ∠PQR. 2} If ∠QTR = 40° find ∠QPR.
Answer: (Note: The bisectors meet at point \( T \), and part 1 proves that \( \angle QTR = \frac{1}{2}\angle QPR \).)
1} In \( \Delta PQR \), side \( QR \) is produced to \( S \). Exterior angle \( \angle PRS \) satisfies:
\( \angle PRS = \angle QPR + \angle PQR \)
Dividing by 2:
\( \frac{1}{2}\angle PRS = \frac{1}{2}\angle QPR + \frac{1}{2}\angle PQR \) ...(1)
Because ray \( RT \) bisects \( \angle PRS \), \( \frac{1}{2}\angle PRS = \angle TRS \).
Because ray \( QT \) bisects \( \angle PQR \), \( \frac{1}{2}\angle PQR = \angle TQR \).
Substituting these into equation (1):
\( \angle TRS = \frac{1}{2}\angle QPR + \angle TQR \) ...(2)
Now examine \( \Delta TQR \). Exterior angle \( \angle TRS \) satisfies:
\( \angle TRS = \angle QTR + \angle TQR \) ...(3)
Equating relations (2) and (3):
\( \angle QTR + \angle TQR = \frac{1}{2}\angle QPR + \angle TQR \)
Subtracting \( \angle TQR \) from both sides:
\( \angle QTR = \frac{1}{2}\angle QPR \)
Hence proved.

2} We are given that \( \angle QTR = 40^\circ \):
Using the proven relation:
\( 40^\circ = \frac{1}{2}\angle QPR \)
\( \implies \angle QPR = 2 \times 40^\circ = 80^\circ \)
In simple words: Write the exterior angle rule for both triangles. Comparing them shows angle QTR is half of angle QPR, so if QTR is 40 degrees, QPR must be 80 degrees.

Exam Tip: This is NCERT Theorem 6.8 / Exercise 6.3 Q6; set the two exterior angle expressions for \( \angle TRS \) equal to eliminate \( \angle TQR \) directly.

Question 14. AOB is a straight line. If a - b = 40° find a and b.
Answer: Because \( AOB \) is a straight line, ray \( OC \) stands on it to form a linear pair:
\( a + b = 180^\circ \) ...(1)
We are given:
\( a - b = 40^\circ \) ...(2)
Adding equation (1) and equation (2):
\( (a + b) + (a - b) = 180^\circ + 40^\circ \)
\( \implies 2a = 220^\circ \)
\( \implies a = 110^\circ \)
Substituting \( a = 110^\circ \) into equation (1):
\( 110^\circ + b = 180^\circ \)
\( \implies b = 180^\circ - 110^\circ = 70^\circ \)
Therefore, \( a = 110^\circ \) and \( b = 70^\circ \).
In simple words: The two angles add up to 180 degrees and their difference is 40 degrees. Solving the two equations gives 110 degrees for a and 70 degrees for b.

Exam Tip: Add the sum equation \( a + b = 180^\circ \) and the difference equation \( a - b = 40^\circ \) to solve for \( a \) without needing substitution.

Question 15. AB||CD. ∠APO = 60° ∠CQO = 80°. Find reflex ∠POQ.
Answer: Through vertex \( O \), construct an auxiliary ray \( OE \) parallel to line \( AB \) directed toward the left (so that \( OE \parallel AB \parallel CD \)).
Since \( OE \parallel AB \) and ray \( PA \) points left while ray \( OE \) points left, transversal \( PO \) forms alternate interior angles:
\( \angle POE = \angle APO = 60^\circ \)
Similarly, since \( OE \parallel CD \) and ray \( QC \) points left while ray \( OE \) points left, transversal \( QO \) forms alternate interior angles:
\( \angle QOE = \angle CQO = 80^\circ \)
The total interior angle \( \angle POQ \) is the sum of these two adjacent parts:
\( \angle POQ = \angle POE + \angle QOE = 60^\circ + 80^\circ = 140^\circ \)
The problem requests the reflex angle:
\( \text{Reflex } \angle POQ = 360^\circ - \angle POQ = 360^\circ - 140^\circ = 220^\circ \)
In simple words: Draw a parallel line through point O to split the bend into 60 degrees and 80 degrees, giving an inside angle of 140 degrees. Subtracting from 360 gives 220 degrees for the reflex angle.

Exam Tip: Be mindful of the word "reflex"; finding only the acute/obtuse interior angle \( 140^\circ \) without subtracting from \( 360^\circ \) costs a full mark.

Question 16. POQ is a straight line. Find the value of x.
Answer: Because \( POQ \) is a straight line, all adjacent angles formed on one side of this line add up to \( 180^\circ \):
\( \angle POR + \angle ROS + \angle SOQ = 180^\circ \)
Substituting the given expressions:
\( (3x + 10^\circ) + 2x + (4x - 10^\circ) = 180^\circ \)
Combine like algebraic terms:
\( (3x + 2x + 4x) + (10^\circ - 10^\circ) = 180^\circ \)
\( \implies 9x = 180^\circ \)
\( \implies x = 20^\circ \)
Therefore, the value of \( x \) is \( 20^\circ \).
In simple words: The three angles sit together on a straight line and total 180 degrees. The plus 10 and minus 10 cancel out, leaving 9x equals 180, so x is 20 degrees.

Exam Tip: Cancel out opposite constants like \( +10^\circ \) and \( -10^\circ \) early to streamline your calculation and avoid arithmetic errors.

Question 17. AB||CD, ∠APQ = x + 20 and ∠DQP = 3x - 60. Find x.
Answer: Lines \( AB \) and \( CD \) are parallel, and segment \( PQ \) serves as a transversal intersecting them.
Angles \( \angle APQ \) and \( \angle DQP \) are alternate interior angles.
Since alternate interior angles are equal when lines are parallel:
\( \angle APQ = \angle DQP \)
\( x + 20^\circ = 3x - 60^\circ \)
Rearranging the terms:
\( 3x - x = 20^\circ + 60^\circ \)
\( \implies 2x = 80^\circ \)
\( \implies x = 40^\circ \)
Thus, the value of \( x \) is \( 40^\circ \).
In simple words: The two angles are alternate interior angles, so they are equal. Setting x plus 20 equal to 3x minus 60 gives 2x equal to 80, meaning x is 40 degrees.

Exam Tip: State "alternate interior angles are equal" as your geometric justification to secure the method mark before setting up the equation.

Question 18. PS is the bisector of ∠QPR. Prove that 1} ∠QPS + ∠PSR = ∠PRT 2} ∠PQS + ∠PRT = 2∠PSR
Answer: In \( \Delta PQR \), side \( QR \) is extended to point \( T \), and \( PS \) is the angle bisector of \( \angle QPR \), meaning:
\( \angle QPS = \angle RPS \) ...(1)

1} Consider \( \Delta PSR \), where side \( QS \) is produced to \( T \).
By the exterior angle theorem, an exterior angle equals the sum of the two opposite interior angles:
\( \angle PRT = \angle RPS + \angle PSR \)
Using the bisector property (1), replace \( \angle RPS \) with \( \angle QPS \):
\( \implies \angle PRT = \angle QPS + \angle PSR \)
\( \implies \angle QPS + \angle PSR = \angle PRT \)
Hence proved.

2} Now consider \( \Delta PQS \). Side \( QS \) is extended to \( R \), so \( \angle PSR \) is an exterior angle to \( \Delta PQS \):
\( \angle PSR = \angle PQS + \angle QPS \)
Rearranging this gives:
\( \angle QPS = \angle PSR - \angle PQS \) ...(2)
From part 1}, we already established:
\( \angle PRT = \angle QPS + \angle PSR \)
Substitute equation (2) for \( \angle QPS \) into this relation:
\( \angle PRT = (\angle PSR - \angle PQS) + \angle PSR \)
\( \angle PRT = 2\angle PSR - \angle PQS \)
Adding \( \angle PQS \) to both sides:
\( \implies \angle PQS + \angle PRT = 2\angle PSR \)
Hence proved.
In simple words: Apply the exterior angle rule to the right triangle to prove part 1. Then use the exterior angle rule on the left triangle to replace angle QPS, which proves that the outer sum equals twice angle PSR.

Exam Tip: Apply the exterior angle theorem separately to \( \Delta PSR \) and \( \Delta PQS \); substituting one result into the other connects both parts directly.

Question 19. In ΔABC, 6∠A = 3∠B = 2∠C. Find the angles.
Answer: Let the given continuous equality be set to a common variable \( k \):
\( 6\angle A = 3\angle B = 2\angle C = k \)
Express each angle in terms of \( k \):
\( \angle A = \frac{k}{6}, \quad \angle B = \frac{k}{3}, \quad \angle C = \frac{k}{2} \)
By the angle sum property of a triangle, the interior angles sum to \( 180^\circ \):
\( \angle A + \angle B + \angle C = 180^\circ \)
\( \frac{k}{6} + \frac{k}{3} + \frac{k}{2} = 180^\circ \)
Finding a common denominator of 6:
\( \frac{k + 2k + 3k}{6} = 180^\circ \)
\( \frac{6k}{6} = 180^\circ \)
\( \implies k = 180^\circ \)
Now determine the measure of each individual angle:
\( \angle A = \frac{180^\circ}{6} = 30^\circ \)
\( \angle B = \frac{180^\circ}{3} = 60^\circ \)
\( \angle C = \frac{180^\circ}{2} = 90^\circ \)
Therefore, the three angles of the triangle are \( 30^\circ \), \( 60^\circ \), and \( 90^\circ \).
In simple words: Setting the ratios equal to k shows the angles are k/6, k/3, and k/2. Since their sum is 180 degrees, k equals 180, giving angles of 30, 60, and 90 degrees.

Exam Tip: Setting equal ratios to a single constant \( k \) is the cleanest method to express all three angles with one variable.

Question 20. AOB is a straight line. If OP and OQ are bisectors of ∠AOC and ∠BOC respectively. Prove that ∠POQ = 90°.
Answer: Since \( AOB \) is a straight line, ray \( OC \) stands on it to form a linear pair of angles:
\( \angle AOC + \angle BOC = 180^\circ \)
Multiplying the entire equation by \( \frac{1}{2} \):
\( \frac{1}{2}\angle AOC + \frac{1}{2}\angle BOC = 90^\circ \)
We are given that ray \( OP \) bisects \( \angle AOC \) and ray \( OQ \) bisects \( \angle BOC \):
\( \angle POC = \frac{1}{2}\angle AOC \)
\( \angle QOC = \frac{1}{2}\angle BOC \)
Substituting these expressions into the halved linear pair equation:
\( \angle POC + \angle QOC = 90^\circ \)
From the geometry of adjacent angles sharing ray \( OC \):
\( \angle POQ = \angle POC + \angle QOC \)
\( \implies \angle POQ = 90^\circ \)
Hence proved.
In simple words: The two big angles make a straight line of 180 degrees. Taking half of each big angle gives half of 180, so the combined angle between their bisectors is exactly 90 degrees.

Exam Tip: The angle between the bisectors of a linear pair is always a right angle; state \( \angle POQ = \frac{1}{2}(\angle AOC + \angle BOC) = \frac{1}{2}(180^\circ) = 90^\circ \) clearly.