NCERT Solutions for Class 9 Maths Lines and Angles Exercise 6.3

Core Geometric Theorems in Exercise 6.3

To solve any problem in lines and angles ex 6 3 class 9 ncert, you must master two key theorems:

1. Angle Sum Property of a Triangle (Theorem 6.7): This theorem states that the sum of all interior angles of a triangle is equal to $180^\circ$. If we have $\Delta ABC$, then $\angle A + \angle B + \angle C = 180^\circ$.

1. Exterior Angle Theorem (Theorem 6.8): If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles. For example, if side $BC$ of $\Delta ABC$ is extended to $D$, then the exterior angle $\angle ACD = \angle CAB + \angle ABC$.

These theorems allow us to translate geometric configurations into simple algebraic equations. Every question in Exercise 6.3 is a direct or indirect application of these two principles, combined with previously learned concepts like linear pairs and vertically opposite angles.

Proof of the Angle Sum Property (Theorem 6.7)

1. State the Given and to Prove — Given: A triangle $PQR$ with interior angles $\angle 1$, $\angle 2$, and $\angle 3$. To Prove: $\angle 1 + \angle 2 + \angle 3 = 180^\circ$.
2. Perform Construction — Draw a line $XPY$ parallel to side $QR$ passing through the vertex $P$. Let the angles formed by line $XY$ with sides $PQ$ and $PR$ be $\angle 4$ and $\angle 5$ respectively.
3. Apply Parallel Line Axioms — Since $XPY$ is a straight line, the sum of angles on a straight line at point $P$ is $180^\circ$. Therefore, $\angle 4 + \angle 1 + \angle 5 = 180^\circ$ (Equation 1). Now, since line $XPY \parallel QR$ and $PQ$, $PR$ act as transversals: $\angle 4 = \angle 2$ (Alternate interior angles) and $\angle 5 = \angle 3$ (Alternate interior angles).
4. Substitute and Conclude — Substitute the values of $\angle 4$ and $\angle 5$ into Equation 1. This gives: $\angle 2 + \angle 1 + \angle 3 = 180^\circ$, which simplifies directly to $\angle 1 + \angle 2 + \angle 3 = 180^\circ$. Hence proved!

Common Exam Mistakes & Tips for Ex 6.3

• Confusing Exterior Angles: Always make sure you identify the correct opposite interior angles. For an exterior angle at vertex $C$, the interior opposite angles are at vertices $A$ and $B$, not the adjacent interior angle at $C$.
• Missing Reasons: CBSE board examiners deduct marks if you do not write reasons like 'Linear Pair Axiom', 'Angle Sum Property', or 'Alternate Interior Angles' in brackets next to your equations. Always state your reasons!
• Incorrect Angle Bisector Relations: If $OX$ bisects $\angle YXZ$, then $\angle OXY = \angle OXZ = \frac{1}{2} \angle YXZ$. Do not assume it bisects any other angle or side unless explicitly stated.

Practice Questions with Solutions

• Q: In $\Delta PQR$, sides $QP$ and $RQ$ are produced to points $S$ and $T$ respectively. If $\angle SPR = 135^\circ$ and $\angle PQT = 110^\circ$, find $\angle PRQ$. A: Step 1: Identify linear pairs. The ray $QP$ stands on line $QS$. Therefore, $\angle SPR + \angle QPR = 180^\circ$ (Linear pair). Step 2: Calculate $\angle QPR$. We have $135^\circ + \angle QPR = 180^\circ \implies \angle QPR = 180^\circ - 135^\circ = 45^\circ$. Step 3: Similarly, ray $QP$ stands on line $TR$. Therefore, $\angle PQT + \angle PQR = 180^\circ$ (Linear pair). Step 4: Calculate $\angle PQR$. We have $110^\circ + \angle PQR = 180^\circ \implies \angle PQR = 180^\circ - 110^\circ = 70^\circ$. Step 5: Apply the Angle Sum Property in $\Delta PQR$. $\angle QPR + \angle PQR + \angle PRQ = 180^\circ$. Step 6: Substitute the known values: $45^\circ + 70^\circ + \angle PRQ = 180^\circ \implies 115^\circ + \angle PRQ = 180^\circ \implies \angle PRQ = 180^\circ - 115^\circ = 65^\circ$. Final answer: $\angle PRQ = 65^\circ$.
• Q: In $\Delta XYZ$, $\angle X = 62^\circ$ and $\angle XYZ = 54^\circ$. If $YO$ and $ZO$ are the bisectors of $\angle XYZ$ and $\angle XZY$ respectively of $\Delta XYZ$, find $\angle OZY$ and $\angle YOZ$. A: Step 1: Use the Angle Sum Property on the larger triangle $\Delta XYZ$. We know $\angle X + \angle XYZ + \angle XZY = 180^\circ$. Step 2: Substitute values: $62^\circ + 54^\circ + \angle XZY = 180^\circ \implies 116^\circ + \angle XZY = 180^\circ \implies \angle XZY = 180^\circ - 116^\circ = 64^\circ$. Step 3: Since $YO$ bisects $\angle XYZ$, we get $\angle OZY = \frac{1}{2} \angle XYZ$ (Wait, $ZO$ bisects $\angle XZY$). Let's find both: $\angle OYZ = \frac{1}{2} \times 54^\circ = 27^\circ$, and $\angle OZY = \frac{1}{2} \angle XZY = \frac{1}{2} \times 64^\circ = 32^\circ$. Step 4: Use the Angle Sum Property in the smaller triangle $\Delta YOZ$. $\angle YOZ + \angle OYZ + \angle OZY = 180^\circ$. Step 5: Substitute the values: $\angle YOZ + 27^\circ + 32^\circ = 180^\circ \implies \angle YOZ + 59^\circ = 180^\circ \implies \angle YOZ = 180^\circ - 59^\circ = 121^\circ$. Final answer: $\angle OZY = 32^\circ$ and $\angle YOZ = 121^\circ$.
• Q: If lines $PQ$ and $RS$ intersect at point $T$, such that $\angle PRT = 40^\circ$, $\angle RPT = 95^\circ$ and $\angle TSQ = 75^\circ$, find $\angle SQT$. A: Step 1: Apply the Angle Sum Property in $\Delta PRT$. $\angle PRT + \angle RPT + \angle PTR = 180^\circ$. Step 2: Substitute the known values: $40^\circ + 95^\circ + \angle PTR = 180^\circ \implies 135^\circ + \angle PTR = 180^\circ \implies \angle PTR = 180^\circ - 135^\circ = 45^\circ$. Step 3: Use vertically opposite angles. Since lines $PQ$ and $RS$ intersect at $T$, $\angle STQ = \angle PTR = 45^\circ$. Step 4: Apply the Angle Sum Property in $\Delta TSQ$. $\angle STQ + \angle TSQ + \angle SQT = 180^\circ$. Step 5: Substitute values: $45^\circ + 75^\circ + \angle SQT = 180^\circ \implies 120^\circ + \angle SQT = 180^\circ \implies \angle SQT = 180^\circ - 120^\circ = 60^\circ$. Final answer: $\angle SQT = 60^\circ$.
• Q: In $\Delta PQR$, side $QR$ is produced to a point $S$. If the bisectors of $\angle PQR$ and $\angle PRS$ meet at point $T$, prove that $\angle QTR = \frac{1}{2} \angle QPR$. A: Step 1: Identify exterior angles. In $\Delta QTR$, exterior $\angle TRS = \angle TQR + \angle QTR$ (Theorem 6.8). Therefore, $\angle QTR = \angle TRS - \angle TQR$ (Equation 1). Step 2: Similarly, in $\Delta PQR$, exterior $\angle PRS = \angle PQR + \angle QPR$ (Theorem 6.8). Therefore, $\angle QPR = \angle PRS - \angle PQR$ (Equation 2). Step 3: Use the angle bisector properties. We are given that $QT$ and $RT$ are bisectors of $\angle PQR$ and $\angle PRS$ respectively. Thus, $\angle PQR = 2 \angle TQR$ and $\angle PRS = 2 \angle TRS$. Step 4: Substitute these bisector values into Equation 2: $\angle QPR = 2 \angle TRS - 2 \angle TQR = 2(\angle TRS - \angle TQR)$. Step 5: Since $(\angle TRS - \angle TQR) = \angle QTR$ from Equation 1, substitute it here: $\angle QPR = 2 \angle QTR$. Step 6: Rearrange the equation: $\angle QTR = \frac{1}{2} \angle QPR$. Hence Proved. Final answer: Hence proved that $\angle QTR = \frac{1}{2} \angle QPR$.

Frequently Asked Questions