NCERT Class 9 Maths Exercise 6.1 Solutions

📘 Concept & Theory Concept Used ›

Linear Pair: Two adjacent angles forming a straight line add up to \(\small 180^\circ\).
Angle Addition Property: A larger angle can be expressed as the sum of smaller adjacent angles.
Reflex Angle: Reflex angle \(\small =360^\circ-\text{smaller angle}\)

🗺️ Solution Roadmap Step-by-step Plan ›

Use linear pair property to determine \(\small \angle COE\).
Use straight line property on line CD to determine \(\small \angle EOD\).
Use angle addition property to calculate \(\angle BOE\).
Find reflex \(\angle COE\).

📊 Graph / Figure Graph / Figure ›

Fig. 6.1 : Intersecting lines and ray OE

✏️ Solution Complete Solution ›

Step-by-step Solution · 33 steps

Given
\[\small \angle AOC+\angle BOE=70^\circ\]
\[\small \angle BOD=40^\circ\]
To Find
\[\small \angle BOE\] and reflex \[\small \angle COE\]
Solution
Since AB is a straight line, \(\small \angle AOC\) and \(\small \angle COB\) form a linear pair.
\[\small \angle AOC+\angle COB=180^\circ \tag{1}\]
Now ray OE lies inside angle COB.
\[\small \angle COB=\angle COE+\angle BOE \tag{2}\]
Substitute equation (2) into equation (1):
\[\small \angle AOC+\angle COE+\angle BOE=180^\circ\]
But
\[\small \angle AOC+\angle BOE=70^\circ \text{: Given}\]
Therefore,
\[\small 70^\circ+\angle COE=180^\circ\]
\[\small \angle COE=180^\circ-70^\circ\]
\[\small \angle COE=110^\circ\]
Now C, O and D lie on a straight line. Therefore \(\small \angle COE\) and \(\small \angle EOD\) form a linear pair.
\[\small \angle COE+\angle EOD=180^\circ\]
Substitute \(\small \angle COE=110^\circ\):
\[\small 110^\circ+\angle EOD=180^\circ\]
\[\small \angle EOD=180^\circ-110^\circ\]
\[\small \angle EOD=70^\circ\]
Angle EOD is made up of two adjacent angles:
\[\small \angle EOD=\angle BOE+\angle BOD\]
Substitute \(\angle BOD=40^\circ\):
\[\small 70^\circ=\angle BOE+40^\circ\]
\[\small \angle BOE=70^\circ-40^\circ\]
\[\small \angle BOE=30^\circ\]
Now find reflex \(\small \angle COE\)
\[\small \text{Reflex }\angle COE = 360^\circ-110^\circ\]
\[\small =250^\circ\]

💡 Answer Final Answer ›

Final Answer: \[ \boxed{\angle BOE=30^\circ} \] \[ \boxed{\text{Reflex }\angle COE=250^\circ} \]

🎯 Exam Significance Exam Significance ›

This question strengthens understanding of linear pairs and angle relationships.
Such problems are frequently asked in CBSE Board examinations.
Intersecting line problems form the foundation for advanced geometry chapters like triangles and parallel lines.
Competitive exams such as NTSE, Olympiads and Polytechnic entrance tests often include angle reasoning questions of this type.

📘 Concept & Theory Concept used ›

Vertically Opposite Angles: When two lines intersect, vertically opposite angles are equal.
Linear Pair: Adjacent angles on a straight line add up to \(\small 180^\circ\).
Ratio Distribution: If two quantities are in the ratio \(\small 2:3\), then total parts \(\small =2+3=5\).

🗺️ Solution Roadmap Step-by-step Plan ›

Use linear pair property to form an equation involving \(\small a\) and \(\small b\).
Use the given ratio \(\small a:b=2:3\) to calculate values of \(\small a\) and \(\small b\).
Use angle addition to calculate angle \(c\).

📊 Graph / Figure Graph / Figure ›

Fig. 6.2 : Intersecting lines with a right angle

✏️ Solution Complete Solution ›

Step-by-step Solution · 31 steps

Given
\[\small \angle POY=90^\circ\]
\[\small a:b=2:3\]
To Find
\[c\]
Solution
Lines XY and MN intersect each other at point O.
Therefore vertically opposite angles are equal.
\[\small \angle XON=\angle MOY\]
Now \(\small\angle MOY\) and \(\small\angle XOM\) form a linear pair.
\[\small \angle MOY+\angle XOM=180^\circ \tag{1}\]
Angle MOY is made up of two adjacent angles:
\[\small \angle MOY=\angle MOP+\angle POY\]
Substitute into equation (1):
\[\small \angle MOP+\angle POY+\angle XOM=180^\circ\]
\[\small \angle POY=90^\circ \text{: Given}\]
Therefore,
\[\small \angle MOP+90^\circ+\angle XOM=180^\circ\]
\[\small \angle MOP+\angle XOM=180^\circ-90^\circ\]
\[\small \angle MOP+\angle XOM=90^\circ\]
From the figure:
\[\small \angle MOP=a\] and \[\small \angle XOM=b\]
Therefore, \[\small a+b=90^\circ \tag{2}\]
Given ratio:\[\small a:b=2:3\]
Total parts \[\small 2+3=5\]
Therefore, \[\small \begin{aligned}a&=\frac{2}{5}\times90^\circ\\a&=36^\circ\end{aligned}\]
Similarly,\[\small \begin{aligned}b&=\frac{3}{5}\times90^\circ\\b&=54^\circ\end{aligned}\]
Now angle \(\small c\) is vertically opposite to angle XON.
\[\small c=\angle XON\]
Angle XON consists of:\[\small \angle XON=\angle MOP+\angle POY\]
Substitute values:\[\small \begin{aligned}c&=36^\circ+90^\circ\\c&=126^\circ\end{aligned}\]

💡 Answer Final Answer ›

Final Answer: \( \boxed{c=126^\circ} \)

🎯 Exam Significance Exam Significance ›

This problem develops understanding of vertically opposite angles and linear pair relationships.
Questions involving angle ratios are commonly asked in CBSE Board examinations.
Such geometry reasoning questions are important for NTSE, Olympiads and other competitive entrance exams.
This question also improves algebraic handling of ratio-based geometry problems.

📘 Concept & Theory Concept Used ›

Linear Pair Axiom: If two adjacent angles form a straight line, then their sum is \(180^\circ\).
Equality Property: If equal quantities are subtracted from equal quantities, the remaining quantities are also equal.
This question connects interior angles and exterior angles.

🗺️ Solution Roadmap Step-by-step Plan ›

Use linear pair property at point Q.
Use linear pair property at point R.
Compare both equations.
Use the given condition \(\small\angle PQR=\angle PRQ\) to prove the required result.

📊 Graph / Figure Graph / Figure ›

Fig. 6.3 : Equal base angles and exterior angles

✏️ Solution Complete Solution ›

Step-by-step Solution · 16 steps

Given\[\small \angle PQR=\angle PRQ\]
To Prove\[\small \angle PQS=\angle PRT\]
Proof
Points S, Q, R and T lie on the same straight line.
At point Q, line PQ stands on straight line SQR. Therefore, \(\small \angle PQS\) and \(\angle PQR\) form a linear pair.
\[\small \angle PQS+\angle PQR=180^\circ \tag{1} \]
Similarly, at point R, line PR stands on straight line QRT. Therefore, \(\small \angle PRT\) and \(\small \angle PRQ\) form a linear pair.
\[\small \angle PRT+\angle PRQ=180^\circ \tag{2}\]
Since the right-hand side of equations (1) and (2) are equal to \(\small 180^\circ\), equate the left-hand sides:
\[\small \angle PQS+\angle PQR = \angle PRT+\angle PRQ \]
\[\small \angle PQR=\angle PRQ\quad\text{Given}\]
Substitute equal angles:
\[\small \angle PQS+\angle PQR = \angle PRT+\angle PQR \]
Subtract \(\angle PQR\) from both sides:
\[\small \angle PQS=\angle PRT\]
Hence Proved

🎯 Exam Significance Exam Significance ›

This is an important geometry proof question based on linear pair properties.
Proof-based questions are frequently asked in CBSE Board examinations.
The problem strengthens logical reasoning and theorem application skills.
Similar proofs are commonly seen in Olympiads, NTSE and other competitive entrance exams.
Understanding such proofs helps in advanced topics involving triangles, parallel lines and transversals.

📘 Concept & Theory Concept Used ›

Angles Around a Point: The sum of all angles around a point is \(\small 360^\circ\).
Straight Angle: If two adjacent angles together measure \(\small 180^\circ\), then they form a straight line.
Algebraic Reasoning in Geometry: Unknown angle expressions can be solved using equations.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the property that the sum of angles around a point is \(\small 360^\circ\).
Use the given condition \(\small x+y=w+z\).
Represent both sums by a common variable.
Show that \(\small x+y=180^\circ\).
Conclude that AOB is a straight line.

📊 Graph / Figure Graph / Figure ›

Fig. 6.4 : Angles around point O

✏️ Solution Complete Solution ›

Step-by-step Solution · 13 steps

Given\[\small x+y=w+z\]
To Prove
AOB is a straight line.

For AOB to be a straight line, we must prove:
\[\small x+y=180^\circ \]
Proof
The sum of all angles around point O is \(\small 360^\circ\).
\[\small x+y+w+z=360^\circ \tag{1}\]
\[\small x+y=w+z\quad\text{Given}\]
Let\[\small x+y=w+z=k\]
Substitute into equation (1): \[\begin{aligned}\small k+k&=360^\circ\\ 2k&&=360^\circ\\ k&=\frac{360^\circ}{2}\\ k=180^\circ\end{aligned}\]
Since, \[\small x+y=k\]
therefore, \[\small x+y=180^\circ\]
Also,\[\small w+z=180^\circ\]
Thus the angle AOB is a straight angle.
Hence Proved

🎯 Exam Significance Exam Significance ›

This problem develops understanding of angles around a point.
It combines geometry with algebraic equations, which is a key skill in mathematics.
Such proof-based questions are frequently asked in CBSE Board examinations.
Similar reasoning problems appear in NTSE, Olympiads and other entrance examinations.
Understanding straight angle conditions is useful in advanced geometry and theorem proving.

Q5

NUMERIC2 marks

In Fig. 6.5, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

\[\small \angle ROS= \frac{1}{2} \left( \angle QOS-\angle POS \right) \]

📘 Concept & Theory concpet Used ›

Perpendicular Lines: If two lines are perpendicular, they form a right angle of \(\small 90^\circ\).
Angle Addition Property: A larger angle can be expressed as the sum of adjacent smaller angles.
Linear Pair: Angles on a straight line add up to \(\small 180^\circ\).

🗺️ Solution Roadmap Step-by-step Plan ›

Use perpendicular property to determine right angles.
Express \(\small 90^\circ\) using angles \(\small \angle ROS\) and \(\small \angle POS\).
Express another \(90^\circ\) using \(\small \angle QOS\) and \(\small \angle ROS\).
Equate both expressions and simplify.

📊 Graph / Figure Graph / Figure ›

Fig. 6.5 : Perpendicular rays and angle relations

✏️ Solution Complete Solution ›

Step-by-step Solution · 16 steps

Given
POQ is a straight line.
\[ OR\perp PQ \]
Ray OS lies between OP and OR.
To Prove \[\small \angle ROS= \frac{1}{2} \left( \angle QOS-\angle POS \right) \]
Proof
Since \(\small OR\perp PQ\), therefore \(\small \angle ROP\) and \(\small \angle ROQ\) are right angles.
\[\small \angle ROP=90^\circ \] and \[\small \angle ROQ=90^\circ \]
Ray OS lies between OP and OR. Therefore,\[\small \angle ROS+\angle POS=\angle ROP\]
Substitute \(\small\angle ROP=90^\circ\): \[\small \angle ROS+\angle POS=90^\circ \tag{1} \]
Also, ray OR lies between OQ and OS.\[\small \angle QOS=\angle QOR+\angle ROS\]
Since \[\small \angle QOR=90^\circ \]
therefore, \[\small \angle QOS=90^\circ+\angle ROS \]
Rearranging: \[\small \angle QOS-\angle ROS=90^\circ \tag{2} \]
From equations (1) and (2): \[\small \angle ROS+\angle POS = \angle QOS-\angle ROS \]
Add \(\small\angle ROS\) to both sides: \[\small 2\angle ROS+\angle POS=\angle QOS \]
Subtract \(\small \angle POS\) from both sides: \[\small 2\angle ROS = \angle QOS-\angle POS \]
Divide both sides by \(2\): \[\small \angle ROS= \frac{1}{2} \left( \angle QOS-\angle POS \right) \]
Hence Proved.

🎯 Exam Significance Exam Significance ›

This question develops understanding of perpendicular lines and angle relationships.
It strengthens proof-writing skills required in CBSE Board examinations.
The problem demonstrates how algebraic manipulation is applied in geometry proofs.
Similar proof-based geometry problems are commonly asked in NTSE, Olympiads and entrance examinations.
Mastery of such angle identities is useful in higher geometry and trigonometry.

Q6

NUMERIC2 marks

It is given that \(\small \angle XYZ = 64^\circ\) and \(\small XY\) is produced to point \(\small P\). Draw a figure from the given information. If ray \(\small YQ\) bisects \(\small \angle ZYP\), find \(\small \angle XYQ\) and reflex \(\small \angle QYP\).

📘 Concept & Theory Concpet Used ›

Linear Pair: Adjacent angles on a straight line add up to \(180^\circ\).
Angle Bisector: A bisector divides an angle into two equal parts.
Reflex Angle: Reflex angle \(=360^\circ-\text{smaller angle}\).

🗺️ Solution Roadmap Step-by-step Plan ›

Use linear pair property to calculate \(\small \angle ZYP\).
Since ray \(\small YQ\) bisects \(\small \angle ZYP\), divide the angle into two equal parts.
Calculate \(\small\angle XYQ\).
Find reflex \(\small\angle QYP\).

📊 Graph / Figure Graph / Figure ›

Fig. 6.6 : Angle bisector and reflex angle

✏️ Solution Complete Solution ›

Step-by-step Solution · 15 steps

Given \[\small \angle XYZ=64^\circ \]
Line \(\small XY\) is produced to point \(\small P\).

Ray \(\small YQ\) bisects \(\small \angle ZYP\).
To Find \[\small \angle XYQ \] and reflex \[\small \angle QYP \]
Solution Since line \(\small XY\) is produced to point \(\small P\), angles \(\small \angle XYZ\) and \(\small \angle ZYP\) form a linear pair.
\[\small \angle XYZ+\angle ZYP=180^\circ\]
Substitute \(\small\angle XYZ=64^\circ\):
\[\small \begin{aligned} 64^\circ+\angle ZYP&=180^\circ\\ \angle ZYP&=180^\circ-64^\circ\\ \angle ZYP=&116^\circ \end{aligned} \]
Ray \(\small YQ\) bisects \(\small \angle ZYP\). Therefore,\[\small \angle ZYQ=\angle QYP\]
Let\[\small \angle ZYQ=\angle QYP=x\]
Then, \[\small \begin{aligned} x+x&=116^\circ\\ 2x&=116^\circ\\ x&=\frac{116^\circ}{2}\\ x&=58^\circ \end{aligned} \]
Therefore, \[\small \angle ZYQ=58^\circ \] and \[\small \angle QYP=58^\circ \]
Now, \(\small \angle XYQ\) consists of \(\small \angle XYZ\) and \(\angle ZYQ\).
\[\small \angle XYQ = \angle XYZ+\angle ZYQ \]
Substitute values: \[\small \angle XYQ = 64^\circ+58^\circ \] \[\small \angle XYQ=122^\circ \]
Finding Reflex \(\small\angle QYP\)
\[\small\begin{aligned} \text{Reflex }\angle QYP &=360^\circ-58^\circ\\ &=302^\circ\end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \( \boxed{\angle XYQ=122^\circ} \) \( \boxed{\text{Reflex }\angle QYP=302^\circ} \)

🎯 Exam Significance Exam Significance ›

This problem strengthens understanding of linear pairs, angle bisectors and reflex angles.
Such angle-based numerical problems are common in CBSE Board examinations.
It improves diagram interpretation and logical geometry reasoning.
Similar questions are frequently asked in NTSE, Olympiads and entrance examinations.
Understanding angle bisectors is important for higher geometry topics involving triangles and constructions.

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