RBSE Solutions Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Exercise 5.1

Question 1. If angles of magnitude \( (2x + 4) \) and \( (x - 1) \) form a linear pair, find these angles.
Answer: According to the problem, the sum of a linear pair of angles is 180 degrees.
So, \( (2x + 4)° + (x - 1)° = 180° \) (This is based on the linear pair axiom.)
\( 3x + 3 = 180 \)
\( 3x = 180 - 3 \)
\( 3x = 177 \)
\( x = \frac { 177 }{ 3 } \)
\( x = 59 \)
Now, we find the magnitude of each angle:
First angle: \( (2 \times 59 + 4)° = (118 + 4)° = 122° \)
Second angle: \( (59 - 1)° = 58° \)
So, the two angles are \( 122° \) and \( 58° \). These two angles add up to 180 degrees, confirming they form a linear pair.
In simple words: When two angles form a straight line, they add up to 180 degrees. We use this fact to find the value of 'x' and then calculate each angle.

🎯 Exam Tip: Remember that a linear pair of angles always sums to 180 degrees. This property is crucial for solving such problems.

 

Question 2. From the given figure:
(i) Find the magnitude of \( \angle BOD \)
(ii) Find the magnitude of \( \angle AOD \)
(iii) Write the pair of vertically opposite angles.
(iv) Name the adjacent supplementary angles of \( \angle AOC \)

Answer:
(i) From the figure, we can see that \( \angle COB \) and \( \angle AOD \) are vertically opposite angles. Also, \( \angle COB \) and \( \angle AOD \) are supplementary angles forming a linear pair with \( \angle AOC \) and \( \angle BOD \) respectively. Since \( \angle COB = 52° \) (given in figure), and \( \angle COB \) and \( \angle AOD \) are vertically opposite angles, then \( \angle AOD = \angle COB \).
So, \( \angle AOD = 52° \).
Angles \( \angle COB \) and \( \angle BOD \) form a linear pair, so their sum is \( 180° \).
\( \angle COB + \angle BOD = 180° \)
\( 52° + \angle BOD = 180° \)
\( \angle BOD = 180° - 52° \)
\( \angle BOD = 128° \)
(ii) As explained above, \( \angle AOD \) and \( \angle COB \) are vertically opposite angles. Vertically opposite angles are equal.
So, \( \angle AOD = \angle COB = 52° \)
(iii) The pairs of vertically opposite angles are: \( (\angle AOC, \angle BOD) \) and \( (\angle AOD, \angle BOC) \).
(iv) The adjacent supplementary angles of \( \angle AOC \) are \( \angle AOD \) and \( \angle BOC \). These pairs form a straight line, meaning they add up to 180 degrees.
In simple words: Vertically opposite angles are always equal. Adjacent angles that form a straight line are supplementary and add up to 180 degrees. We use these rules to find missing angles and identify pairs.

🎯 Exam Tip: Clearly identify linear pairs and vertically opposite angles from the figure as this is key to solving problems involving intersecting lines.

 

Question 3. In the given figure, if \( \angle PQR = \angle PRQ \) then prove that \( \angle PQS = \angle PRT \).

Answer: We are given that \( \angle PQR = \angle PRQ \) ...(i)
From the figure, we observe the following relationships:
Angles \( \angle PQS \) and \( \angle PQR \) form a linear pair on the straight line SQT.
Therefore, \( \angle PQS + \angle PQR = 180° \) (This is the linear pair axiom.) ...(ii)
Similarly, angles \( \angle PRT \) and \( \angle PRQ \) form a linear pair on the straight line SQT.
Therefore, \( \angle PRT + \angle PRQ = 180° \) (This is also the linear pair axiom.) ...(iii)
Now, we can compare equations (ii) and (iii):
\( \angle PQS + \angle PQR = \angle PRT + \angle PRQ \)
Since we are given \( \angle PQR = \angle PRQ \) from equation (i), we can substitute or subtract \( \angle PQR \) from both sides:
\( \angle PQS = \angle PRT \)
Hence, the proof is complete. This shows that if the base angles of an isosceles triangle are equal, then the exterior angles formed by extending the base are also equal.
In simple words: If the inside angles of a triangle are equal, and you extend the sides to make outside angles, those outside angles will also be equal. This works because angles on a straight line always add up to 180 degrees.

🎯 Exam Tip: When proving angle relationships, always clearly state the geometric axioms or theorems you are using, such as "linear pair axiom" or "vertically opposite angles."

 

Question 4. In figure OP, OQ, OR and OS are four rays. Prove that \( \angle POQ + \angle QOR + \angle SOR + \angle POS = 360° \).

Answer: To prove that the sum of angles around a point is 360°, we can draw a line extending one of the rays.
Let's produce ray OQ backwards to a point T, so that TOQ becomes a straight line, as shown in the figure. A straight line always forms an angle of 180 degrees.
Now, consider the ray OP standing on the line TOQ.
Therefore, \( \angle TOP + \angle POQ = 180° \) (This is based on the linear pair axiom.) ...(i)
Similarly, consider the ray OS standing on the line TOQ.
Therefore, \( \angle TOS + \angle SOQ = 180° \) (This is also the linear pair axiom.) ...(ii)
From the figure, we can see that \( \angle SOQ \) is made up of two smaller angles: \( \angle SOR \) and \( \angle QOR \).
So, \( \angle SOQ = \angle SOR + \angle QOR \).
Substitute this into equation (ii):
\( \angle TOS + (\angle SOR + \angle QOR) = 180° \) ...(iii)
Now, add equations (i) and (iii):
\( (\angle TOP + \angle POQ) + (\angle TOS + \angle SOR + \angle QOR) = 180° + 180° \)
\( \angle TOP + \angle POQ + \angle TOS + \angle SOR + \angle QOR = 360° \) ...(iv)
Observe from the figure that \( \angle TOP + \angle TOS \) together form \( \angle POS \).
So, \( \angle POS = \angle TOP + \angle TOS \).
Substitute this into equation (iv):
\( \angle POQ + \angle QOR + \angle SOR + \angle POS = 360° \)
Hence, it is proven that the sum of all angles around a point is 360 degrees.
In simple words: If you add up all the angles that meet at a single point, they will always sum up to 360 degrees, just like a full circle. We can show this by drawing a line through the point and using the rule that angles on a straight line add to 180 degrees.

🎯 Exam Tip: When proving angle sums around a point, constructing an auxiliary line is often helpful. Remember that angles on a straight line sum to 180 degrees.

 

Question 5. In the given figure, if \( \angle x + \angle y = \angle p + \angle q \), then prove that AOB is a line.

Answer: We need to prove that AOB is a straight line. A line forms a 180-degree angle.
We know that the sum of all angles around a point is 360 degrees.
So, in the given figure, for the point O:
\( \angle x + \angle y + \angle p + \angle q = 360° \)
We are given the condition: \( \angle x + \angle y = \angle p + \angle q \).
Substitute this condition into the sum of angles equation:
\( (\angle x + \angle y) + (\angle x + \angle y) = 360° \)
\( 2 (\angle x + \angle y) = 360° \)
Divide both sides by 2:
\( \angle x + \angle y = \frac{360°}{2} \)
\( \angle x + \angle y = 180° \)
Since the sum of adjacent angles \( \angle x \) and \( \angle y \) is \( 180° \), it means they form a linear pair. This implies that the line segment AOB must be a straight line. If angles add up to 180 degrees, they sit on a straight line.
In simple words: If the angles on one side of a point add up to the same as the angles on the other side, and all angles around the point make 360 degrees, then the parts forming those two sets of angles must make a straight line. This is because a straight line is 180 degrees.

🎯 Exam Tip: To prove that a ray or segment is a line, you need to show that the sum of angles forming it is 180 degrees. The sum of angles around a point (360 degrees) is a fundamental concept to use here.