GSEB Class 9 Maths Solutions Chapter 6 Lines and Angles Exercise 6.3

Question 1. In figure sides QP and RQ of \( \triangle PQR \) are produced to points S and T respectively. If \( \angle SPR = 135^\circ \) and \( \angle PQT = 110^\circ \), find \( \angle PRQ \).
Answer:

We know that \( \angle PQT + \angle PQR = 180^\circ \) because they form a linear pair.
\( \implies 110^\circ + \angle PQR = 180^\circ \)
\( \implies \angle PQR = 180^\circ - 110^\circ \)
\( \implies \angle PQR = 70^\circ \)
Also, we use the exterior angle property of a triangle, which states that an exterior angle is equal to the sum of its two interior opposite angles.
So, \( \angle RPS = \angle PQR + \angle PRQ \)
\( \implies 135^\circ = 70^\circ + \angle PRQ \)
\( \implies \angle PRQ = 135^\circ - 70^\circ \)
\( \implies \angle PRQ = 65^\circ \)
In simple words: We find the angle inside the triangle next to 110 degrees first. Then, we use the rule that an outside angle equals the sum of the two opposite inside angles to find the missing angle.

Exam Tip: Remember the properties of linear pairs and the exterior angle theorem. Drawing the figure and marking the given angles helps visualize the problem.

Question 2. In figure \( \angle X = 62^\circ, \angle XYZ = 54^\circ \). If YO and ZO are the bisectors of \( \angle XYZ \) and \( \angle XZY \) respectively of \( \triangle XYZ \), find \( \angle OZY \) and \( \angle YOZ \).
Answer:

In \( \triangle XYZ \), the sum of angles is \( 180^\circ \) (Angle sum property).
So, \( \angle YXZ + \angle XYZ + \angle XZY = 180^\circ \)
\( \implies 62^\circ + 54^\circ + \angle XZY = 180^\circ \)
\( \implies 116^\circ + \angle XZY = 180^\circ \)
\( \implies \angle XZY = 180^\circ - 116^\circ \)
\( \implies \angle XZY = 64^\circ \)
Since ZO is the bisector of \( \angle XZY \), it divides the angle into two equal parts.
\( \angle OZY = \frac{\angle XZY}{2} \)
\( \implies \angle OZY = \frac{64^\circ}{2} \)
\( \implies \angle OZY = 32^\circ \)
Since YO is the bisector of \( \angle XYZ \).
\( \angle OYZ = \frac{\angle XYZ}{2} \)
\( \implies \angle OYZ = \frac{54^\circ}{2} \)
\( \implies \angle OYZ = 27^\circ \)
Now, consider \( \triangle OYZ \). The sum of angles in a triangle is \( 180^\circ \).
\( \angle OYZ + \angle OZY + \angle YOZ = 180^\circ \)
\( \implies 27^\circ + 32^\circ + \angle YOZ = 180^\circ \)
\( \implies 59^\circ + \angle YOZ = 180^\circ \)
\( \implies \angle YOZ = 180^\circ - 59^\circ \)
\( \implies \angle YOZ = 121^\circ \)
In simple words: First, we find the third angle of the big triangle using the angle sum property. Then, we split the two known angles in half because of the bisectors. Finally, we use the angle sum property again in the smaller triangle to find the last missing angle.

Exam Tip: Remember that an angle bisector divides an angle into two equal halves. The sum of angles in any triangle is always \( 180^\circ \).

Question 3. In figure AB || DE, \( \angle BAC = 35^\circ \) and \( \angle CDE = 53^\circ \), find \( \angle DCE \).
Answer:

Given that AB || DE, we know that alternate interior angles are equal.
So, \( \angle DEC = \angle BAC \)
\( \implies \angle DEC = 35^\circ \)
Now, consider \( \triangle CDE \). The sum of angles in a triangle is \( 180^\circ \) (Angle sum property).
\( \angle DCE + \angle CDE + \angle DEC = 180^\circ \)
\( \implies \angle DCE + 53^\circ + 35^\circ = 180^\circ \)
\( \implies \angle DCE + 88^\circ = 180^\circ \)
\( \implies \angle DCE = 180^\circ - 88^\circ \)
\( \implies \angle DCE = 92^\circ \)
In simple words: Since two lines are parallel, we can use the rule about alternate interior angles to find one missing angle. Then, we use the fact that all angles inside a triangle add up to 180 degrees to find the last angle.

Exam Tip: Clearly identify parallel lines and transversals to correctly apply angle properties like alternate interior angles or corresponding angles. Always state the property used for each step.

Question 4. In figure, 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 \( \angle SQT \).
Answer:

In \( \triangle PRT \), the sum of angles is \( 180^\circ \) (Angle sum property).
\( \angle PRT + \angle RPT + \angle PTR = 180^\circ \)
\( \implies 40^\circ + 95^\circ + \angle PTR = 180^\circ \)
\( \implies 135^\circ + \angle PTR = 180^\circ \)
\( \implies \angle PTR = 180^\circ - 135^\circ \)
\( \implies \angle PTR = 45^\circ \)
When two lines intersect, vertically opposite angles are equal.
So, \( \angle STQ = \angle PTR \)
\( \implies \angle STQ = 45^\circ \)
Now, in \( \triangle SQT \), the sum of angles is \( 180^\circ \) (Angle sum property).
\( \angle STQ + \angle SQT + \angle TSQ = 180^\circ \)
\( \implies 45^\circ + \angle SQT + 75^\circ = 180^\circ \)
\( \implies \angle SQT + 120^\circ = 180^\circ \)
\( \implies \angle SQT = 180^\circ - 120^\circ \)
\( \implies \angle SQT = 60^\circ \)
In simple words: First, we find the third angle in the top triangle using the 180-degree rule. Since the lines cross, the angle we just found is equal to its opposite angle. Then, we use the 180-degree rule again in the bottom triangle to find the final missing angle.

Exam Tip: Remember that the sum of angles in a triangle is \( 180^\circ \) and vertically opposite angles are always equal when lines intersect. These are key concepts for solving such problems.

Question 5. In figure, if PQ \( \perp \) PS, PQ || SR, \( \angle SQR = 28^\circ \) and \( \angle QRT = 65^\circ \), then find the values of x and y.
Answer:

We are given that PQ || SR. Also, we can consider ST as a transversal line.
Since PQ || ST, we know that alternate interior angles are equal.
So, \( \angle PQR = \angle QTR \)
The angle \( \angle PQR \) is made of two parts: \( x \) and \( 28^\circ \).
\( \implies x + 28^\circ = 65^\circ \)
\( \implies x = 65^\circ - 28^\circ \)
\( \implies x = 37^\circ \)
Now, consider \( \triangle PQS \). The sum of angles in a triangle is \( 180^\circ \) (Angle sum property).
We are given that PQ \( \perp \) PS, which means \( \angle QPS = 90^\circ \).
So, \( \angle PQS + \angle PSQ + \angle SPQ = 180^\circ \)
\( \implies x + y + 90^\circ = 180^\circ \)
Substitute the value of \( x = 37^\circ \):
\( \implies 37^\circ + y + 90^\circ = 180^\circ \)
\( \implies 127^\circ + y = 180^\circ \)
\( \implies y = 180^\circ - 127^\circ \)
\( \implies y = 53^\circ \)
In simple words: First, we use the parallel lines property to find the value of x. Alternate interior angles are equal here. Then, we use the fact that the angles in the triangle PQS add up to 180 degrees, knowing that one angle is 90 degrees, to find the value of y.

Exam Tip: Always remember that the angle in a right-angled triangle at the right angle is \( 90^\circ \). Clearly identify parallel lines and their transversals to apply the correct angle properties.

Question 6. In \( \triangle PQR \), side QR is produced to a point S. If the bisectors of \( \angle PQR \) and \( \angle PRS \) meet at point T, then prove that \( \angle QTR = \frac {1}{2} \angle QPR \).
Answer:

Given a \( \triangle PQR \) where side QR is extended to S. The bisectors of \( \angle PQR \) and \( \angle PRS \) meet at point T.
**Proof:**
Consider \( \triangle TQR \). The exterior angle \( \angle TRS \) is equal to the sum of the two interior opposite angles.
\( \implies \angle TRS = \angle TQR + \angle QTR \)
\( \implies \angle QTR = \angle TRS - \angle TQR \) ... (1)
Now, consider \( \triangle PQR \). The exterior angle \( \angle PRS \) is equal to the sum of the two interior opposite angles.
\( \implies \angle PRS = \angle PQR + \angle QPR \)
We know that RT bisects \( \angle PRS \), so \( \angle TRS = \frac{1}{2} \angle PRS \).
Also, QT bisects \( \angle PQR \), so \( \angle TQR = \frac{1}{2} \angle PQR \).
Substitute these into the exterior angle property for \( \triangle PQR \):
\( 2 \angle TRS = 2 \angle TQR + \angle QPR \)
\( \implies \angle QPR = 2 \angle TRS - 2 \angle TQR \)
\( \implies \angle QPR = 2 (\angle TRS - \angle TQR) \)
From equation (1), we know that \( \angle QTR = \angle TRS - \angle TQR \).
Substitute this into the equation above:
\( \implies \angle QPR = 2 \angle QTR \)
\( \implies \angle QTR = \frac{1}{2} \angle QPR \)
Hence Proved.
In simple words: We use the exterior angle property for both the small triangle TQR and the big triangle PQR. Because QT and RT cut angles in half, we can link the angles of the big triangle to the angles of the small triangle. This helps us show that the angle at T is half of the angle at P.

Exam Tip: When proving geometric theorems, clearly state the properties being used, such as the exterior angle property of a triangle and the definition of an angle bisector. Label your steps and equations for clarity.