Class 9 Mathematics
Chapter 6: Lines and Angles
Exercise 6.2 – Stepwise Solutions
Important Concepts Used
1. Corresponding angles are equal when two parallel lines are cut by a transversal.
2. Alternate interior angles are equal when two parallel lines are cut by a transversal.
3. Interior angles on the same side of a transversal are supplementary.
4. Angles on a straight line sum to 180°.
5. Sum of angles of a triangle = 180°.
Q1. In Fig. 6.23, if AB || CD, CD || EF and y : z = 3 : 7, find x.
Given: AB || CD and CD || EF, so all three lines are parallel.
Angles y and z are interior angles on the same side of the transversal between CD and EF.
Therefore,
y + z = 180°
Given y : z = 3 : 7
Let y = 3k and z = 7k
So,
3k + 7k = 180°
10k = 180°
k = 18°
Hence,
y = 3 × 18 = 54°
z = 7 × 18 = 126°
Now x and y are corresponding angles since AB || CD.
Therefore,
x = y = 54°
Answer: x = 54°
Q2. In Fig. 6.24, if AB || CD, EF ⟂ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Given: AB || CD, EF ⟂ CD and ∠GED = 126°
Step 1: Find ∠GEC
Since C, E and D are on a straight line,
∠GEC + ∠GED = 180°
∠GEC = 180° − 126° = 54°
Step 2: Find ∠AGE
AB || CD and GE is a transversal.
∠AGE and ∠GEC are corresponding angles.
So,
∠AGE = 54°
Step 3: Find ∠GEF
EF ⟂ CD, so ∠FED = 90°
Given ∠GED = 126°
Therefore,
∠GEF = ∠GED − ∠FED
= 126° − 90°
= 36°
Step 4: Find ∠FGE
In triangle GEF,
∠GFE = 90° (since EF ⟂ AB and AB || CD)
Now,
∠FGE + ∠GEF + ∠GFE = 180°
∠FGE + 36° + 90° = 180°
∠FGE = 54°
Answers:
∠AGE = 54°
∠GEF = 36°
∠FGE = 54°
Q3. In Fig. 6.25, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
Given: PQ || ST
Step 1: Find angle made by QR with a line through R parallel to ST
Draw a line through R parallel to ST.
Since PQ || ST and the new line through R is also parallel to ST, it is parallel to PQ.
At Q, ∠PQR = 110°
So the acute angle between QR and the parallel line through R is
180° − 110° = 70°
Step 2: Find angle made by RS with the parallel line through R
Given ∠RST = 130°
So the acute angle between RS and the parallel line through R is
180° − 130° = 50°
Step 3: Find ∠QRS
∠QRS = 70° + 50° = 120°
Answer: ∠QRS = 120°
Q4. In Fig. 6.26, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Step 1: Find x
AB || CD and PQ is a transversal.
∠APQ and angle PQR are alternate interior angles.
Therefore,
x = ∠PQR = 50°
Step 2: Find interior angle at R of triangle PQR
Given exterior angle ∠PRD = 127°
So interior angle ∠PRQ = 180° − 127° = 53°
Step 3: Find y
In triangle PQR,
x + y + ∠PRQ = 180°
50° + y + 53° = 180°
y = 180° − 103° = 77°
Answers:
x = 50°
y = 77°
Q5. In Fig. 6.27, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes mirror PQ at B, the reflected ray moves along BC and strikes mirror RS at C and again reflects back along CD. Prove that AB || CD.
Given: PQ || RS
By the law of reflection, angle of incidence = angle of reflection.
At point B on mirror PQ,
the angle made by AB with PQ = angle made by BC with PQ
At point C on mirror RS,
the angle made by BC with RS = angle made by CD with RS
Since PQ || RS, the angle made by BC with PQ is equal to the angle made by BC with RS.
Therefore, the angle made by AB with PQ = angle made by CD with RS.
As PQ || RS, equal corresponding angles imply the lines are parallel.
Hence,
AB || CD