Lines Angles

Exercise 6.2

Question 1: In the following figure find the value of x and y, the show that AB║CD.

line and angles 1

Answer: It is clear that ∠APM + ∠APN = 180° (Linear pair)
Or, ∠APN = 180° - 50° = 130° = x………………..(1)
Now, ∠CQN = ∠DQM (Opposite angles)
Or, ∠DQM = 130° = y
As we have seen ∠APN = ∠CQN
So by the theorem of corresponding angles on one side of the transversal it is clear that AB║CD


Question 2: In the following figure if AB║CD and CD║EF and y:z=3:7, find the value of x

line and angles 3

Answer: As CD and EF are parallel lines, so ∠FQP = ∠DPO (corresponding angles)
Now, ∠DPO + ∠CPO = 180°
Or, 3x + 7x = 180°
Or, 10x = 180°
Or, x = 18°

Putting the value of x in the given ratio we get following values:
∠DPO = 126° and ∠CPO = 54°

Now, it is given that AB || CD,
So, ∠DPO = ∠AOP = 126° = x


Question 3: In the following figure if AB || CD, EF || CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

line and angles 6

Answer: ∠GEF = ∠GED - ∠FED
Or, ∠GEF = 126° - 90° = 36°

As AB || CD,
So, ∠EFG = ∠FED = 90°
Or, ∠FGE = 180°- (90° + 36°) = 54°
Now, ∠AGE + ∠FGE = 180° (Linear pair)
Or, ∠AGE = 180° - 54° = 126°

Question 4: In the following figure PQ ||ST, values of ∠PQR = 110° and ∠RST = 130°, find the value of ∠QRS.

line and angles 9

Answer: Let us draw another line AB which is parallel to PQ and ST.
Now, ∠RST + ∠BRS = 180°
And, ∠PQR + ∠ARQ = 180°
(Because, internal angles on the one side of transversal are complementary angles.)
Hence, ∠BRS = 180° - 130° = 50°
∠ARQ = 180° - 110° = 70°
Now, it is clear that ∠ARQ + ∠QRS + ∠BRS = 180°
Or, 70° + ∠QRS + 50° = 180°Or, ∠QRS = 60°

Question 5: In the following figure AB || CD, ∠APQ = 50°and ∠PRD = 127°, find values of x and y.

line and angles 12

Answer: ∠BPR + ∠PRD = 180° (Internal angles on one side of transversal)
Or, ∠BPR = 180° - 127° = 53°
On the line CD, ∠PRD + ∠PRQ = 180°
Or, ∠PRQ = 180° - 127° = 53°
On the line AB, ∠APQ + ∠QPR + ∠BPR = 180°
Or, ∠QPR = 180° - (50° + 53°) = 77°
In ΔPQR, ∠PQR + ∠QPR + ∠PRQ = 180° (angle sum of triangle)
Or, ∠PQR = 180° - (77° + 53°) = 50°
Or, x = 50° and y = 77°

Question 6: In the given figure PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ, the reflected ray moves along the path BC and strikes the mirror RS. The second mirror reflects the ray along CD. Prove that AB║CD.

line and angles 14

Answer: From the theory of reflection in Physics we know that angle of incidence is equal to angle of reflection.

Here, in case of mirror PQ< angle of incidence i = ∠ABP and angle of reflection r = ∠QBC.
In case of mirror RS, angle of incidence i = ∠BCR and angle of reflection r = ∠SCD
Required evidence to prove AB || CD
We need to check if ∠ABC = ∠BCD (alternate angles)

On the line PQ, ∠ABP + ∠ABC + ∠QBC = 180°
Or, i + ∠ABC + r = 180°
Or, i + i + ∠ABC = 180° …………(1)
Similarly, on the line RS it can be observed that
i + i + ∠BCD = 180° …………..(2)
from the question it is given that PQ ||RS
Hence, ∠QBC = ∠BCR (alternate angles)
Hence, values of angles of incidence for both mirrors are same
Correlating this finding with equations (1) and (2) it is clear that
∠ABC = ∠BCD
So, AB||CD proved



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