# Lines Angles

## Exercise 6.2

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

**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

**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.

**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.

**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.

**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.

**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