### Chapter 14 Exercise 14.2

**Q1) In Figure, line n is a transversal to line l and m. Identify the following:**

**i) Alternate and corresponding angles in Fig. 58 (i)**

**Solution:** The alternate angles are:

âˆ EGA and âˆ FHD

âˆ EGB and âˆ CHF

âˆ AGH and âˆ GHD

âˆ BGH and âˆ CHG

Corresponding angles:

âˆ EGA and âˆ CHG

âˆ AGH and âˆ CHF

âˆ EGB and âˆ DHG

âˆ BGH and âˆ DHF

**ii) Angles alternate to âˆ d and âˆ g and angles corresponding to angle âˆ f and âˆ h in Fig. 58 (ii)**

**Solutions:** The angle alternate to âˆ d is âˆ e.

The angle alternate to âˆ g is âˆ b.

The angle corresponding to âˆ f is âˆ c.

The angle corresponding to âˆ h is âˆ a.

**(iii) Angle alternate to âˆ PQR, angle corresponding to âˆ RQF and angle alternate to **

**âˆ PQE in Fig. 58 (iii)**

**Solutions:** Angle alternate to âˆ PQR is âˆ ARQ

Angle corresponding to âˆ RQF is âˆ BRA

Angle alternate to âˆ PQE is âˆ BRA

**(iv) Pairs of interior and exterior angles on the same side of the transversal in Fig. 58 (ii)**

**Solution:** Pairs of Interior angles are:

âˆ a , âˆ e

âˆ d, âˆ f

Pairs of Exterior angles are:

âˆ c, âˆ g

âˆ b, âˆ h

**Q2) In Figure 59, AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If âˆ CMQ = 60 ^{0}, find all other angles in the figure.**

**Solutions:** âˆ CMQ = 60^{0} is given in the figure.

âˆ CMQ = âˆ PLB = 60^{0} (Alternate exterior angle)

âˆ CMQ = âˆ LMD = 60^{0} (Vertically opposite angles)

âˆ CMQ = âˆ ALM = 60^{0 }(Corresponding angles)

âˆ PLB and âˆ ALP is a linear pair.

So âˆ PLB + âˆ ALP = 180^{0}

60^{0} + âˆ ALP = 180^{0}

âˆ ALP = 120^{0}

âˆ LMD and âˆ LMC is a linear pair

âˆ LMD + âˆ LMC = 180^{0}

60^{0} + âˆ LMC = 180^{0}

âˆ LMC = 120^{0}

âˆ CMQ and âˆ DMQ is a linear pair

âˆ CMQ + âˆ DMQ = 180^{0}

60^{0} + âˆ DMQ = 180^{0}

âˆ DMQ = 120^{0}

âˆ DMQ = âˆ MLB = 120^{0} (Corresponding angles)

**Q3) In Fig. 60, AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If âˆ LMD = 35 ^{0} Find âˆ ALM and âˆ PLA.**

Solution: It is given that âˆ LMD = 35^{0}

âˆ LMD and âˆ LMC is a linear pair.

âˆ LMD + âˆ LMC = 180^{0}

35^{0} + âˆ LMC = 180^{0}

âˆ LMC = 145^{0}

âˆ LMC = âˆ PLA = 145^{0} (Corresponding angles)

âˆ LMC = âˆ MLB = 145^{0} (Alternate Interior angles)

âˆ ALM and âˆ MLB is a linear pair

âˆ ALM + âˆ MLB = 180^{0}

âˆ ALM + 145^{0} = 180^{0}

âˆ ALM = 35^{0}

**Q4) The line n is transversal to line l and m in Fig. 61. Identify the angle alternate to âˆ 13, angle corresponding to âˆ 15, and angle alternate to âˆ 15.**

**Solution: **Angle alternate to âˆ 13 is âˆ 7.

Angle corresponding to âˆ 15 is 7.

Angle alternate to âˆ 15 is 5.

**Q5) In Fig. 62, line l || m and n is transversal. If âˆ 1 = 40 ^{0}, find all the angles and check that all corresponding angles and alternate angles are equal.**

**Solution:** It is given that âˆ 1 = 40^{0}

As we can see in the figure that âˆ 1 and âˆ 2 is a linear pair.

âˆ 1 + âˆ 2 = 180^{0}

40^{0} + âˆ 2 = 180^{0}

âˆ 2 = 140^{0}

âˆ 2 = âˆ 6 = 140^{0} (Corresponding angles)

âˆ 6 and âˆ 5 is a linear pair.

âˆ 6 + âˆ 5 = 180^{0}

140^{0} + âˆ 5 = 180^{0}

âˆ 5 = 40^{0}

âˆ 3 = âˆ 5 = 40^{0} (Alternate interior angles)

âˆ 3 = âˆ 7 = 40^{0} (Corresponding angles)

âˆ 3 and âˆ 4 is a linear pair.

âˆ 3 + âˆ 4 = 180^{0}

40^{0} + âˆ 4 = 180^{0}

âˆ 4 = 140^{0}

âˆ 4 = âˆ 8 = 140^{0} (Corresponding angles)

**Q6) In Fig. 63, line l || m and a transversal n cuts them P and Q respectively. If âˆ 1 = 75 ^{0}, find all other angles.**

**Solution:** It is given that âˆ 1 = 75^{0}

As we can see in the figure that âˆ 1 and âˆ 2 is a linear pair.

âˆ 1 + âˆ 2 = 180^{0}

75^{0} + âˆ 2 = 180^{0}

âˆ 2 = 105^{0}

âˆ 2 = âˆ 6 = 105^{0} (Corresponding angles)

âˆ 6 and âˆ 5 is a linear pair.

âˆ 6 + âˆ 5 = 180^{0}

105^{0} + âˆ 5 = 180^{0}

âˆ 5 = 75^{0}

âˆ 3 = âˆ 5 = 75^{0} (Alternate interior angles)

âˆ 3 = âˆ 7 = 75^{0} (Corresponding angles)

âˆ 3 and âˆ 4 is a linear pair.

âˆ 3 + âˆ 4 = 180^{0}

75^{0} + âˆ 4 = 180^{0}

âˆ 4 = 105^{0}

âˆ 4 = âˆ 8 = 105^{0} (Corresponding angles)

**Q7) In Fig. 64, AB || CD and a transversal PQ cuts at L and M respectively. If âˆ QMD = 100 ^{0}, find all the other angles.**

**Solution:** It is given that âˆ QMD = 100^{0}

âˆ QMD = âˆ MLB = 100^{0} (Corresponding angles)

âˆ MLB and âˆ MLA is a linear angles.

âˆ MLB + âˆ MLA = 180^{0}

100^{0} + âˆ MLA = 180^{0}

âˆ MLA = 80^{0}

âˆ QMD = âˆ PLA = 100^{0} (Alternate exterior angles)

We can see from the figure, âˆ PLA and PLB is a linear pair.

âˆ PLA + âˆ PLB = 180^{0}

100^{0} + âˆ PLB = 180^{0}

âˆ PLB = 80^{0}

âˆ PLB = âˆ LMD = 80^{0} (Corresponding angles)

âˆ LMD and âˆ LMC is a linear pair.

âˆ LMD + âˆ LMC = 180^{0}

80^{0} + âˆ LMC = 180^{0}

âˆ LMC = 100^{0}

As from the figure, âˆ QMD and âˆ QMC is a linear pair.

âˆ QMD + âˆ QMC = 180^{0}

100^{0} + âˆ QMC = 180^{0}

âˆ QMC = 80^{0}

**Q8) In Fig. 65, l || m and p || q. Find the values of x,y,z,t.**

**Solutions:** The angle given the figure is 80^{0}

âˆ z and 80^{0} are vertically opposite angles.

So, âˆ z = 80^{0}

âˆ z = âˆ y = 80^{0} (Corresponding angles)

âˆ z = âˆ t = 80^{0} (Corresponding angles)

âˆ y = âˆ x = 80^{0} (Corresponding angles)

**Q9) In Fig. 66, line l || m, âˆ 1 = 120 ^{0} and âˆ 2= 100^{0}, find out âˆ 3 and âˆ 4.**

Solutions: It is given in that,

âˆ 1 = 120^{0}, âˆ 2 = 100^{0}

âˆ 2 = âˆ 6 = 100^{0} (Corresponding angles)

âˆ 6 and âˆ 3 is a linear pair.

âˆ 6 + âˆ 3 = 180^{0}

100 + âˆ 3 = 180^{0}

âˆ 3 = 80^{0}

âˆ 1 and âˆ 5 is a linear pair.

âˆ 1 +âˆ 5 = 180^{0}

120^{0} + âˆ 5 = 180^{0}

âˆ 5 = 60^{0}

By using angle sum property;

âˆ 5 + âˆ 4 + âˆ 3 = 180^{0}

60^{0} + âˆ 4 + 80^{0} = 180^{0}

âˆ 4 = 180^{0} – 140^{0}

âˆ 4 = 40^{0.}

**Q10) In Fig. 67, l || m. Find the values of a, b, c, d. Give reasons.**

**Solution:** As it is given in that l||m

âˆ a = 110^{0} (Vertically opposite angles)

âˆ d = 85^{0} (Vertically opposite angles)

From the figure;

âˆ a = âˆ b = 110^{0 }(Corresponding angles)

âˆ c = âˆ d = 85^{0} (Corresponding angles)

**Q11) In Fig. 68, AB || CD and âˆ 1 and âˆ 2 are in the ratio of 3 : 2. Determine all angles from 1 to 8.**

**Solution: **As given in the question that;

âˆ 1 and âˆ 2 are in 3:2.

Let the âˆ 1 = 3a and âˆ 2 = 2a

âˆ 1 and âˆ 2 is a linear pair.

âˆ 1 + âˆ 2 = 180^{0}

3a + 2a = 180^{0}

5a = 180^{0}

a = 36^{0}

So, âˆ 1 = 3a = 3 Ã— 36^{0} = 108^{0}

âˆ 2 = 2a = 2 Ã— 36^{0} = 72^{0}

As we can see in the figure;

âˆ 1 = âˆ 5 = 108^{0} (Corresponding angles)

âˆ 2 = âˆ 6 = 72^{0} (Corresponding angles)

âˆ 5 = âˆ 3 = 108^{0} (Alternate interior angles)

âˆ 6 = âˆ 4 = 72^{0} (Alternate interior angles)

âˆ 2 = âˆ 8 = 72^{0} (Alternate exterior angles)

âˆ 1 = âˆ 7 = 108^{0} (Alternate exterior angles)

Thus the angles are:

âˆ 1 = 108^{0}, âˆ 2 = 72^{0}, âˆ 3 = 108^{0}, âˆ 4 = 72^{0}, âˆ 5 = 108^{0}, âˆ 6 = 72^{0}, âˆ 7 = 108^{0}, âˆ 8 = 72^{0}

**Q12) In Fig. 69 l, m and n are parallel lines intersected by a transversal p at X, Y and Z respectively. Find âˆ 1, âˆ 2 and âˆ 3.**

**Solution:** âˆ 4 and 60^{0} is a linear pair.

âˆ 4 + 60^{0} = 180^{0}

âˆ 4 = 120^{0}

As we can see in the figure;

âˆ 4 = âˆ 1 = 120^{0} (Corresponding angles)

âˆ 1 = âˆ 2 = 120^{0} (Corresponding angles)

âˆ 2 = âˆ 3 = 120^{0} (Vertically opposite angles)

Thus; âˆ 1 = 120^{0}, âˆ 2 = 120^{0}, âˆ 3 = 120^{0}

**Q13) In Fig. 70, if l || m || n and âˆ 1 = 60 ^{0}, find âˆ 2.**

**Solution:** As we can see in the figure:

âˆ 1 = âˆ 3 (Corresponding angles)

It is given that âˆ 1 = 60^{0}

So, âˆ 3 = 60^{0}

âˆ 3 and âˆ 4 are linear pair

âˆ 3 + âˆ 4 = 180^{0}

60^{0} + âˆ 4 = 180^{0}

âˆ 4 = 120^{0}

âˆ 4 and âˆ 2 are alternate interior angles.

So, âˆ 2 = 120^{0}

**Q14) In Fig. 71, if AB || CD and CD || EF, find âˆ ACE.**

**Solution:** It is given in the figure that âˆ CEF = 130^{0}

As per the property of sum of interior angles:

âˆ CEF + âˆ ECD = 180^{0}

130^{0} + âˆ ECD = 180^{0}

âˆ ECD = 50^{0}

âˆ ACD and âˆ BAC are Alternate angles.

So, âˆ ACD = âˆ BAC = 70^{0}

âˆ ACE = âˆ ACD – âˆ ECD

âˆ ACE = 70^{0} – 50^{0}

âˆ ACE = 20^{0}

**Q15) In Fig. 72, if l || m, n || p and âˆ 1 = 85 ^{0}, find âˆ 2.**

**Solution:** It is given that âˆ 1 = 85^{0}

âˆ 1 = âˆ 3 = 85^{0} (Corresponding angles)

âˆ 3 and âˆ 2 are Interior angles.

âˆ 3 + âˆ 2 = 180^{0} (Sum of interior angle)

85^{0} + âˆ 2 = 180^{0}

âˆ 2 = 95^{0}

**Q16) In Fig. 73, a transversal n cuts two lines l and m. If âˆ 1 = 70 ^{0} and âˆ 7 = 80^{0}, is l || m?**

**Solution: **It is given that âˆ 1 =70^{0} and âˆ 7 = 80^{0}

As we can see in the figure that;

âˆ 1 and âˆ 7 are Alternate exterior angles.

So, âˆ 1 should be equal to âˆ 7. But both the angles have different values.

Therefore, line l is not parallel to m (Lines are parallel only if alternate exterior angles are equal).

**Q17) In Fig. 74, a transversal n cuts two lines l and m such that âˆ 2 = 65 ^{0} and âˆ 8 = 65^{0}. Are the lines parallel?**

**Solution:** It is given that;

âˆ 2 = 65^{0} and âˆ 8 = 65^{0 }

From the figure we can see that

âˆ 2 and âˆ 8 are Alternate Exterior angles.

So, âˆ 2 = âˆ 8 = 180^{0}

As the value for both angles are given equal in the question itself. Therefore, line l and m are parallel to each other (Two lines are parallel if the alternate angles formed with the transverse are equal).

**Q18) In Fig. 75, Show that AB || EF.**

**Solution:** It is given that;

âˆ BAC = 57^{0}, âˆ ACE = 22^{0}, âˆ DCE = 35^{0}, âˆ CEF = 145^{0}

As we can see;

âˆ ACE + âˆ DCE = âˆ ACD

22^{0} + 35^{0} = âˆ ACD

âˆ ACD = 57^{0}

As the âˆ BAC and âˆ ACD are alternate angles and both are equal to 57^{0}.

Therefore, line AB || CD (Two lines are parallel if the alternate angles are equal)…â€¦â€¦.(i)

Now,

âˆ DCE + âˆ CEF

= 35^{0} + 145^{0}

= 180^{0}

This shows that the sum of the alternate interior angles on the same side of the transverse line CE is 180^{0}. So, the âˆ DCE and âˆ CEF are supplementary angles.

Thus, line CD || EFâ€¦â€¦â€¦â€¦â€¦..(ii)

From equations i and ii

AB || CD || EF

Therefore, AB || EF.

**Q19) In Fig. 76, AB || CD. Find the values of x, y, z.**

**Solution:** âˆ x and 125^{0} are linear pairs.

âˆ x + 125^{0} = 180^{0}

âˆ x = 55^{0}

The sum of interior angles on the same side of the transverse line AC is 180^{0}.

âˆ x + âˆ z = 180^{o}

55^{0} + âˆ z = 180^{0}

âˆ z = 125^{0}

The sum of interior angles is 180^{0}.

âˆ x + âˆ y = 180^{o}

55^{0} + âˆ y = 180^{0}

âˆ y = 125^{0}

**Q20) In Fig. 77, find out âˆ PXR, if PQ || RS.**

**Solution: **Look at the diagram below;

âˆ QPX = âˆ 1 = 70^{0} (Alternate angles)

âˆ SRX = âˆ 2 = 50^{0} (Alternate angles)

âˆ PXR = âˆ 1 + âˆ 2

âˆ PXR = 70^{0} + 50^{0}

âˆ PXR = 120^{0}

**Q21) In Fig. 78, we have**

**i) âˆ MLY = 2âˆ LMQ, find âˆ LMQ**

**Solution**: âˆ MLY and âˆ LMQ are interior angles.

âˆ MLY + âˆ LMQ = 180^{0}

2âˆ LMQ + âˆ LMQ = 180^{0}

3âˆ LMQ = 180^{0}

âˆ LMQ = 60^{0}

**ii) âˆ XLM = (2x âˆ’ 10) ^{0} and âˆ LMQ= x + 30^{0}, find x.**

**Solution:** âˆ XLM = âˆ LMQ (Alternate angles)

2x – 10^{0} = x + 30^{0}

X = 40^{0}

**iii) âˆ XLM = âˆ PML, find âˆ ALY**

**Solution:** âˆ XLM and âˆ PML are interior angles.

âˆ XLM + âˆ PML = 180^{0}

âˆ XLM + âˆ XML = 180^{0}

2âˆ XML = 180^{0}

âˆ XML = 90^{0}

âˆ ALY = âˆ XML (Vertically opposite angles)

So, âˆ ALY = 90^{0}

**iv) âˆ ALY = (2x âˆ’ 15) ^{0}, âˆ LMQ = (x + 40)^{0}, find x**

**Solution:** âˆ ALY = âˆ LMQ (Corresponding angles)

2x – 15^{0} = x + 40^{0}

X = 55^{0}

**Q22) In Fig. 79, DE || BC. Find the values of x and y.**

**Solution:** As we can see from the figure that;

âˆ DAB = âˆ ABC (Alternate interior angles)

x = 40^{0}

âˆ EAC = CAD (Alternate interior angles)

y = 55^{0}

**Q23) In Fig. 80, line AC || line DE and âˆ ABD= 32 ^{0}, Find out the angles x and y if âˆ E = 122^{0}.**

**Solution: **As we can see in the figure.

y + 32^{0} = 180^{o} (Sum of interior angles)

y = 148^{0}

Now,

y^{ }and âˆ BDE are linear pairs

148^{0 }+ âˆ BDE = 180^{0}

âˆ BDE = 32^{0.}

âˆ BED + 122^{0 }=180^{0} (Linear pair angles)

âˆ BED = 58^{0}

By using angle sum property of a triangle;

âˆ BDE + âˆ BED + x= 180^{0}

32^{0} + 58^{0} + x = 180^{0}

x = 180^{0} – 90^{0}

x = 90^{0}

**Q24) In Fig. 81, side BC of Î”ABC has been produced to D and CE || BA. If **

**âˆ ABC = 65 ^{0}, âˆ BAC = 55^{0}, find âˆ ACE, âˆ ECD, âˆ ACD.**

**Solution:** By using angle sum property of a triangle:

65^{0} + 55^{0} + âˆ ACB = 180^{0}

âˆ ACB = 180^{0} – 120^{0}

âˆ ACB = 60^{0}

âˆ BAC = âˆ ACE = 55^{0} (Alternate interior angles)

âˆ ACB + âˆ ACE + âˆ ECD = 180^{0 }(Linear pair angles)

60^{0} + 55^{0} + âˆ ECD = 180^{0}

âˆ ECD = 65^{0}

âˆ ACD = âˆ ACE +âˆ ECD

âˆ ACD = 55^{0} + 65^{0}

âˆ ACD = 120^{0}

**Q25) In Fig. 82, line CA âŠ¥ AB || line CR and line PR || line BD. Find âˆ x, âˆ y, âˆ z.**

**Solution: **As it is given that

CA** **âŠ¥ AB

So, x = 90^{0}

By using angle sum property in triangle APQ;

x + âˆ AQP + âˆ APQ = 180^{0}

90^{0} + 20^{0} + âˆ APQ = 180^{0}

âˆ APQ = 70^{0}

âˆ APQ = y = 70^{0 }(Corresponding angles)

âˆ APQ + z = 180^{0} (Interior angles)

70^{0 }+ z = 180^{0}

Z = 110^{0}

**Q26) In Fig. 83, PQ || RS. Find the value of x.**

**Solution:** âˆ RCB + 130^{0} = 180^{0} (Linear pairs)

âˆ RCB = 50^{0}

Applying angle sum property in triangle BRC.

âˆ RCB + 55^{0} + x = 180^{0}

50^{0} + 55^{0} + x = 180^{0}

x = 75^{0}

**Q27) In Fig. 84, AB || CD and AE || CF, âˆ FCG = 90 ^{0} and âˆ BAC = 120^{0}. Find the value of x, y and z.**

**Solution:** âˆ BAC and âˆ ACG are Alternate interior angles.

So, âˆ BAC = âˆ ACG = 120^{0}

âˆ ACG = y + 90^{0}

120^{0} = y + 90^{0}

y = 30^{0}

x + âˆ ACG = 180^{0} (Linear pairs)

x + 120^{0} = 180^{0}

x = 60^{0}

âˆ DCF = x + y

âˆ DCF = 60^{0} + 30^{0}

âˆ DCF = 90^{0}

âˆ BAE = âˆ DCF = 90^{0}

âˆ BAE + z + 120^{0} = 360^{0}

90^{0} + z + 120^{0} = 360^{0 }

z = 150^{0}

**Q28) In Fig. 85, AB || CD and AC || BD. Find the values of x, y, z.**

**Solution:** **(i)** As it is given AB || CD and AC || BD.

So, ABCD is a parallelogram.

Sum of the adjacent angles of a parallelogram is equal to 180^{0}.

65^{0} + z = 180^{0}

Z = 115^{0}

z = x = 115^{0} (Opposite angles of a parallelogram are equal)

y = 65^{0} (Opposite angles of a parallelogram are equal)

**(ii)** x = 40^{0} (Alternate interior angles)

y = 35^{0} (Alternate interior angles)

**Q29) In Fig. 86, state which lines are parallel and why? **

**Solution:** As the alternate angles are equals;

âˆ ACD = âˆ EDC = 100^{0}

Therefore the line AC || DE

**Q30) In Fig. 87, the corresponding arms of âˆ ABC and âˆ DEF are parallel. If âˆ ABC = 75 ^{0}, find âˆ DEF.**

**Solution:** The line BC and DE intersect each other at point G as shown in the figure below.

So, AB || DE and BC || EF

Thus, âˆ ABC = âˆ DEF = 75^{0} (Corresponding angles)