The materials of RD Sharma Solutions for Class 7 Maths Exercise 14.2 of Chapter 14, in the form of PDF, is available here. Students who want to score high in Maths can download these materials from the given links. Our expert team has formulated RD Sharma Solutions for Class 7 to help and fulfil the dreams of students. This exercise has thirty questions based on parallel lines and transversal lines. Let us have a look at some of the important concepts that are being discussed in this exercise.

- Definition and meaning of parallel lines
- Parallel rays
- Parallel segments

- Transversals
- Angle made by a transversal with two lines
- Exterior and interior angles
- Corresponding angles
- Alternate interior and interior angles
- Angles made by a transversal to two parallel lines

## Download the PDF of RD Sharma Solutions For Class 7 Chapter 14 – Lines And Angles Exercise 14.2

### Access answers to Maths RD Sharma Solutions For Class 7 Chapter 14 – Lines And Angles Exercise 14.2

**1. In Fig. 58,Â line n is a transversal to line l and m. Identify the following:**

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

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

**(iii) Angle alternate to âˆ PQR, angle corresponding to âˆ RQF and angle alternate to âˆ PQE in Fig. 58 (iii)**

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

** **

**Solution:**

(i) A pair of angles in which one arm of both the angles is on the same side of the transversal and their other arms are directed in the same sense is called a pair of corresponding angles.

In Figure (i) Corresponding angles are

âˆ EGB andÂ âˆ GHD

âˆ HGB andÂ âˆ FHD

âˆ EGA andÂ âˆ GHC

âˆ AGH andÂ âˆ CHF

A pair of angles in which one arm of each of the angle is on opposite sides of the transversal and whose other arms include the one segment is called a pair of alternate angles.

The alternate angles are:

âˆ EGB andÂ âˆ CHF

âˆ HGB andÂ âˆ CHG

âˆ EGA andÂ âˆ FHD

âˆ AGH andÂ âˆ GHD

(ii) In Figure (ii)

The alternate angle to âˆ d is âˆ e.

The alternate angle to âˆ g is âˆ b.

The corresponding angle to âˆ f is âˆ c.

The corresponding angle to âˆ h is âˆ a.

(iii) In Figure (iii)

Angle alternate to âˆ PQR is âˆ QRA.

Angle corresponding to âˆ RQF is âˆ ARB.

Angle alternate to âˆ POE is âˆ ARB.

(iv) In Figure (ii)

Pair of interior angles are

âˆ a is âˆ e.

âˆ d is âˆ f.

Pair of exterior angles are

âˆ b is âˆ h.

âˆ c is âˆ g.

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

** **

**Â **

**Solution:**

A pair of angles in which one arm of both the angles is on the same side of the transversal and their other arms are directed in the same sense is called a pair of corresponding angles.

Therefore corresponding angles are

âˆ ALM = âˆ CMQ = 60^{o} [given]

Vertically opposite angles are

âˆ LMD = âˆ CMQ = 60^{o }[given]

Vertically opposite angles are

âˆ ALM = âˆ PLB = 60^{o}

Here, âˆ CMQ + âˆ QMD = 180^{o} are the linear pair

On rearranging we get

= âˆ QMD = 180^{o} â€“ 60^{o}

= 120^{o}

Corresponding angles are

âˆ QMD = âˆ MLB = 120^{o}

Vertically opposite angles

âˆ QMD = âˆ CML = 120^{o}

Vertically opposite angles

âˆ MLB = âˆ ALP = 120^{o}

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

**Solution:**

Given that, âˆ LMD = 35^{o}

From the figure we can write

âˆ LMD and âˆ LMC is a linear pair

âˆ LMD + âˆ LMC = 180^{o }[sum of angles in linear pair = 180^{o}]

On rearranging, we get

=Â âˆ LMC = 180^{o} â€“ 35^{o}

= 145^{o}

So, âˆ LMC = âˆ PLA = 145^{o}

And, âˆ LMC = âˆ MLB = 145^{o}

âˆ MLB and âˆ ALM is a linear pair

âˆ MLB + âˆ ALM = 180^{o} [sum of angles in linear pair = 180^{o}]

= âˆ ALM = 180^{o} â€“ 145^{o}

= âˆ ALM = 35^{0}

Therefore, âˆ ALM = 35^{o}, âˆ PLA = 145^{o}.

**4. 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:

Given that, l âˆ¥ m

From the figure the angle alternate to âˆ 13 is âˆ 7

From the figure the angle corresponding to âˆ 15 is âˆ 7 [A pair of angles in which one arm of both the angles is on the same side of the transversal and their other arms are directed in the same sense is called a pair of corresponding angles.]

Again from the figure angle alternate to âˆ 15 is âˆ 5

**5. In Fig. 62, line l âˆ¥ m and n is transversal. If âˆ 1 = 40Â°, find all the angles and check that all corresponding angles and alternate angles are equal.**

** **

**Solution:**

Given that, âˆ 1 = 40^{o}

âˆ 1 and âˆ 2 is a linear pair [from the figure]

âˆ 1 + âˆ 2 = 180^{o}

âˆ 2 = 180^{o} â€“ 40^{o}

âˆ 2 = 140^{o}

Again from the figure we can say that

âˆ 2 and âˆ 6Â is a corresponding angle pair

So, âˆ 6 = 140^{o}

âˆ 6 and âˆ 5 is a linear pair [from the figure]

âˆ 6 + âˆ 5 = 180^{o}

âˆ 5 = 180^{o} â€“ 140^{o}

âˆ 5 = 40^{o}

From the figure we can write as

âˆ 3 and âˆ 5Â are alternative interior angles

So, âˆ 5 = âˆ 3 = 40^{o}

âˆ 3 and âˆ 4 is a linear pair

âˆ 3 + âˆ 4 = 180^{o}

âˆ 4 = 180^{o} â€“ 40^{o}

âˆ 4 = 140^{o}

Now, âˆ 4 and âˆ 6Â are a pair interior angles

So, âˆ 4 = âˆ 6 = 140^{o}

âˆ 3 and âˆ 7 are pair of corresponding angles

So, âˆ 3 = âˆ 7 = 40^{o}

Therefore, âˆ 7 = 40^{o}

âˆ 4 and âˆ 8Â are a pair corresponding angles

So, âˆ 4 = âˆ 8 = 140^{o}

Therefore, âˆ 8 = 140^{o}

Therefore, âˆ 1 = 40^{o}, âˆ 2 = 140^{o}, âˆ 3 = 40^{o}, âˆ 4 = 140^{o}, âˆ 5 = 40^{o}, âˆ 6 = 140^{o}, âˆ 7 = 40^{o} and âˆ 8 = 140^{o}

**6. In Fig.63, line l âˆ¥ m and a transversal n cuts them P and Q respectively. If âˆ 1 = 75Â°, find all other angles.**

**Solution:**

Given that, l âˆ¥ m and âˆ 1 = 75^{o}

We know that, from the figure

âˆ 1 + âˆ 2 = 180^{o} is a linear pair

âˆ 2 = 180^{o} â€“ 75^{o}

âˆ 2 = 105^{o}

Here, âˆ 1 = âˆ 5 = 75^{o} are corresponding angles

âˆ 5 = âˆ 7 = 75^{o} are vertically opposite angles.

âˆ 2 = âˆ 6 = 105^{o} are corresponding angles

âˆ 6 = âˆ 8 = 105^{o} are vertically opposite angles

âˆ 2 = âˆ 4 = 105^{o} are vertically opposite angles

So, âˆ 1 = 75^{o}, âˆ 2 = 105^{o}, âˆ 3 = 75^{o}, âˆ 4 = 105^{o}, âˆ 5 = 75^{o}, âˆ 6 = 105^{o}, âˆ 7 = 75^{o} and âˆ 8 = 105^{o}

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

** **

**Solution:**

Given that, AB âˆ¥ CD and âˆ QMD = 100^{o}

We know that, from the figure âˆ QMD + âˆ QMC = 180^{o} is a linear pair,

âˆ QMC = 180^{o} â€“ âˆ QMD

âˆ QMC = 180^{o} â€“ 100Â°

âˆ QMC = 80^{o}

Corresponding angles are

âˆ DMQ = âˆ BLM = 100^{o}

âˆ CMQ = âˆ ALM = 80^{o}

Vertically Opposite angles are

âˆ DMQ = âˆ CML = 100^{o}

âˆ BLM = âˆ PLA = 100^{o}

âˆ CMQ = âˆ DML = 80^{o}

âˆ ALM = âˆ PLB = 80^{o}

**8. In Fig. 65, l âˆ¥ m and p âˆ¥ q. Find the values of x, y, z, t.**

**Solution:**

Given that one of the angle is 80^{o}

âˆ z and 80^{o} are vertically opposite angles

Therefore âˆ z = 80^{o}

âˆ z and âˆ t are corresponding angles

âˆ z = âˆ t

Therefore, âˆ t = 80^{o}

âˆ z and âˆ y are corresponding angles

âˆ z = âˆ y

Therefore, âˆ y = 80^{o}

âˆ x and âˆ y are corresponding angles

âˆ y = âˆ x

Therefore, âˆ x = 80^{o}

**9. In Fig. 66, line l âˆ¥ m, âˆ 1 = 120 ^{o} and âˆ 2 = 100^{o}, find out âˆ 3 and âˆ 4.**

**Solution:**

Given that, âˆ 1 = 120^{o} and âˆ 2 = 100^{o}

From the figure âˆ 1 and âˆ 5 is a linear pair

âˆ 1 + âˆ 5 = 180^{o}

âˆ 5 = 180^{o} â€“ 120^{o}

âˆ 5 = 60^{o}

Therefore, âˆ 5 = 60^{o}

âˆ 2 and âˆ 6 are corresponding angles

âˆ 2 = âˆ 6 =Â 100^{o}

Therefore,Â âˆ 6 =Â 100^{o}

âˆ 6 and âˆ 3 a linear pair

âˆ 6 + âˆ 3 = 180^{o}

âˆ 3 = 180^{o} â€“ 100^{o}

âˆ 3 = 80^{o}

Therefore, âˆ 3 = 80^{o}

By, angles of sum property

âˆ 3 + âˆ 5 + âˆ 4Â = 180^{o}

âˆ 4 = 180^{o} â€“ 80^{o} â€“ 60^{o}

âˆ 4 = 40^{o}

Therefore, âˆ 4 = 40^{o}

**10. Â In Fig. 67, l âˆ¥ m. Find the values of a, b, c, d. Give reasons.**

**Solution:**

Given l âˆ¥ m

From the figure vertically opposite angles,

âˆ a = 110^{o}

Corresponding angles, âˆ a = âˆ b

Therefore, âˆ b = 110^{o}

Vertically opposite angle,

âˆ d = 85^{o}

Corresponding angles, âˆ d = âˆ c

Therefore, âˆ c = 85^{o}

Hence, âˆ a = 110^{o}, âˆ b = 110^{o}, âˆ c = 85^{o}, âˆ d = 85^{o}

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

**Solution:**

Given âˆ 1 and âˆ 2 are in the ratio 3: 2

Let us take the angles as 3x, 2x

âˆ 1 and âˆ 2 are linear pair [from the figure]

3x + 2x = 180^{o}

5x = 180^{o}

x = 180^{o}/5

x = 36^{o}

Therefore, âˆ 1 = 3x = 3(36) = 108^{o}

âˆ 2 = 2x = 2(36) = 72^{o}

âˆ 1 and âˆ 5 are corresponding angles

Therefore âˆ 1 = âˆ 5

Hence, âˆ 5 = 108^{o}

âˆ 2 and âˆ 6 are corresponding angles

So âˆ 2 = âˆ 6

Therefore, âˆ 6 = 72^{o}

âˆ 4 and âˆ 6 are alternate pair of angles

âˆ 4 = âˆ 6 =Â 72^{o}

Therefore, âˆ 4 = 72^{o}

âˆ 3 and âˆ 5 are alternate pair of angles

âˆ 3 = âˆ 5 =Â 108^{o}

Therefore, âˆ 5 = 108^{o}

âˆ 2 and âˆ 8 are alternate exterior of angles

âˆ 2 = âˆ 8 =Â 72^{o}

Therefore, âˆ 8 = 72^{o}

âˆ 1 and âˆ 7 are alternate exterior of angles

âˆ 1 = âˆ 7 =Â 108^{o}

Therefore, âˆ 7 = 108^{o}

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

Â

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

**Â **

**Solution:**

Given l, m and n are parallel lines intersected by transversal p at X, Y and Z

Therefore linear pair,

âˆ 4 + 60^{o} = 180^{o}

âˆ 4 = 180^{o} â€“ 60^{o}

âˆ 4 = 120^{o}

From the figure,

âˆ 4 and âˆ 1 are corresponding angles

âˆ 4 = âˆ 1

Therefore, âˆ 1 = 120^{o}

âˆ 1 and âˆ 2 are corresponding angles

âˆ 2 = âˆ 1

Therefore, âˆ 2 = 120^{o}

âˆ 2 and âˆ 3 are vertically opposite angles

âˆ 2 = âˆ 3

Therefore, âˆ 3 = 120^{0}

Â

**13. In Fig. 70, if l âˆ¥ m âˆ¥ n and âˆ 1 = 60 ^{o}, find âˆ 2**

**Solution:**

Given that l âˆ¥ m âˆ¥ n

From the figure Corresponding angles are

âˆ 1 = âˆ 3

âˆ 1 = 60^{o}

Therefore, âˆ 3 = 60^{o}

âˆ 3 and âˆ 4 are linear pair

âˆ 3 + âˆ 4 = 180^{o}

âˆ 4 = 180^{o} â€“ 60^{o}

âˆ 4 = 120^{o}

âˆ 3 and âˆ 4 are alternative interior angles

âˆ 4 = âˆ 2

Therefore, âˆ 2 = 120^{o}

**14. In Fig. 71, if AB âˆ¥ CD and CD âˆ¥ EF, find âˆ ACE**

**Solution:**

Given that, AB âˆ¥ CD and CD âˆ¥ EF

Sum of the interior angles,

âˆ CEF + âˆ ECD = 180^{o}

130^{o} + âˆ ECD = 180^{o}

âˆ ECD = 180^{o} â€“ 130^{o}

âˆ ECD = 50^{o}

We know that alternate angles are equal

âˆ BAC = âˆ ACD

âˆ BAC = âˆ ECD + âˆ ACE

âˆ ACE = 70^{o} â€“ 50^{o}

âˆ ACE = 20^{o}

Therefore, âˆ ACE = 20^{o}

**15. In Fig. 72, if l âˆ¥ m, n âˆ¥ p and âˆ 1 = 85 ^{o}, find âˆ 2.**

**Solution:**

Given that, âˆ 1 = 85^{o}

âˆ 1 and âˆ 3 are corresponding angles

So,Â âˆ 1 = âˆ 3

âˆ 3 = 85^{o}

Sum of the interior angles is 180^{o}

âˆ 3 + âˆ 2 = 180^{o}

âˆ 2 = 180^{o} â€“ 85^{o}

âˆ 2 = 95^{o}

**16. In Fig. 73, a transversal n cuts two lines l and m. If âˆ 1 = 70 ^{o} and âˆ 7 = 80^{o}, is l âˆ¥ m?**

**Solution:**

Given âˆ 1 = 70^{o} and âˆ 7 = 80^{o}

We know that if the alternate exterior angles of the two lines are equal, then the lines are parallel.

Here,Â âˆ 1 and âˆ 7 are alternate exterior angles, but they are not equal

âˆ 1 â‰ âˆ 7 â‰ 80^{o}

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

**Solution:**

From the figure âˆ 2 = âˆ 3 are vertically opposite angels,

âˆ 2 = âˆ 3 = 65^{o}

âˆ 8 = âˆ 6 = 65^{o}

Therefore, âˆ 3 = âˆ 6

Hence, l âˆ¥ m

**18. In Fig. 75, Show that AB âˆ¥ EF.**

**Solution:**

We know that,

âˆ ACD = âˆ ACE + âˆ ECD

âˆ ACD = 35^{o} + 22^{o}

âˆ ACD = 57^{o} = âˆ BAC

Thus, lines BA and CD are intersected by the line AC such that, âˆ ACD = âˆ BAC

So, the alternate angles are equal

Therefore, AB âˆ¥ CDÂ â€¦â€¦1

Now,

âˆ ECD + âˆ CEF = 35^{o} + 45^{o} = 180^{o}

This, shows that sum of the angles of the interior angles on the same side of the transversal CE is 180^{o}

So, they are supplementary angles

Therefore, EF âˆ¥ CDÂ â€¦â€¦.2

From equation 1 and 2

We conclude that, AB âˆ¥ EF

**19. In Fig. 76, AB âˆ¥ CD. Find the values of x, y, z.**

**Solution:**

Given that AB âˆ¥ CD

Linear pair,

âˆ xÂ + 125^{o} = 180^{o}

âˆ xÂ = 180^{o} â€“ 125^{o}

âˆ x = 55^{o}

Corresponding angles

âˆ z = 125^{o}

Adjacent interior angles

âˆ x + âˆ z = 180^{o}

âˆ x + 125^{o} = 180^{o}

âˆ x = 180^{o} â€“ 125^{o}

âˆ x = 55^{o}

Adjacent interior angles

âˆ x + âˆ y = 180^{o}

âˆ y + 55^{o} = 180^{o}

âˆ y = 180^{o} â€“ 55^{o}

âˆ y = 125^{o}

**20. In Fig. 77, find out âˆ PXR, if PQ âˆ¥ RS.**

**Solution:**

Given PQ âˆ¥ RS

We need to find âˆ PXR

âˆ XRS = 50^{o}

âˆ XPR = 70^{o}

Given, that PQ âˆ¥ RS

âˆ PXR = âˆ XRS + âˆ XPR

âˆ PXR = 50^{o} + 70^{o}

âˆ PXR = 120^{o}

Therefore, âˆ PXR = 120^{o}

**21. In Figure, we have**

**(i) âˆ MLY = 2âˆ LMQ**

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

**(iii) âˆ XLM = âˆ PML, find âˆ ALY**

**(iv) âˆ ALY = (2x – 15) ^{o}, âˆ LMQ = (x + 40)^{o }, find x.**

**Solution:**

(i) âˆ MLY and âˆ LMQ are interior angles

âˆ MLY + âˆ LMQ = 180^{o}

2âˆ LMQ + âˆ LMQ = 180^{o}

3âˆ LMQ = 180^{o}

âˆ LMQ = 180^{o}/3

âˆ LMQ = 60^{o}

(ii) âˆ XLM = (2x – 10)^{o} and âˆ LMQ = (x + 30)^{o}, find x.

âˆ XLM = (2x – 10)^{o} and âˆ LMQ = (x + 30)^{o}

âˆ XLM and âˆ LMQ are alternate interior angles

âˆ XLM = âˆ LMQ

(2x – 10)^{o} = (x + 30)^{o}

2x â€“ x = 30^{o} + 10^{o}

x = 40^{o}

Therefore, x = 40Â°

(iii) âˆ XLM = âˆ PML, find âˆ ALY

âˆ XLM = âˆ PML

Sum of interior angles is 180 degrees

âˆ XLM + âˆ PML = 180^{o}

âˆ XLM + âˆ XLM = 180^{o}

2âˆ XLM = 180^{o}

âˆ XLM = 180^{o}/2

âˆ XLM = 90^{o}

âˆ XLM and âˆ ALY are vertically opposite angles

Therefore, âˆ ALY = 90^{o}

(iv) âˆ ALY = (2x – 15)^{o}, âˆ LMQ = (x + 40)^{o}, find x.

âˆ ALY and âˆ LMQ are corresponding angles

âˆ ALY = âˆ LMQ

(2x – 15)^{o }= (x + 40)^{o}

2x â€“ x = 40^{o} + 15^{o}

x = 55^{o}Â

Therefore, x = 55^{o}

**22. In Fig. 79, DE âˆ¥ BC. Find the values of x and y.**

**Solution:**

We know that,

ABC, DAB are alternate interior angles

âˆ ABC = âˆ DAB

So, x = 40^{o}

And ACB, EAC are alternate interior angles

âˆ ACB = âˆ EAC

So, y = 40^{o}

**23. In Fig. 80, line AC âˆ¥ line DE and âˆ ABD = 32 ^{o}, Find out the angles x and y if âˆ E = 122^{o}.**

**Solution:**

Given line AC âˆ¥ line DE and âˆ ABD = 32^{o}

âˆ BDE = âˆ ABD = 32^{o} â€“ Alternate interior angles

âˆ BDE + y = 180^{o}â€“ linear pair

32^{o }+ y = 180^{o}

y = 180^{o} â€“ 32^{o}

y = 148^{o}

âˆ ABE = âˆ E = 32^{o} â€“ Alternate interior angles

âˆ ABD + âˆ DBE = 122^{o}

32^{o} + x = 122^{o}

x = 122^{o} â€“ 32^{o}

x = 90^{o}

**24. In Fig. 81, side BC of Î”ABC has been produced to D and CE âˆ¥ BA. If âˆ ABC = 65 ^{o}, âˆ BAC = 55^{o}, find âˆ ACE, âˆ ECD, âˆ ACD.**

**Solution:**

Given âˆ ABC = 65^{o}, âˆ BAC = 55^{o}

Corresponding angles,

âˆ ABC = âˆ ECD = 55^{o}

Alternate interior angles,

âˆ BAC = âˆ ACE = 65^{o}

Now, âˆ ACD = âˆ ACE + âˆ ECD

âˆ ACD = 55^{o} + 65^{o}

= 120^{o}

**25. In Fig. 82, line CA âŠ¥ AB âˆ¥ line CR and line PR âˆ¥ line BD. FindÂ âˆ x, âˆ y, âˆ z.**

**Solution:**

Given that, CA âŠ¥ AB

âˆ CAB = 90^{o}

âˆ AQP = 20^{o}

By, angle of sum property

In Î”APD

âˆ CAB + âˆ AQP + âˆ APQ = 180^{o}

âˆ APQ = 180^{o} â€“ 90^{o} â€“ 20^{o}

âˆ APQ = 70^{o}

y and âˆ APQ are corresponding angles

y = âˆ APQ = 70^{o}

âˆ APQ and âˆ z are interior angles

âˆ APQ + âˆ z = 180^{o}

âˆ z = 180^{o} â€“ 70^{o}

âˆ z = 110^{o}

**26. In Fig. 83, PQ âˆ¥ RS. Find the value of x.**

**Â **

**Solution:**

Given, linear pair,

âˆ RCD + âˆ RCB = 180^{o}

âˆ RCB = 180^{o} â€“ 130^{o}

50^{o}

In Î”ABC,

âˆ BAC + âˆ ABC + âˆ BCA = 180^{o}

By, angle sum property

âˆ BAC = 180^{o} â€“ 55^{o} â€“ 50^{o}

âˆ BAC = 75^{o}

**27. In Fig. 84, AB âˆ¥ CD and AE âˆ¥ CF, âˆ FCG = 90 ^{o} and âˆ BAC = 120^{o}. Find the value of x, y and z.**

**Solution:**

Alternate interior angle

âˆ BAC = âˆ ACG = 120^{o}

âˆ ACF + âˆ FCG = 120^{o}

So, âˆ ACF = 120^{o} â€“ 90^{o}

= 30^{o}

Linear pair,

âˆ DCA + âˆ ACG = 180^{o}

âˆ x = 180^{o} â€“ 120^{o}

= 60^{o}

âˆ BAC + âˆ BAE + âˆ EAC = 360^{o}

âˆ CAE = 360^{o} â€“ 120^{o} â€“ (60^{o} + 30^{o})

= 150^{o}

**28. In Fig. 85, AB âˆ¥ CD and AC âˆ¥ BD. Find the values of x, y, z.**

**Solution:**

(i)Â Since, AC âˆ¥ BD and CD âˆ¥ AB, ABCD is a parallelogram

Adjacent angles of parallelogram,

âˆ CAD + âˆ ACD = 180^{o}

âˆ ACD = 180^{o} â€“ 65^{o}

= 115^{o}

Opposite angles of parallelogram,

âˆ CAD = âˆ CDB = 65^{o}

âˆ ACD = âˆ DBA = 115^{o}

(ii)Â Here,

AC âˆ¥ BD and CD âˆ¥ AB

Alternate interior angles,

âˆ DCA = x = 40^{o}

âˆ DAB = y = 35^{o}

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

**Solution:**

Let, F be the point of intersection of the line CD and the line passing through point E.

Here, âˆ ACD and âˆ CDE are alternate and equal angles.

So, âˆ ACD = âˆ CDEÂ = 100^{o}

Therefore, AC âˆ¥ EF

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

**Solution:**

Let, G be the point of intersection of the lines BC and DE

Since, AB âˆ¥ DE and BC âˆ¥ EF

The corresponding angles are,

âˆ ABC = âˆ DGC = âˆ DEF = 100^{o}