# RD Sharma Solutions Class 7 Lines And Angles Exercise 14.2

## RD Sharma Solutions Class 7 Chapter 14 Exercise 14.2

### RD Sharma Class 7 Solutions Chapter 14 Ex 14.2 PDF Free Download

#### 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)

(ii) Angles alternate to $\angle d$ and $\angle g$ and angles corresponding to $\angle RQF$ and angle alternate to $\angle PQE$ in Fig. 58 (ii)

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

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

Sol:

(i) Figure (i)

Corresponding angles :

$\angle EGB$ and  $\angle GHD$

$\angle HGB$ and  $\angle FHD$

$\angle EGA$ and  $\angle GHC$

$\angle AGH$ and  $\angle CHF$

The alternate angles are :

$\angle EGB$ and  $\angle CHF$

$\angle HGB$ and  $\angle CHG$

$\angle EGA$ and  $\angle FHD$

$\angle AGH$ and  $\angle GHD$

(ii) Figure (ii)

The alternate angle to $\angle d$ is $\angle e$.

The alternate angle to $\angle g$ is $\angle b$.

The corresponding angle to $\angle f$ is $\angle c$.

The corresponding angle to $\angle h$ is $\angle a$.

(iii) Figure (iii)

Angle alternate to $\angle PQR$ is $\angle QRA$.

Angle corresponding to $\angle RQF$ is $\angle ARB$.

Angle alternate to $\angle POE$ is $\angle ARB$.

(iv) Figure (ii)

Pair of interior angles are

$\angle a$ is $\angle e$.

$\angle d$ is $\angle f$.

Pair of exterior angles are

$\angle b$ is $\angle h$.

$\angle c$ is $\angle g$.

Q2. In Figure, AB and CD are parallel lines intersected by a transversal PQ at L and M respectively, If  $\angle CMQ$ = $60^{\circ}$, find all other angles in the figure.

Sol:

Corresponding angles :

$\angle ALM$ = $\angle CMQ$ = $60^{\circ}$

Vertically opposite angles :

$\angle LMD$ = $\angle CMQ$ = $60^{\circ}$

Vertically opposite angles :

$\angle ALM$ = $\angle PLB$ = $60^{\circ}$

Here,

$\angle CMQ$ + $\angle QMD$ = $180^{\circ}$ are the linear pair

= $\angle QMD$ = $180^{\circ}$$60^{\circ}$

= $120^{\circ}$

Corresponding angles :

$\angle QMD$ = $\angle MLB$ = $120^{\circ}$

Vertically opposite angles

$\angle QMD$ = $\angle CML$ = $120^{\circ}$

Vertically opposite angles

$\angle MLB$ = $\angle ALP$ = $120^{\circ}$

Q3. In Fig. 60, AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M respectively. If $\angle LMD$ = $35^{\circ}$ find $\angle ALM$ and $\angle PLA$.

Sol:

Given that,

$\angle LMD$ = $35^{\circ}$

$\angle LMD$ and $\angle LMC$ is a linear pair

$\angle LMD$ + $\angle LMC$ = $180^{\circ}$

=      $\angle LMC$ = $180^{\circ}$$35^{\circ}$

= $145^{\circ}$

So, $\angle LMC$ = $\angle PLA$ = $145^{\circ}$

And, $\angle LMC$ = $\angle MLB$ = $145^{\circ}$

$\angle MLB$ and $\angle ALM$ is a linear pair

$\angle MLB$ + $\angle ALM$ = $180^{\circ}$

=         $\angle ALM$ = $180^{\circ}$$145^{\circ}$

=         $\angle ALM$ = $35^{\circ}$

Therefore, $\angle ALM$ = $35^{\circ}$, $\angle PLA$ = $145^{\circ}$.

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

Sol:

Given that, l || m

So,

The angle alternate to $\angle 13$ is $\angle 7$

The angle corresponding to $\angle 15$ is $\angle 7$

The angle alternate to $\angle 15$ is $\angle 5$

Q5. In Fig. 62, line l || m and n is transversal. If $\angle 1$ = $40^{\circ}$, find all the angles and check that all corresponding angles and alternate angles are equal.

Sol:

Given that,

$\angle 1$ = $40^{\circ}$

$\angle 1$ and $\angle 2$ is a linear pair

= $\angle 1$ + $\angle 2$ = $180^{\circ}$

= $\angle 2$ = $180^{\circ}$$40^{\circ}$

= $\angle 2$ = $140^{\circ}$

$\angle 2$ and $\angle 6$  is a corresponding angle pair

So, $\angle 6$ = $140^{\circ}$

$\angle 6$ and $\angle 5$ is a linear pair

= $\angle 6$ + $\angle 5$ = $180^{\circ}$

= $\angle 5$ = $180^{\circ}$$140^{\circ}$

= $\angle 5$ = $40^{\circ}$

$\angle 3$ and $\angle 5$  are alternative interior angles

So, $\angle 5$ = $\angle 3$ = $40^{\circ}$

$\angle 3$ and $\angle 4$ is a linear pair

= $\angle 3$ + $\angle 4$ = $180^{\circ}$

= $\angle 4$ = $180^{\circ}$$40^{\circ}$

= $\angle 4$ = $140^{\circ}$

$\angle 4$ and $\angle 6$  are a pair interior angles

So, $\angle 4$ = $\angle 6$ = $140^{\circ}$

$\angle 3$ and $\angle 7$ are pair of corresponding angles

So, $\angle 3$ = $\angle 7$ = $40^{\circ}$

Therefore, $\angle 7$ = $40^{\circ}$

$\angle 4$ and $\angle 8$  are a pair corresponding angles

So, $\angle 4$ = $\angle 8$ = $140^{\circ}$

Therefore, $\angle 8$ = $140^{\circ}$

So, $\angle 1$ = $40^{\circ}$, $\angle 2$ = $140^{\circ}$, $\angle 3$ = $40^{\circ}$, $\angle 4$ = $140^{\circ}$, $\angle 5$ = $40^{\circ}$, $\angle 6$ = $140^{\circ}$, $\angle 7$ = $40^{\circ}$, $\angle 8$ = $140^{\circ}$

Q6. In Fig. 63, line l || m and a transversal n cuts them P and Q respectively. If $\angle 1$ = $75^{\circ}$, find all other angles.

Sol:

Given that, l || m and $\angle 1$ = $75^{\circ}$

We know that,

$\angle 1$ + $\angle 2$ = $180^{\circ}$ —- (linear pair)

= $\angle 2$ = $180^{\circ}$$75^{\circ}$

= $\angle 2$ = $105^{\circ}$

here, $\angle 1$ = $\angle 5$ = $75^{\circ}$ are corresponding angles

$\angle 5$ = $\angle 7$ = $75^{\circ}$ are vertically opposite angles.

$\angle 2$ = $\angle 6$ = $105^{\circ}$ are corresponding angles

$\angle 6$ = $\angle 8$ = $105^{\circ}$ are vertically opposite angles

$\angle 2$ = $\angle 4$ = $105^{\circ}$ are vertically opposite angles

So, $\angle 1$ = $75^{\circ}$, $\angle 2$ = $105^{\circ}$, $\angle 3$ = $75^{\circ}$, $\angle 4$ = $105^{\circ}$, $\angle 5$ = $75^{\circ}$, $\angle 6$ = $105^{\circ}$, $\angle 7$ = $75^{\circ}$, $\angle 8$ = $105^{\circ}$

Q7. In Fig. 64, AB || CD and a transversal PQ cuts at L and M respectively. If $\angle QMD$ = $100^{\circ}$, find all the other angles.

Sol:

Given that, AB || CD and $\angle QMD$ = $100^{\circ}$

We know that,

Linear pair,

$\angle QMD$ + $\angle QMC$ = $180^{\circ}$

= $\angle QMC$ = $180^{\circ}$$\angle QMD$

= $\angle QMC$ = $180^{\circ}$$100^{\circ}$

= $\angle QMC$ = $80^{\circ}$

Corresponding angles,

$\angle DMQ$ = $\angle BLM$ = $100^{\circ}$

$\angle CMQ$ = $\angle ALM$ = $80^{\circ}$

Vertically Opposite angles,

$\angle DMQ$ = $\angle CML$ = $100^{\circ}$

$\angle BLM$ = $\angle PLA$ = $100^{\circ}$

$\angle CMQ$ = $\angle DML$ = $80^{\circ}$

$\angle ALM$ = $\angle PLB$ = $80^{\circ}$

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

Sol:

Give that , angle is $80^{\circ}$

$\angle z$ and $80^{\circ}$ are vertically opposite angles

= $\angle z$ = $80^{\circ}$

$\angle z$ and $\angle t$ are corresponding angles

= $\angle z$ = $\angle t$

Therefore, $\angle t$ = $80^{\circ}$

$\angle z$ and $\angle y$ are corresponding angles

= $\angle z$ = $\angle y$

Therefore, $\angle y$ = $80^{\circ}$

$\angle x$ and $\angle y$ are corresponding angles

= $\angle y$ = $\angle x$

Therefore, $\angle x$ = $80^{\circ}$

Q9. In Fig. 66, line l || m, $\angle 1$ = $120^{\circ}$ and $\angle 2$ = $100^{\circ}$, find out $\angle 3$ and $\angle 4$.

Sol:

Given that, $\angle 1$ = $120^{\circ}$ and $\angle 2$ = $100^{\circ}$

$\angle 1$ and $\angle 5$ a linear pair

= $\angle 1$ + $\angle 5$ = $180^{\circ}$

= $\angle 5$ = $180^{\circ}$$120^{\circ}$

$\angle 5$ = $60^{\circ}$

Therefore, $\angle 5$ = $60^{\circ}$

$\angle 2$ and $\angle 6$ are corresponding angles

= $\angle 2$ = $\angle 6$$100^{\circ}$

Therefore,  $\angle 6$$100^{\circ}$

$\angle 6$ and $\angle 3$ a linear pair

= $\angle 6$ + $\angle 3$ = $180^{\circ}$

= $\angle 3$ = $180^{\circ}$$100^{\circ}$

$\angle 3$ = $80^{\circ}$

Therefore, $\angle 3$ = $80^{\circ}$

By, angles of sum property

= $\angle 3$ + $\angle 5$ + $\angle 4$  = $180^{\circ}$

= $\angle 4$ = $180^{\circ}$$80^{\circ}$$60^{\circ}$

= $\angle 4$ = $40^{\circ}$

Therefore, $\angle 4$ = $40^{\circ}$

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

Sol:

Given that, l || m

Vertically opposite angles,

$\angle a$ = $110^{\circ}$

Corresponding angles,

$\angle a$ = $\angle b$

Therefore, $\angle b$ = $110^{\circ}$

Vertically opposite angle,

$\angle d$ = $85^{\circ}$

Corresponding angles,

$\angle d$ = $\angle c$

Therefore, $\angle c$ = $85^{\circ}$

Hence, $\angle a$ = $110^{\circ}$, $\angle b$ = $110^{\circ}$, $\angle c$ = $85^{\circ}$, $\angle d$ = $85^{\circ}$

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

Sol:

Given that,

$\angle 1$ and $\angle 2$ are 3 : 2

Let us take the angles as 3x, 2x

$\angle 1$ and $\angle 2$ are linear pair

= 3x + 2x = $180^{\circ}$

= 5x = $180^{\circ}$

=  x = $\frac{180^{\circ}}{5}$

=  x = $36^{\circ}$

Therefore, $\angle 1$ = 3x = 3(36) = $108^{\circ}$

$\angle 2$ = 2x = 2(36) = $72^{\circ}$

$\angle 1$ and $\angle 5$ are corresponding angles

= $\angle 1$ = $\angle 5$

Therefore, $\angle 5$ = $108^{\circ}$

$\angle 2$ and $\angle 6$ are corresponding angles

= $\angle 2$ = $\angle 6$

Therefore, $\angle 6$ = $72^{\circ}$

$\angle 4$ and $\angle 6$ are alternate pair of angles

= $\angle 4$ = $\angle 6$$72^{\circ}$

Therefore, $\angle 4$ = $72^{\circ}$

$\angle 3$ and $\angle 5$ are alternate pair of angles

= $\angle 3$ = $\angle 5$$108^{\circ}$

Therefore, $\angle 5$ = $108^{\circ}$

$\angle 2$ and $\angle 8$ are alternate exterior of angles

= $\angle 2$ = $\angle 8$$72^{\circ}$

Therefore, $\angle 8$ = $72^{\circ}$

$\angle 1$ and $\angle 7$ are alternate exterior of angles

= $\angle 1$ = $\angle 7$$108^{\circ}$

Therefore, $\angle 7$ = $108^{\circ}$

Hence, $\angle 1$ = $108^{\circ}$, $\angle 2$ = $72^{\circ}$, $\angle 3$ = $108^{\circ}$, $\angle 4$ = $72^{\circ}$, $\angle 5$ = $108^{\circ}$, $\angle 6$ = $72^{\circ}$, $\angle 7$ = $108^{\circ}$, $\angle 8$ = $72^{\circ}$

Q12. In Fig. 69 l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find $\angle 1$, $\angle 2$ and $\angle 3$.

Sol:

Linear pair,

= $\angle 4$ + $60^{\circ}$ = $180^{\circ}$

= $\angle 4$ = $180^{\circ}$$60^{\circ}$

= $\angle 4$ = $120^{\circ}$

$\angle 4$ and $\angle 1$ are corresponding angles

= $\angle 4$ = $\angle 1$

Therefore, $\angle 1$ = $120^{\circ}$

$\angle 1$ and $\angle 2$ are corresponding angles

= $\angle 2$ = $\angle 1$

Therefore, $\angle 2$ = $120^{\circ}$

$\angle 2$ and $\angle 3$ are vertically opposite angles

= $\angle 2$ = $\angle 3$

Therefore, $\angle 3$ = $120^{\circ}$

Q13. In Fig. 70, if l || m || n and $\angle 1$ = $60^{\circ}$, find $\angle 2$

Sol:

Given that,

Corresponding angles :

$\angle 1$ = $\angle 3$

= $\angle 1$ = $60^{\circ}$

Therefore, $\angle 3$ = $60^{\circ}$

$\angle 3$ and $\angle 4$ are linear pair

= $\angle 3$ + $\angle 4$ = $180^{\circ}$

= $\angle 4$ = $180^{\circ}$$60^{\circ}$

= $\angle 4$ = $120^{\circ}$

$\angle 3$ and $\angle 4$ are alternative interior angles

= $\angle 4$ = $\angle 2$

Therefore, $\angle 2$ = $120^{\circ}$

Q14. In Fig. 71, if AB || CD and CD|| EF, find $\angle ACE$.

Sol :

Given that,

Sum of the interior angles,

= $\angle CEF$ + $\angle ECD$ = $180^{\circ}$

= $130^{\circ}$ + $\angle ECD$ = $180^{\circ}$

= $\angle ECD$ = $180^{\circ}$$130^{\circ}$

= $\angle ECD$ = $50^{\circ}$

We know that alternate angles are equal

= $\angle BAC$ = $\angle ACD$

= $\angle BAC$ = $\angle ECD$ + $\angle ACE$

= $\angle ACE$ = $70^{\circ}$$50^{\circ}$

= $\angle ACE$ = $20^{\circ}$

Therefore, $\angle ACE$ = $20^{\circ}$

Q15. In Fig. 72, if l || m, n || p and $\angle 1$ = $85^{\circ}$, find $\angle 2$.

Sol:

Given that, $\angle 1$ = $85^{\circ}$

$\angle 1$ and $\angle 3$ are corresponding angles

So,  $\angle 1$ = $\angle 3$

$\angle 3$ = $85^{\circ}$

Sum of the interior angles

= $\angle 3$ + $\angle 2$ = $180^{\circ}$

= $\angle 2$ = $180^{\circ}$$85^{\circ}$

= $\angle 2$ = $95^{\circ}$

Q16. In Fig. 73, a transversal n cuts two lines l and m. If $\angle 1$ = $70^{\circ}$ and $\angle 7$ = $80^{\circ}$, is l || m?

Sol:

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

Here,  $\angle 1$ and $\angle 7$ are alternate exterior angles , but they are not equal

= $\angle 1$$\angle 7$$80^{\circ}$

Q17.  In Fig. 74, a transversal n cuts two lines l and m such that $\angle 2$ = $65^{\circ}$ and $\angle 8$ = $65^{\circ}$. Are the lines parallel?

Sol:

vertically opposite angels,

$\angle 2$ = $\angle 3$ = $65^{\circ}$

$\angle 8$ = $\angle 6$ = $65^{\circ}$

Therefore, $\angle 3$ = $\angle 6$

Hence , l || m

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

Sol:

We know that,

$\angle ACD$ = $\angle ACE$ + $\angle ECD$

= $\angle ACD$ = $35^{\circ}$ + $22^{\circ}$

= $\angle ACD$ = $57^{\circ}$ = $\angle BAC$

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

So, the alternate angles are equal

Therefore, AB || CD   ——— 1

Now,

$\angle ECD$ + $\angle CEF$ = $35^{\circ}$ + $45^{\circ}$ = $180^{\circ}$

This, shows that sum of the angles of the interior angles on the same side of the transversal CE is 180 degrees

So, they are supplementary angles

Therefore, EF || CD           ————- 2

From eq 1 and 2

We can say that, AB || EF

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

Sol:

Linear pair,

= $\angle x$  + $125^{\circ}$ = $180^{\circ}$

$\angle x$  = $180^{\circ}$$125^{\circ}$

$\angle x$  = $55^{\circ}$

Corresponding angles

= $\angle z$ = $125^{\circ}$

= $\angle x$ + $\angle z$ = $180^{\circ}$

= $\angle x$ + $125^{\circ}$ = $180^{\circ}$

= $\angle x$ = $180^{\circ}$$125^{\circ}$

= $\angle x$ = $55^{\circ}$

= $\angle x$ + $\angle y$ = $180^{\circ}$

= $\angle y$ + $55^{\circ}$ = $180^{\circ}$

= $\angle y$ = $180^{\circ}$$55^{\circ}$

= $\angle y$ = $125^{\circ}$

Q20. In Fig. 77, find out $\angle PXR$, if PQ || RS.

Sol:

We need to find $\angle PXR$

$\angle XRS$ = $50^{\circ}$

$\angle XPR$ = $70^{\circ}$

Given, that PQ || RS

$\angle PXR$ = $\angle XRS$ + $\angle XPR$

$\angle PXR$ = $50^{\circ}$ + $70^{\circ}$

$\angle PXR$ = $120^{\circ}$

Therefore, $\angle PXR$ = $120^{\circ}$

Q21. In Fig. 78, we have

(i) $\angle MLY$ = 2$\angle LMQ$

Sol:

$\angle MLY$ and $\angle LMQ$ are interior angles

= $\angle MLY$ + $\angle LMQ$ = $180^{\circ}$

= 2$\angle LMQ$ + $\angle LMQ$ = $180^{\circ}$

= 3$\angle LMQ$ = $180^{\circ}$

= $\angle LMQ$ = $\frac{180^{\circ}}{3}$

= $\angle LMQ$ = $60^{\circ}$

(ii) $\angle XLM$ = $(2x-10)^{\circ}$ and $\angle LMQ$ = $(x+30)^{\circ}$, find x.

Sol:

$\angle XLM$ = $(2x-10)^{\circ}$ and $\angle LMQ$ = $(x+30)^{\circ}$

$\angle XLM$ and $\angle LMQ$ are alternate interior angles

= $\angle XLM$ = $\angle LMQ$

= $(2x-10)^{\circ}$ = $(x+30)^{\circ}$

= 2x – x = $30^{\circ}$ + $10^{\circ}$

=      x = $40^{\circ}$

Therefore, x = $40^{\circ}$

(iii) $\angle XLM$ = $\angle PML$, find $\angle ALY$

Sol:

$\angle XLM$ = $\angle PML$

Sum of interior angles is 180 degrees

$\angle XLM$ + $\angle PML$ = $180^{\circ}$

$\angle XLM$ + $\angle XLM$ = $180^{\circ}$

=  2$\angle XLM$ = $180^{\circ}$

= $\angle XLM$ = $\frac{180^{\circ}}{2}$

= $\angle XLM$ = $90^{\circ}$

$\angle XLM$ and $\angle ALY$ are vertically opposite angles

Therefore, $\angle ALY$ = $90^{\circ}$

(iv) $\angle ALY$ = $(2x-15)^{\circ}$, $\angle LMQ$ = $(x+40)^{\circ}$, find x.

Sol:

$\angle ALY$ and $\angle LMQ$ are corresponding angles

= $\angle ALY$ = $\angle LMQ$

= $(2x-15)^{\circ}$ = $(x+40)^{\circ}$

= 2x – x = $40^{\circ}$ + $15^{\circ}$

=          x = $55^{\circ}$

Therefore, x = $55^{\circ}$

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

Sol:

We know that, ABC, DAB are alternate interior angles

$\angle ABC$ = $\angle DAB$

So, x = $40^{\circ}$

And ACB, EAC are alternate interior angles

$\angle ACB$ = $\angle EAC$

So, y = $40^{\circ}$

Q23. In Fig. 80, line AC || line DE and $\angle ABD$ = $32^{\circ}$, Find out the angles x and y if $\angle E$ = $122^{\circ}$.

Sol:

$\angle BDE$ = $\angle ABD$ = $32^{\circ}$ – alternate interior angles

= $\angle BDE$ + y = $180^{\circ}$         – linear pair

= $32^{\circ}$ + y = $180^{\circ}$

=     y = $180^{\circ}$$32^{\circ}$

=      y = $148^{\circ}$

$\angle ABE$ = $\angle E$ = $32^{\circ}$ – alternate interior angles

= $\angle ABD$ + $\angle DBE$ = $122^{\circ}$

= $32^{\circ}$ + x = $122^{\circ}$

=     x = $122^{\circ}$$32^{\circ}$

=      x = $90^{\circ}$

Q24. In Fig. 81, side BC of $\Delta$ABC has been produced to D and CE || BA. If $\angle ABC$ = $65^{\circ}$, $\angle BAC$ = $55^{\circ}$, find $\angle ACE$, $\angle ECD$, $\angle ACD$.

Sol:

Corresponding angles,

$\angle ABC$ = $\angle ECD$ = $55^{\circ}$

Alternate interior angles,

$\angle BAC$ = $\angle ACE$ = $65^{\circ}$

Now, $\angle ACD$ = $\angle ACE$ + $\angle ECD$

= $\angle ACD$ = $55^{\circ}$ +$65^{\circ}$

= $120^{\circ}$

Q25. In Fig. 82, line CA $\perp$ AB || line CR and line PR || line BD. Find  $\angle x$, $\angle y$, $\angle z$.

Sol:

Given that, CA $\perp$ AB

= $\angle CAB$ = $90^{\circ}$

= $\angle AQP$ = $20^{\circ}$

By, angle of sum property

In $\Delta$APD

= $\angle CAB$ + $\angle AQP$ + $\angle APQ$ = $180^{\circ}$

= $\angle APQ$ = $180^{\circ}$$90^{\circ}$$20^{\circ}$

= $\angle APQ$ = $70^{\circ}$

y and $\angle APQ$ are corresponding angles

= y = $\angle APQ$ = $70^{\circ}$

$\angle APQ$ and $\angle z$ are interior angles

= $\angle APQ$ + $\angle z$ = $180^{\circ}$

= $\angle z$ = $180^{\circ}$$70^{\circ}$

= $\angle z$ = $110^{\circ}$

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

Sol:

Given,

Linear pair,

$\angle RCD$ + $\angle RCB$ = $180^{\circ}$

= $\angle RCB$ = $180^{\circ}$$130^{\circ}$

= $50^{\circ}$

In $\Delta$ABC,

$\angle BAC$ + $\angle ABC$ + $\angle BCA$ = $180^{\circ}$

By, angle sum property

= $\angle BAC$ = $180^{\circ}$$55^{\circ}$$50^{\circ}$

= $\angle BAC$ = $75^{\circ}$

Q27. In Fig. 84, AB || CD and AE || CF, $\angle FCG$ = $90^{\circ}$ and $\angle BAC$ = $120^{\circ}$. Find the value of x, y and z.

Sol:

Alternate interior angle

$\angle BAC$ = $\angle ACG$ = $120^{\circ}$

=   $\angle ACF$ + $\angle FCG$ = $120^{\circ}$

So, $\angle ACF$ = $120^{\circ}$$90^{\circ}$

= $30^{\circ}$

Linear pair,

$\angle DCA$ + $\angle ACG$ = $180^{\circ}$

= $\angle x$ = $180^{\circ}$$120^{\circ}$

= $60^{\circ}$

$\angle BAC$ + $\angle BAE$ + $\angle EAC$ = $360^{\circ}$

$\angle CAE$ = $360^{\circ}$$120^{\circ}$ – ($60^{\circ}$ + $30^{\circ}$)

= $150^{\circ}$

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

Sol:

(i)

Since, AC || BD and CD || AB, ABCD is a parallelogram

$\angle CAD$ + $\angle ACD$ = $180^{\circ}$

= $\angle ACD$ = $180^{\circ}$$65^{\circ}$

= $115^{\circ}$

Opposite angles of parallelogram,

= $\angle CAD$ = $\angle CDB$ = $65^{\circ}$

= $\angle ACD$ = $\angle DBA$ = $115^{\circ}$

(ii)

here,

AC || BD and CD || AB

Alternate interior angles,

$\angle DCA$ = x = $40^{\circ}$

$\angle DAB$ = y = $35^{\circ}$

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

Sol:

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

Here, $\angle ACD$ and $\angle CDE$ are alternate and equal angles.

So, $\angle ACD$ = $\angle CDE$  = $100^{\circ}$

Therefore, AC || EF

Q30. In Fig. 87, the corresponding arms of $\angle ABC$ and $\angle DEF$ are parallel. If $\angle ABC$ = $75^{\circ}$, find $\angle DEF$.

Sol:

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

Since, AB || DE and BC || EF

The corresponding angles,

= $\angle ABC$ = $\angle DGC$ = $\angle DEF$ = $100^{\circ}$

#### Practise This Question

The equation ax+by+c=0 completely describes a linear equation in two variables where a, b, and c are real numbers.