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:

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(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.

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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 \(\angle PLA\).

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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 \(\angle 15\).

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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.

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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.

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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.

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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.

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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 \(\angle 4\).

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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.

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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.

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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 \(\angle 3\).

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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 \(\angle 2\)

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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 \(\angle ACE\).

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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 \(\angle 2\).

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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?

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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?

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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.

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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.

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Sol:

Linear pair,

= \(\angle x\)  + \(125^{\circ}\) = \(180^{\circ}\)

\(\angle x\)  = \(180^{\circ}\)\(125^{\circ}\)

\(\angle x\)  = \(55^{\circ}\)

Corresponding angles

= \(\angle z\) = \(125^{\circ}\)

Adjacent interior angles

= \(\angle x\) + \(\angle z\) = \(180^{\circ}\)

= \(\angle x\) + \(125^{\circ}\) = \(180^{\circ}\)

= \(\angle x\) = \(180^{\circ}\)\(125^{\circ}\)

= \(\angle x\) = \(55^{\circ}\)

Adjacent interior angles

= \(\angle x\) + \(\angle y\) = \(180^{\circ}\)

= \(\angle y\) + \(55^{\circ}\) = \(180^{\circ}\)

= \(\angle y\) = \(180^{\circ}\)\(55^{\circ}\)

= \(\angle y\) = \(125^{\circ}\)

 

 

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

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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\(\angle LMQ\)

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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.

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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 \(\angle E\) = \(122^{\circ}\).

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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 \(\angle ACE\), \(\angle ECD\), \(\angle ACD\).

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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  \(\angle x\), \(\angle y\), \(\angle z\).

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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.

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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.

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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)

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

Adjacent angles of 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)

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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?

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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 \(\angle DEF\).

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Sol:

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

Since, AB || DE and BC || EF

The corresponding angles,

= \(\angle ABC\) = \(\angle DGC\) = \(\angle DEF\) = \(100^{\circ}\)