#### 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 \(\angle d\)

(iii) Angle alternate to \(\angle PQR\)

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

Sol:

(i) Figure (i)

Corresponding angles :

\(\angle EGB\)

\(\angle HGB\)

\(\angle EGA\)

\(\angle AGH\)

The alternate angles are :

\(\angle EGB\)

\(\angle HGB\)

\(\angle EGA\)

\(\angle AGH\)

(ii) Figure (ii)

The alternate angle to \(\angle d\)

The alternate angle to \(\angle g\)

The corresponding angle to \(\angle f\)

The corresponding angle to \(\angle h\)

(iii) Figure (iii)

Angle alternate to \(\angle PQR\)

Angle corresponding to \(\angle RQF\)

Angle alternate to \(\angle POE\)

(iv) Figure (ii)

Pair of interior angles are

\(\angle a \)

\(\angle d\)

Pair of exterior angles are

\(\angle b\)

\(\angle c\)

Q2. In Figure, AB and CD are parallel lines intersected by a transversal PQ at L and M respectively, If \(\angle CMQ\)

Sol:

Corresponding angles :

\(\angle ALM\)

Vertically opposite angles :

\(\angle LMD\)

Vertically opposite angles :

\(\angle ALM\)

Here,

\(\angle CMQ\)

= \(\angle QMD\)

= \(120^{\circ}\)

Corresponding angles :

\(\angle QMD\)

Vertically opposite angles

\(\angle QMD\)

Vertically opposite angles

\(\angle MLB\)

Q3. In Fig. 60, AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M respectively. If \(\angle LMD\)

Sol:

Given that,

\(\angle LMD\)

\(\angle LMD\)

\(\angle LMD\)

= \(\angle LMC\)

= \(145^{\circ}\)

So, \(\angle LMC\)

And, \(\angle LMC\)

\(\angle MLB\)

\(\angle MLB\)

= \(\angle ALM\)

= \(\angle ALM\)

Therefore, \(\angle ALM\)

Q4. The line n is transversal to line l and m in Fig. 61. Identify the angle alternate to \(\angle 13\)

Sol:

Given that, l || m

So,

The angle alternate to \(\angle 13\)

The angle corresponding to \(\angle 15\)

The angle alternate to \(\angle 15\)

Q5. In Fig. 62, line l || m and n is transversal. If \(\angle 1\)

Sol:

Given that,

\(\angle 1\)

\(\angle 1\)

= \(\angle 1\)

= \(\angle 2\)

= \(\angle 2\)

\(\angle 2\)

So, \(\angle 6\)

\(\angle 6\)

= \(\angle 6\)

= \(\angle 5\)

= \(\angle 5\)

\(\angle 3\)

So, \(\angle 5\)

\(\angle 3\)

= \(\angle 3\)

= \(\angle 4\)

= \(\angle 4\)

\(\angle 4\)

So, \(\angle 4\)

\(\angle 3\)

So, \(\angle 3\)

Therefore, \(\angle 7\)

\(\angle 4\)

So, \(\angle 4\)

Therefore, \(\angle 8\)

So, \(\angle 1\)

Q6. In Fig. 63, line l || m and a transversal n cuts them P and Q respectively. If \(\angle 1\)

Sol:

Given that, l || m and \(\angle 1\)

We know that,

\(\angle 1\)

= \(\angle 2\)

= \(\angle 2\)

here, \(\angle 1\)

\(\angle 5\)

\(\angle 2\)

\(\angle 6\)

\(\angle 2\)

So, \(\angle 1\)

Q7. In Fig. 64, AB || CD and a transversal PQ cuts at L and M respectively. If \(\angle QMD\)

Sol:

Given that, AB || CD and \(\angle QMD\)

We know that,

Linear pair,

\(\angle QMD\)

= \(\angle QMC\)

= \(\angle QMC\)

= \(\angle QMC\)

Corresponding angles,

\(\angle DMQ\)

\(\angle CMQ\)

Vertically Opposite angles,

\(\angle DMQ\)

\(\angle BLM\)

\(\angle CMQ\)

\(\angle ALM\)

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

Sol:

Give that , angle is \(80^{\circ}\)

\(\angle z\)

= \(\angle z\)

\(\angle z\)

= \(\angle z\)

Therefore, \(\angle t\)

\(\angle z\)

= \(\angle z\)

Therefore, \(\angle y\)

\(\angle x\)

= \(\angle y\)

Therefore, \(\angle x\)

Q9. In Fig. 66, line l || m, \(\angle 1\)

Sol:

Given that, \(\angle 1\)

\(\angle 1\)

= \(\angle 1\)

= \(\angle 5\)

= \(\angle 5\)

Therefore, \(\angle 5\)

\(\angle 2\)

= \(\angle 2\)

Therefore, \(\angle 6\)

\(\angle 6\)

= \(\angle 6\)

= \(\angle 3\)

= \(\angle 3\)

Therefore, \(\angle 3\)

By, angles of sum property

= \(\angle 3\)

= \(\angle 4\)

= \(\angle 4\)

Therefore, \(\angle 4\)

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

Sol:

Given that, l || m

Vertically opposite angles,

\(\angle a\)

Corresponding angles,

\(\angle a\)

Therefore, \(\angle b\)

Vertically opposite angle,

\(\angle d\)

Corresponding angles,

\(\angle d\)

Therefore, \(\angle c\)

Hence, \(\angle a\)

Q11. In Fig. 68, AB || CD and \(\angle 1\)

Sol:

Given that,

\(\angle 1\)

Let us take the angles as 3x, 2x

\(\angle 1\)

= 3x + 2x = \(180^{\circ}\)

= 5x = \(180^{\circ}\)

= x = \(\frac{180^{\circ}}{5}\)

= x = \(36^{\circ}\)

Therefore, \(\angle 1\)

\(\angle 2\)

\(\angle 1\)

= \(\angle 1\)

Therefore, \(\angle 5\)

\(\angle 2\)

= \(\angle 2\)

Therefore, \(\angle 6\)

\(\angle 4\)

= \(\angle 4\)

Therefore, \(\angle 4\)

\(\angle 3\)

= \(\angle 3\)

Therefore, \(\angle 5\)

\(\angle 2\)

= \(\angle 2\)

Therefore, \(\angle 8\)

\(\angle 1\)

= \(\angle 1\)

Therefore, \(\angle 7\)

Hence, \(\angle 1\)

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

Sol:

Linear pair,

= \(\angle 4\)

= \(\angle 4\)

= \(\angle 4\)

\(\angle 4\)

= \(\angle 4\)

Therefore, \(\angle 1\)

\(\angle 1\)

= \(\angle 2\)

Therefore, \(\angle 2\)

\(\angle 2\)

= \(\angle 2\)

Therefore, \(\angle 3\)

Q13. In Fig. 70, if l || m || n and \(\angle 1\)

Sol:

Given that,

Corresponding angles :

\(\angle 1\)

= \(\angle 1\)

Therefore, \(\angle 3\)

\(\angle 3\)

= \(\angle 3\)

= \(\angle 4\)

= \(\angle 4\)

\(\angle 3\)

= \(\angle 4\)

Therefore, \(\angle 2\)

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

Sol :

Given that,

Sum of the interior angles,

= \(\angle CEF\)

= \(130^{\circ}\)

= \(\angle ECD\)

= \(\angle ECD\)

We know that alternate angles are equal

= \(\angle BAC\)

= \(\angle BAC\)

= \(\angle ACE\)

= \(\angle ACE\)

Therefore, \(\angle ACE\)

Q15. In Fig. 72, if l || m, n || p and \(\angle 1\)

Sol:

Given that, \(\angle 1\)

\(\angle 1\)

So, \(\angle 1\)

= \(\angle 3\)

Sum of the interior angles

= \(\angle 3\)

= \(\angle 2\)

= \(\angle 2\)

Q16. In Fig. 73, a transversal n cuts two lines l and m. If \(\angle 1\)

Sol:

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

Here, \(\angle 1\)

= \(\angle 1\)

Q17. In Fig. 74, a transversal n cuts two lines l and m such that \(\angle 2\)

Sol:

vertically opposite angels,

\(\angle 2\)

\(\angle 8\)

Therefore, \(\angle 3\)

Hence , l || m

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

Sol:

We know that,

\(\angle ACD\)

= \(\angle ACD\)

= \(\angle ACD\)

Thus, lines BA and CD are intersected by the line AC such that, \(\angle ACD\)

So, the alternate angles are equal

Therefore, AB || CD ——— 1

Now,

\(\angle ECD\)

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,

= \(\angle x\)

= \(\angle x\)

= \(\angle x\)

Corresponding angles

= \(\angle z\)

Adjacent interior angles

= \(\angle x\)

= \(\angle x\)

= \(\angle x\)

= \(\angle x\)

Adjacent interior angles

= \(\angle x\)

= \(\angle y\)

= \(\angle y\)

= \(\angle y\)

Q20. In Fig. 77, find out \(\angle PXR\)

Sol:

We need to find \(\angle PXR\)

\(\angle XRS\)

\(\angle XPR\)

Given, that PQ || RS

\(\angle PXR\)

\(\angle PXR\)

\(\angle PXR\)

Therefore, \(\angle PXR\)

Q21. In Fig. 78, we have

(i) \(\angle MLY\)

Sol:

\(\angle MLY\)

= \(\angle MLY\)

= 2\(\angle LMQ\)

= 3\(\angle LMQ\)

= \(\angle LMQ\)

= \(\angle LMQ\)

(ii) \(\angle XLM\)

Sol:

\(\angle XLM\)

\(\angle XLM\)

= \(\angle XLM\)

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

= 2x – x = \(30^{\circ}\)

= x = \(40^{\circ}\)

Therefore, x = \(40^{\circ}\)

(iii) \(\angle XLM\)

Sol:

\(\angle XLM\)

Sum of interior angles is 180 degrees

= \(\angle XLM\)

= \(\angle XLM\)

= 2\(\angle XLM\)

= \(\angle XLM\)

= \(\angle XLM\)

\(\angle XLM\)

Therefore, \(\angle ALY\)

(iv) \(\angle ALY\)

Sol:

\(\angle ALY\)

= \(\angle ALY\)

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

= 2x – x = \(40^{\circ}\)

= 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

\(\angle ABC\)

So, x = \(40^{\circ}\)

And ACB, EAC are alternate interior angles

\(\angle ACB\)

So, y = \(40^{\circ}\)

Q23. In Fig. 80, line AC || line DE and \(\angle ABD\)

Sol:

\(\angle BDE\)

= \(\angle BDE\)

= \(32^{\circ}\)

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

= y = \(148^{\circ}\)

\(\angle ABE\)

= \(\angle ABD\)

= \(32^{\circ}\)

= x = \(122^{\circ}\)

= x = \(90^{\circ}\)

Q24. In Fig. 81, side BC of \(\Delta\)

Sol:

Corresponding angles,

\(\angle ABC\)

Alternate interior angles,

\(\angle BAC\)

Now, \(\angle ACD\)

= \(\angle ACD\)

= \(120^{\circ}\)

Q25. In Fig. 82, line CA \(\perp\)

Sol:

Given that, CA \(\perp\)

= \(\angle CAB\)

= \(\angle AQP\)

By, angle of sum property

In \(\Delta\)

= \(\angle CAB\)

= \(\angle APQ\)

= \(\angle APQ\)

y and \(\angle APQ\)

= y = \(\angle APQ\)

\(\angle APQ\)

= \(\angle APQ\)

= \(\angle z\)

= \(\angle z\)

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

Sol:

Given,

Linear pair,

\(\angle RCD\)

= \(\angle RCB\)

= \(50^{\circ}\)

In \(\Delta\)

\(\angle BAC\)

By, angle sum property

= \(\angle BAC\)

= \(\angle BAC\)

Q27. In Fig. 84, AB || CD and AE || CF, \(\angle FCG\)

Sol:

Alternate interior angle

\(\angle BAC\)

= \(\angle ACF\)

So, \(\angle ACF\)

= \(30^{\circ}\)

Linear pair,

\(\angle DCA\)

= \(\angle x\)

= \(60^{\circ}\)

\(\angle BAC\)

\(\angle CAE\)

= \(150^{\circ}\)

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

Adjacent angles of parallelogram,

\(\angle CAD\)

= \(\angle ACD\)

= \(115^{\circ}\)

Opposite angles of parallelogram,

= \(\angle CAD\)

= \(\angle ACD\)

(ii)

here,

AC || BD and CD || AB

Alternate interior angles,

\(\angle DCA\)

\(\angle DAB\)

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, \(\angle ACD\)

So, \(\angle ACD\)

Therefore, AC || EF

Q30. In Fig. 87, the corresponding arms of \(\angle ABC\)

Sol:

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

Since, AB || DE and BC || EF

The corresponding angles,

= \(\angle ABC\)