RD Sharma solution class 7 chapter 14 Lines and Angles Exercise 14.2

Exercise 14.2

Page-14.20

Question 1:

In Fig., line n is a transversal to lines l and m. Identify the following:

(i) Alternate and corresponding angles in Fig. (i).
(ii) Angles alternate to ∠d and ∠g and angles corresponding to angles ∠f and ∠h in Fig. (ii).
(iii) Angle alternate to ∠PQR, angle corresponding to ∠RQF and angle alternate to ∠PQE in Fig. (iii).
(iv) Pairs of interior and exterior angles on the same side of the transversal in Fig. (ii).

Answer 1:

(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
Alternate angles:
$\angle$EGB and $\angle$CHF
$\angle$HGB and $\angle$CHG
$\angle$EGA and $\angle$FHD
$\angle$AGH and $\angle$GHD

(ii) Figure (ii)
Alternate angle to $\angle$d is $\angle$e.
Alternate angle to $\angle$g is $\angle$b.
Also,
Corresponding angle to $\angle$f is $\angle$c.
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 and $\angle$e
$\angle$d and $\angle$f
Pair of exterior angles are
$\angle$b and $\angle$h
$\angle$c and $\angle$g

Question 2:

In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠CMQ = 60°, find all other angles in the figure.

Answer 2:

$\angle$ALM = $\angle$CMQ = $60°$        (Corresponding angles)
$\angle$LMD = $\angle$CMQ = $60°$        (Vertically opposite angles)
$\angle$ALM = $\angle$PLB = $60°$          (Vertically opposite angles)
Since
$\angle$CMQ + $\angle$QMD = $180°$     (Linear pair)
$\therefore \angle$QMD = $180°-60°=120°$
$\angle$QMD = $\angle$MLB = $120°$        (Corresponding angles)
$\angle$QMD = $\angle$CML = $120°$        (Vertically opposite angles)
$\angle$MLB = $\angle$ALP = $120°$          (Vertically opposite angles)

Question 3:

In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.

Answer 3:

In the given Fig., AB || CD.
$$\begin{gathered} \angle \mathrm{ALM}=\angle \mathrm{LMD}=35° \left(\mathrm{Alternate} \mathrm{interior} \mathrm{angles}\right) \\ \mathrm{Since} \angle \mathrm{PLA}+\angle \mathrm{ALM}=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \therefore \angle \mathrm{PLA}=180°-35°=145° \end{gathered}$$

Question 4:

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

Answer 4:

In this given Fig., line l || m.
Here,
Alternate angle to $\angle$13 is $\angle$7.
Corresponding angle to $\angle$15 is $\angle$7.
Alternate angle to $\angle$15 is $\angle$5.

Page-14.21

Question 5:

In Fig., line l || m and n is a transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.

Answer 5:

In the given figure, l || m.
Here,
$$\begin{gathered} \angle 1+\angle 2=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \therefore \angle 2=180°-\angle 1=180°-40°=140° \\ \angle 5=\angle 1=40° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle 3=\angle 1=40° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle 7=\angle 3=40° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle 7=\angle 5=40° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \end{gathered}$$
Also,
$$\begin{gathered} \angle 2=\angle 6=140° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle 2=\angle 4=140° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle 4=\angle 8=140° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle 8=\angle 6=40° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \end{gathered}$$
Thus,
$\angle 2=\angle 8, \angle 3=\angle 5, \angle 6=\angle 4, \angle 1=\angle 7$
Hence, alternate angles are equal.

Question 6:

In Fig., line l || m and a transversal n cuts them at P and Q respectively. If ∠1 = 75°, find all other angles.

Answer 6:

In the given figure, l || m, n is a transversal line and ∠1 = 75°.
Thus, we have:
$$\begin{gathered} \angle 1+\angle 2=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \Rightarrow \angle 2=180°-\angle 1=180°-75°=105° \\ \therefore \angle 1=\angle 5=75° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle 1=\angle 3=75° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle 5=\angle 7=75° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \mathrm{Now}, \\ \angle 2=\angle 6=105° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle 6=\angle 8=105° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle 2=\angle 4=105° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \end{gathered}$$

Question 7:

In Fig., AB || CD and a transversal PQ cuts them at L and M respectively. If ∠QMD = 100°, find all other angles.

Answer 7:

In the given figure, AB || CD, PQ is a transversal line and $\angle$QMD = 100°.
Thus, we have:
$\angle$DMQ + $\angle$QMC = 180°    (Linear pair)
$\therefore \angle \mathrm{QMC}=180°-\angle \mathrm{DMQ}=180°-100°=80°$
Thus,
$\angle$DMQ = $\angle$BLM = 100°         (Corresponding angles)
$\angle$DMQ = $\angle$CML = 100°         (Vertically opposite angles)
$\angle$BLM = $\angle$PLA = 100°           (Vertically opposite angles)
Also,
$\angle$CMQ = $\angle$ALM = 80°         (Corresponding angles)
$\angle$CMQ = $\angle$DML = 80°         (Vertically opposite angles)
$\angle$ALM = $\angle$PLB = 80°           (Vertically opposite angles)

Question 8:

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

Answer 8:

In the given figure, l || m and p || q.
Thus, we have:
$\angle z=80°$                (Vertically opposite angles)
$\angle z=\angle t=80°$       (Corresponding angles)
$\angle z=\angle y=80°$       (Corresponding angles)
$\angle x=\angle y=80°$       (Corresponding angles)
 

Question 9:

In Fig., line l || m, ∠1 = 120° and ∠2 = 100°, find out ∠3 and ∠4.

Answer 9:

In the given figure, ∠1 = 120° and ∠2 =100°.
Since l || m, so
$$\begin{gathered} \angle 2=\angle 5=100° \left(\mathrm{Alternate} \mathrm{interior} \mathrm{angles}\right) \\ \angle 5+\angle 3=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \Rightarrow \angle 3=180°-\angle 5=180°-100°=80° \end{gathered}$$
Also,
$$\begin{gathered} \angle 1+\angle 6=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \Rightarrow \angle 6=180°-\angle 1=180°-120°=60° \end{gathered}$$
We know that the sum of all the angles of triangle is 180°.
$$\begin{gathered} \therefore \angle 6+\angle 3+\angle 4=180° \\ \Rightarrow 60°+80°+\angle 4=180° \\ \Rightarrow 140°+\angle 4=180° \\ \Rightarrow \angle 4=180°-140°=40° \end{gathered}$$

Question 10:

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

Answer 10:

In the given figure, line l || m.
Thus, we have:
$$\begin{gathered} \angle a=110° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle b=\angle a=110° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle d=85° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle c=\angle d=85° \left(\mathrm{Corresponding} \mathrm{angles}\right) \end{gathered}$$

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Question 11:

In Fig., AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.

Answer 11:

In the given figure, AB || CD and t is a transversal line.
Now, let:
$$\begin{gathered} \angle 1=3x \\ \angle 2=2x \end{gathered}$$
Thus, we have:
$$\begin{gathered} \angle 1+\angle 2=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \therefore 3x+2x=180° \\ \Rightarrow 5x=180° \\ \Rightarrow x=\frac{180°}{5}=36° \\ \mathrm{Thus}, \\ \angle 1=3\times 36°=108° \\ \angle 2=2\times 36°=72° \end{gathered}$$
Now,
$$\begin{gathered} \angle 1=\angle 5=108° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle 1=\angle 3=108° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle 5=\angle 7=108° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle 2=\angle 6=72° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle 4=\angle 2=72° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle 8=\angle 6=72° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \end{gathered}$$

Question 12:

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

Answer 12:

In the given figure, l || m || n and p is a transversal line.
Thus, we have:
$$\begin{gathered} \angle 4+60°=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \Rightarrow \angle 4=180°-60°=120° \\ \angle 4=\angle 1=120° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle 1=\angle 2=120° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle 3=\angle 2=120° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \mathrm{Thus}, \\ \angle 1=\angle 2=\angle 3=120° \end{gathered}$$

Question 13:

In Fig., if l || m || n and ∠1 = 60°, find ∠2.

Answer 13:

In the given figure, l || m || n and ∠1 = 60°.
Thus, we have:
$$\begin{gathered} \angle 3=\angle 1=60° \left(\mathrm{Corresponding} \mathrm{angle}\right) \\ \mathrm{Now}, \\ \angle 3+\angle 4=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \angle 4=180°-\angle 3=180°-60°=120° \\ \angle 2=\angle 4=120° \left(\mathrm{Alternate} \mathrm{interior} \mathrm{angles}\right) \end{gathered}$$

Question 14:

In Fig., if AB || CD and CD || EF, find ∠ACE.

Answer 14:

In the given figure, AB || CD and CD || EF.
Extend line CE to E^'.

Thus, we have:
$$\begin{gathered} \angle \mathrm{BAC}=\angle \mathrm{ACD}=70° \left(\mathrm{Alternate} \mathrm{angles}\right) \\ \mathrm{Now}, \\ \angle 3+\angle \mathrm{CEF}=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \Rightarrow \angle 3=180°-\angle \mathrm{CEF}=180°-130°=50° \\ \mathrm{Since} \mathrm{CD}\left|\right|\mathrm{EF}, \mathrm{then} \\ \angle 2=\angle 3=50° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle \mathrm{ACE}=\angle \mathrm{ACD}-\angle 2=70°-50°=20° \end{gathered}$$

Question 15:

In Fig., if l || m, n || p and ∠1 = 85°, find ∠2.

Answer 15:

In the given figure, l || m, n || p and ∠1 = 85°.
Now, let ∠4 be the adjacent angle of ∠2.
Thus, we have:
$\angle 3=\angle 1=85° \left(\mathrm{Corresponding} \mathrm{angles}\right)$
$\angle 3+\angle 2=180°$       (Sum of interior angles on the same side of the transversal)
$\therefore \angle 2=180°-\angle 3=180°-85°=95°$

Question 16:

In Fig., a transversal n cuts two lines l and m. If ∠1 = 70° and ∠7 = 80°, is l || m?

Answer 16:

We know that if the alternate exterior angles of two lines are equal, then the lines are parallel.
In the given figure, $\angle 1 \mathrm{and} \angle 7$ are alternate exterior angles, but they are not equal.
 $$\begin{gathered} \angle 1 \ne \angle 7 \\ 70°\ne 80° \end{gathered}$$

Therefore, lines l and m are not parallel.

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Question 17:

In Fig., a transversal n cuts two lines l and m such that ∠2 = 65° and ∠8 = 65°. Are the lines parallel?

Answer 17:

 $\angle$2 = $\angle$3 = 65°        (Vertically opposite angles)   
 $\angle$8 = $\angle$6 = 65°         (Vertically opposite angles) 
∴ $\angle$3 = $\angle$6
⇒ l || m                       (Two lines are parallel if the alternate angles formed with the transversal are equal) 

Question 18:

In Fig., show that AB || EF.

Answer 18:

Extend line CE to E'.


$$\begin{gathered} \angle \mathrm{BAC}=57°=22°+35°=\angle \mathrm{ACE}+\angle \mathrm{ECD} \\ \therefore \mathrm{AB}\left|\right|\mathrm{CD} \\ \mathrm{Here}, \angle \mathrm{E}\text{'}\mathrm{EF}+\angle \mathrm{FEC}=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \Rightarrow \angle \mathrm{E}\text{'}\mathrm{EF}=180°-\angle \mathrm{FEC}=180°-145°=35°=\angle \mathrm{ECD} \\ \therefore \mathrm{EF}\left|\right|\mathrm{CD} \\ \mathrm{Thus}, \mathrm{AB}\left|\right|\mathrm{CD} \left|\right|\mathrm{EF} \end{gathered}$$

Question 19:

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

Answer 19:

$\angle x+125°=180°$             (Linear pair)
$\therefore \angle x=180°-125°=55°$

$\angle z=125°$            (Corresponding angles)
$\angle x+\angle z=180°$   (Sum of adjacent interior angles is $180°$)
$$\begin{gathered} \angle x+125°=180° \\ \Rightarrow \angle x=180°-125°=55° \end{gathered}$$

$\angle x+\angle y=180°$   (Sum of adjacent interior angles is $180°$)
$$\begin{gathered} 55°+\angle y=180° \\ \Rightarrow \angle y=180°-55°=125° \end{gathered}$$

Question 20:

In Fig., find out ∠PXR, if PQ || RS.

Answer 20:

Draw a line parallel to PQ passing through X.


Here,
$\angle \mathrm{PQX}=\angle \mathrm{PXF}=70° \mathrm{and} \angle \mathrm{SRX}=\angle \mathrm{RXF}=50°$      (Alternate interior angles)
∵ PQ || RS || XF
∴ $\angle \mathrm{PXR}=\angle \mathrm{PXF}+\angle \mathrm{FXR}=70°+50°=120°$

Question 21:

In Fig., we have

(i) ∠MLY = 2 ∠LMQ, find ∠LMQ.
(ii) ∠XLM = (2x − 10)° and ∠LMQ = x + 30°, find x.
(iii) ∠XLM = ∠PML, find ∠ALY
(iv) ∠ALY = (2x − 15)°, and ∠LMQ = (x + 40)°, find x

Answer 21:

(i)
$$\begin{gathered} \angle \mathrm{LMQ}=\angle \mathrm{ALY} \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \therefore \angle \mathrm{MLY}+ \angle \mathrm{ALY}=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \Rightarrow 2\angle \mathrm{ALY}+\angle \mathrm{ALY}=180° \\ \Rightarrow 3\angle \mathrm{ALY}=180° \\ \Rightarrow \angle \mathrm{ALY}=\frac{180°}{3}=60° \\ \therefore \angle \mathrm{LMQ}=60° \end{gathered}$$

(ii)
$$\begin{gathered} \angle \mathrm{XLM}=\angle \mathrm{LMQ} \left(\mathrm{Alternate} \mathrm{interior} \mathrm{angles}\right) \\ \Rightarrow \left(2x-10\right)°=\left(x+30\right)° \\ \Rightarrow 2x-x=30°+10° \\ \Rightarrow x=40° \end{gathered}$$

(iii)
$$\begin{gathered} \angle \mathrm{ALX}=\angle \mathrm{LMP} \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle \mathrm{ALX}+\angle \mathrm{XLM}=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \angle \mathrm{XLM}=\angle \mathrm{LMP} \left(\mathrm{Given}\right) \\ \therefore \angle \mathrm{LMP}+\angle \mathrm{LMP}=180° \\ \Rightarrow 2\angle \mathrm{LMP}=180° \\ \Rightarrow \angle \mathrm{LMP}=\frac{180° }{2}=90° \\ \angle \mathrm{XLM}=\angle \mathrm{LMP}=90° \\ \angle \mathrm{ALY}=\angle \mathrm{XLM} \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \therefore \angle \mathrm{ALY}=90° \end{gathered}$$

(iv)
$$\begin{gathered} \angle \mathrm{ALY}=\angle \mathrm{LMQ} \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \therefore \left(2x-15\right)°=\left(x+40\right)° \\ \Rightarrow 2x-x=40°+15° \\ \Rightarrow x=55° \end{gathered}$$

Question 22:

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

Answer 22:

$\angle$ABC = $\angle$DAB       (Alternate interior angles)
$\therefore x=40°$

$\angle$ACB = $\angle$EAC       (Alternate interior angles)
$\therefore y=55°$

Page-14.24

Question 23:

In Fig., line AC || line DE and ∠ABD = 32°. Find out the angles x and y if ∠E = 122°.

 

Answer 23:

$$\begin{gathered} \angle \mathrm{BDE}=\angle \mathrm{ABD}=32° \left(\mathrm{Alternate} \mathrm{interior} \mathrm{angles}\right) \\ \Rightarrow \angle \mathrm{BDE}+y=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \Rightarrow 32°+y=180° \\ \Rightarrow y=180°-32°=148° \end{gathered}$$

$$\begin{gathered} \angle \mathrm{ABE}=\angle \mathrm{E}=122° (\mathrm{Alternate} \mathrm{interior} \mathrm{angle}) \\ \angle \mathrm{ABD}+\angle \mathrm{DBE}=122° \\ 32°+x=122° \\ x=122°-32°=90° \end{gathered}$$

Question 24:

In Fig., side BC of ∆ABC has been produced to D and CE || BA. If ∠ABC = 65°, ∠BAC = 55°, find ∠ACE, ∠ECD and ∠ACD.

Answer 24:

$\angle$ABC = $\angle$ECD = 55°          (Corresponding angles)
$\angle$BAC = $\angle$ACE = 65°          (Alternate interior angles)
Now, $\angle$ACD = $\angle$ACE + $\angle$ECD
⇒ $\angle$ACD = 55° + 65° = 120° 

Question 25:

In Fig., line CA ⊥ AB || line CR and line PR || line BD. Find ∠x, ∠y and ∠z.

Answer 25:

Since CA ⊥ AB,
$\therefore \angle x=90°$
We know that the sum of all the angles of triangle is 180°.
$$\begin{gathered} \mathrm{In} ∆\mathrm{APQ}, \\ \angle \mathrm{QAP}+\angle \mathrm{APQ}+\angle \mathrm{PQA}=180° \\ \Rightarrow 90°+\angle \mathrm{APQ}+20°=180° \\ \Rightarrow 110°+\angle \mathrm{APQ}=180° \\ \Rightarrow \angle \mathrm{APQ}=180°-110°=70° \end{gathered}$$
$\angle$PBC = $\angle$APQ = $70°$            (Corresponding angles)
Since $\angle \mathrm{PRC}+\angle z=180° \left(\mathrm{Linear} \mathrm{pair}\right)$
$\therefore \angle z=180°-70°=110° \left[\angle \mathrm{APQ}=\angle \mathrm{PRC} \left(\mathrm{Alternate} \mathrm{interior} \mathrm{angles}\right)\right]$ 

Question 26:

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

Answer 26:

$$\begin{gathered} \angle \mathrm{RCD}+\angle \mathrm{RCB}=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \Rightarrow \angle \mathrm{RCB}=180°-130°=50° \\ \mathrm{In} △\mathrm{ABC}, \\ \angle \mathrm{BAC}+\angle \mathrm{ABC}+\angle \mathrm{BCA}=180° \left(\mathrm{Angle} \mathrm{sum} \mathrm{property}\right) \\ \Rightarrow \angle \mathrm{BAC}=180°-55°-50°=75° \end{gathered}$$



 

Question 27:

In Fig., AB || CD and AE || CF; ∠FCG = 90° and ∠BAC = 120°. Find the values of x, y and z.

Answer 27:

$\angle$BAC = $\angle$ACG = 120°          (Alternate interior angle)
∴ $\angle$ACF + $\angle$FCG = 120°  
⇒ $\angle$ACF = 120° − 90° = 30°

$\angle$DCA + $\angle$ACG = 180°            (Linear pair)
⇒$\angle$x = 180° − 120° = 60°

$\angle$BAC + $\angle$BAE + $\angle$EAC = 360°
$\angle$CAE = 360° − 120° − (60° + 30°) = 150°             ($\angle$BAE =  $\angle$DCF)

Page-14.25

Question 28:

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

Answer 28:

(i) Since AC || BD and CD || AB, ABCD is a parallelogram.
$\angle$CAB + $\angle$ACD = 180°     (Sum of adjacent angles of a parallelogram)
∴ $\angle$ACD = 180° − 65° = 115°
$\angle$CAD = $\angle$CDB = 65°         (Opposite angles of a parallelogram)
$\angle$ACD = $\angle$DBA = 115°       (Opposite angles of a parallelogram)

(ii) Here,
AC || BD and CD || AB
$\angle$DAC = x = 40°            (Alternate interior angle)
$\angle$DAB = y = 35°            (Alternate interior angle)

Question 29:

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

Answer 29:

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



Since $\angle$ACD and $\angle$CDE are alternate and equal angles, so
$\angle$ACD = 100° = $\angle$CDE
∴ AC || EF

Question 30:

In Fig. 87, the corresponding arms of ∠ABC and ∠DEF are parallel. If ∠ABC = 75°, find ∠DEF.

Answer 30:

   


Construction: Let G be the point of intersection of lines BC and DE.

∵ AB || DE and BC || EF

∴ $\angle \mathrm{ABC}=\angle \mathrm{DGC}=\angle \mathrm{DEF}=75°$  (Corresponding angles)​