Exercise 14.2
Page-14.20Question 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:
∠EGB and ∠GHD
∠HGB and ∠FHD
∠EGA and ∠GHC
∠AGH and ∠CHF
Alternate angles:
∠EGB and ∠CHF
∠HGB and ∠CHG
∠EGA and ∠FHD
∠AGH and ∠GHD
(ii) Figure (ii)
Alternate angle to ∠d is ∠e.
Alternate angle to ∠g is ∠b.
Also,
Corresponding angle to ∠f is ∠c.
Corresponding angle to ∠h is ∠a.
(iii) Figure (iii)
Angle alternate to ∠PQR is ∠QRA.
Angle corresponding to ∠RQF is ∠ARB.
Angle alternate to ∠POE is ∠ARB.
(iv) Figure (ii)
Pair of interior angles are
∠a and ∠e
∠d and ∠f
Pair of exterior angles are
∠b and ∠h
∠c and ∠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:
∠ALM = ∠CMQ = 60° (Corresponding angles)
∠LMD = ∠CMQ = 60° (Vertically opposite angles)
∠ALM = ∠PLB = 60° (Vertically opposite angles)
Since
∠CMQ + ∠QMD = 180° (Linear pair)
∴ ∠QMD = 180°-60°=120°
∠QMD = ∠MLB = 120° (Corresponding angles)
∠QMD = ∠CML = 120° (Vertically opposite angles)
∠MLB = ∠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.
∠ALM=∠LMD=35° (Alternate interior angles)Since ∠PLA+∠ALM=180° (Linear pair)∴∠PLA=180°-35°=145°
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 ∠13 is ∠7.
Corresponding angle to ∠15 is ∠7.
Alternate angle to ∠15 is ∠5.
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,
∠1+∠2=180° (Linear pair)∴ ∠2=180°-∠1=180°-40°=140°∠5=∠1=40° (Corresponding angles)∠3=∠1=40° (Vertically opposite angles)∠7=∠3=40° (Corresponding angles)∠7=∠5=40° (Vertically opposite angles)
Also,
∠2=∠6=140° (Corresponding angles)∠2=∠4=140° (Vertically opposite angles)∠4=∠8=140° (Corresponding angles)∠8=∠6=40° (Vertically opposite angles)
Thus,
∠2=∠8, ∠3=∠5, ∠6=∠4, ∠1=∠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:
∠1+∠2=180° (Linear pair)⇒∠2=180°-∠1=180°-75°=105°∴∠1=∠5=75° (Corresponding angles)∠1=∠3=75° (Vertically opposite angles)∠5=∠7=75° (Vertically opposite angles)Now, ∠2=∠6=105° (Corresponding angles)∠6=∠8=105° (Vertically opposite angles)∠2=∠4=105° (Vertically opposite angles)
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 ∠QMD = 100°.
Thus, we have:
∠DMQ + ∠QMC = 180° (Linear pair)
∴∠QMC=180°-∠DMQ=180°-100°=80°
Thus,
∠DMQ = ∠BLM = 100° (Corresponding angles)
∠DMQ = ∠CML = 100° (Vertically opposite angles)
∠BLM = ∠PLA = 100° (Vertically opposite angles)
Also,
∠CMQ = ∠ALM = 80° (Corresponding angles)
∠CMQ = ∠DML = 80° (Vertically opposite angles)
∠ALM = ∠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:
∠z=80° (Vertically opposite angles)
∠z=∠t=80° (Corresponding angles)
∠z=∠y=80° (Corresponding angles)
∠x=∠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
∠2=∠5=100° (Alternate interior angles)∠5+∠3=180° (Linear pair)⇒∠3=180°-∠5=180°-100°=80°
Also,
∠1+∠6=180° (Linear pair)⇒∠6=180°-∠1=180°-120°=60°
We know that the sum of all the angles of triangle is 180°.
∴∠6+∠3+∠4=180°⇒60°+80°+∠4=180°⇒140°+∠4=180°⇒∠4=180°-140°=40°
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:
∠a=110° (Vertically opposite angles)∠b=∠a=110° (Corresponding angles)∠d=85° (Vertically opposite angles)∠c=∠d=85° (Corresponding angles)
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:
∠1=3x∠2=2x
Thus, we have:
∠1+∠2=180° (Linear pair)∴ 3x+2x=180°⇒5x=180°⇒x=180°5=36°Thus,∠1=3×36°=108°∠2=2×36°=72°
Now,
∠1=∠5=108° (Corresponding angles)∠1=∠3=108° (Vertically opposite angles)∠5=∠7=108° (Vertically opposite angles)∠2=∠6=72° (Corresponding angles)∠4=∠2=72° (Vertically opposite angles)∠8=∠6=72° (Vertically opposite angles)
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:
∠4+60°=180° (Linear pair)⇒∠4=180°-60°=120°∠4=∠1=120° (Corresponding angles)∠1=∠2=120° (Corresponding angles) ∠3=∠2=120° (Vertically opposite angles)Thus,∠1=∠2=∠3=120°
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:
∠3=∠1=60° (Corresponding angle)Now,∠3+∠4=180° (Linear pair)∠4=180°-∠3=180°-60°=120°∠2=∠4=120° (Alternate interior angles)
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:
∠BAC=∠ACD=70° (Alternate angles)Now,∠3+∠CEF=180° (Linear pair)⇒∠3=180°-∠CEF=180°-130°=50°Since CD||EF, then∠2=∠3=50° (Corresponding angles)∠ACE=∠ACD-∠2=70°-50°=20°
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:
∠3=∠1=85° (Corresponding angles)
∠3+∠2=180° (Sum of interior angles on the same side of the transversal)
∴∠2=180°-∠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, ∠1 and ∠7 are alternate exterior angles, but they are not equal.
∠1 ≠∠770°≠80°
Therefore, lines l and m are not parallel.
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:
∠2 = ∠3 = 65° (Vertically opposite angles)
∠8 = ∠6 = 65° (Vertically opposite angles)
∴ ∠3 = ∠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'.
∠BAC=57°=22°+35°=∠ACE+∠ECD∴ AB||CDHere, ∠E'
Question 19:
In Fig., AB || CD. Find the values of x, y, z.
Answer 19:
(Linear pair)
(Corresponding angles)
(Sum of adjacent interior angles is )
(Sum of adjacent interior angles is )
Question 20:
In Fig., find out ∠PXR, if PQ || RS.
Answer 20:
Draw a line parallel to PQ passing through X.
Here,
(Alternate interior angles)
∵ PQ || RS || XF
∴
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)
(ii)
(iii)
(iv)
Question 22:
In Fig., DE || BC. Find the values of x and y.
Answer 22:
ABC = DAB (Alternate interior angles)
ACB = EAC (Alternate interior angles)
Question 23:
In Fig., line AC || line DE and ∠ABD = 32°. Find out the angles x and y if ∠E = 122°.
Answer 23:
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:
ABC = ECD = 55° (Corresponding angles)
BAC = ACE = 65° (Alternate interior angles)
Now, ACD = ACE + ECD
⇒ 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,
We know that the sum of all the angles of triangle is 180°.
PBC = APQ = (Corresponding angles)
Since
Question 26:
In Fig., PQ || RS. Find the value of x.
Answer 26:
Question 27:
In Fig., AB || CD and AE || CF; ∠FCG = 90° and ∠BAC = 120°. Find the values of x, y and z.
Answer 27:
BAC = ACG = 120° (Alternate interior angle)
∴ ACF + FCG = 120°
⇒ ACF = 120° − 90° = 30°
DCA + ACG = 180° (Linear pair)
⇒x = 180° − 120° = 60°
BAC + BAE + EAC = 360°
CAE = 360° − 120° − (60° + 30°) = 150° (BAE = DCF)
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.
CAB + ACD = 180° (Sum of adjacent angles of a parallelogram)
∴ ACD = 180° − 65° = 115°
CAD = CDB = 65° (Opposite angles of a parallelogram)
ACD = DBA = 115° (Opposite angles of a parallelogram)
(ii) Here,
AC || BD and CD || AB
DAC = x = 40° (Alternate interior 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 ACD and CDE are alternate and equal angles, so
ACD = 100° = 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
∴ (Corresponding angles)
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