In fig., m ll n and ∠1=65°. Find ∠5 and ∠8.
3 points

∠5=25° and ∠8=65°
∠5=65° and ∠8=25°
∠5=65° and ∠8=115°
∠5=115° and ∠8=65°
Answers
Step-by-step explanation:
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R.D. Sharma Solutions»R.D. Sharma Class 7 Solutions»Chapter 14 Lines and Angles»Lines and Angles Exercise 14.2
Chapter 14: Lines and Angles Exercise – 14.2
Question: 1
In Figure, line n is a transversal to line l and m. Identify the following:
line n is a transversal to line l and m
(i) Alternate and corresponding angles in Figure.(i)
(ii) Angles alternate to ∠d and ∠g and angles corresponding to ∠RQF and angle alternate to ∠PQE in Figure. (ii)
(iii) Angle alternate to ∠PQR, angle corresponding to ∠RQF and angle alternate to ∠PQE in Figure. (iii)
(iv) Pairs of interior and exterior angles on the same side of the transversal in Figure. (iii)
Solution:
(i) Figure (i)
Corresponding angles:
∠EGB and ∠GHD
∠HGB and ∠FHD
∠EGA and ∠GHC
∠AGH and ∠CHF
The alternate angles are:
∠EGB and ∠CHF
∠HGB and ∠CHG
∠EGA and ∠FHD
∠AGH and ∠GHD
(ii) Figure (ii)
The alternate angle to ∠d is ∠e.
The alternate angle to ∠g is ∠b.
The corresponding angle to ∠f is ∠c.
The 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 is ∠e.
∠d is ∠f.
Pair of exterior angles are
∠b is ∠h.
∠c is ∠g.
Question: 2
In Figure, 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.
AB and CD are parallel lines intersected by a transversal PQ at L and M
Solution:
Corresponding angles:
∠ALM = ∠CMQ = 60°
Vertically opposite angles:
∠LMD = ∠CMQ = 60°
Vertically opposite angles:
∠ALM = ∠PLB = 60°
Here,
∠CMQ + ∠QMD = 180° are the linear pair
= ∠QMD = 180° – 60°
= 120°
Corresponding angles:
∠QMD = ∠MLB = 120°
Vertically opposite angles
∠QMD = ∠CML = 120°
Vertically opposite angles
∠MLB = ∠ALP = 120°
Question: 3
In Figure, AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.
AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M
Solution:
Given that,
∠LMD = 35°
∠LMD and ∠LMC is a linear pair
∠LMD + ∠LMC = 180°
= ∠LMC = 180° – 35°
= 145°
So, ∠LMC = ∠PLA = 145°
And, ∠LMC = ∠MLB = 145°
∠MLB and ∠ALM is a linear pair
∠MLB + ∠ALM = 180°
= ∠ALM = 180° – 145°
= ∠ALM = 35°
Therefore, ∠ALM = 35°, ∠PLA = 145°.
Question: 4
The line n is transversal to line l and m in Figure. Identify the angle alternate to ∠13, angle corresponding to ∠15, and angle alternate to ∠15.
The line n is transversal to line l and m
Solution:
Given that, l ∥ m
So,
The angle alternate to ∠13 is ∠7
The angle corresponding to ∠15 is ∠7
The angle alternate to ∠15 is ∠5
Question: 5
In Figure, line l ∥ m and n is transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.
line l ∥ m and n is transversal
Solution:
Given that,
∠1 = 40°
∠1 and ∠2 is a linear pair
= ∠1 + ∠2 = 180°
= ∠2 = 180° – 40°
= ∠2 = 140°
∠2 and ∠6 is a corresponding angle pair
So, ∠6 = 140°
∠6 and ∠5 is a linear pair
= ∠6 + ∠5 = 180°
= ∠5 = 180° – 140°
= ∠5 = 40°
∠3 and ∠5 are alternative interior angles
So, ∠5 = ∠3 = 40°
∠3 and ∠4 is a linear pair
= ∠3 + ∠4 = 180°
= ∠4 = 180° – 40°
= ∠4 = 140°
∠4 and ∠6 are a pair interior angles
So, ∠4 = ∠6 = 140°
∠3 and ∠7 are pair of corresponding angles
So, ∠3 = ∠7 = 40°
Therefore, ∠7 = 40°
∠4 and ∠8 are a pair corresponding angles
So, ∠4 = ∠8 = 140°
Therefore, ∠8 = 140°
So, ∠1 = 40°, ∠2 = 140°, ∠3 = 40°, ∠4 = 140°, ∠5 = 40°, ∠6 = 140°, ∠7 = 40°, ∠8 = 140°
Question: 6
In Figure, line l ∥ m and a transversal n cuts them P and Q respectively. If ∠1 = 75°, find all other angles.
line l ∥ m and a transversal n cuts them P and Q
Solution:
Given that, l ∥ m and ∠1 = 75∘
We know that,
∠1 + ∠2 = 180° → (linear pair)
= ∠2 = 180° – 75°
= ∠2 = 105°
here, ∠1 = ∠5 = 75° are corresponding angles
∠5 = ∠7 = 75° are vertically opposite angles.
∠2 = ∠6 = 105° are corresponding angles
∠6 = ∠8 = 105° are vertically opposite angles
∠2 = ∠4 = 105° are vertically opposite angles
So, ∠1 = 75°, ∠2 = 105°, ∠3 = 75°, ∠4 = 105°, ∠5 = 75°, ∠6 = 105°, ∠7 = 75°, ∠8 = 105°
Question: 7
In Figure, AB ∥ CD and a transversal PQ cuts at L and M respectively. If ∠QMD = 100°, find all the other angles.
Answer:
Step-by-step explanation:
In Figure, line n is a transversal to line l and m. Identify the following:
line n is a transversal to line l and m
(i) Alternate and corresponding angles in Figure.(i)
(ii) Angles alternate to ∠d and ∠g and angles corresponding to ∠RQF and angle alternate to ∠PQE in Figure. (ii)
(iii) Angle alternate to ∠PQR, angle corresponding to ∠RQF and angle alternate to ∠PQE in Figure. (iii)
(iv) Pairs of interior and exterior angles on the same side of the transversal in Figure. (iii)
Solution:
(i) Figure (i)
Corresponding angles:
∠EGB and ∠GHD
∠HGB and ∠FHD
∠EGA and ∠GHC
∠AGH and ∠CHF
The alternate angles are:
∠EGB and ∠CHF
∠HGB and ∠CHG
∠EGA and ∠FHD
∠AGH and ∠GHD
(ii) Figure (ii)
The alternate angle to ∠d is ∠e.
The alternate angle to ∠g is ∠b.
The corresponding angle to ∠f is ∠c.
The 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 is ∠e.
∠d is ∠f.
Pair of exterior angles are
∠b is ∠h.
∠c is ∠g.
Question: 2
In Figure, 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.
AB and CD are parallel lines intersected by a transversal PQ at L and M
Solution:
Corresponding angles:
∠ALM = ∠CMQ = 60°
Vertically opposite angles:
∠LMD = ∠CMQ = 60°
Vertically opposite angles:
∠ALM = ∠PLB = 60°
Here,
∠CMQ + ∠QMD = 180° are the linear pair
= ∠QMD = 180° – 60°
= 120°
Corresponding angles:
∠QMD = ∠MLB = 120°
Vertically opposite angles
∠QMD = ∠CML = 120°
Vertically opposite angles
∠MLB = ∠ALP = 120°
Question: 3
In Figure, AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.
AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M
Solution:
Given that,
∠LMD = 35°
∠LMD and ∠LMC is a linear pair
∠LMD + ∠LMC = 180°
= ∠LMC = 180° – 35°
= 145°
So, ∠LMC = ∠PLA = 145°
And, ∠LMC = ∠MLB = 145°
∠MLB and ∠ALM is a linear pair
∠MLB + ∠ALM = 180°
= ∠ALM = 180° – 145°
= ∠ALM = 35°
Therefore, ∠ALM = 35°, ∠PLA = 145°.
Question: 4
The line n is transversal to line l and m in Figure. Identify the angle alternate to ∠13, angle corresponding to ∠15, and angle alternate to ∠15.
The line n is transversal to line l and m
Solution:
Given that, l ∥ m
So,
The angle alternate to ∠13 is ∠7
The angle corresponding to ∠15 is ∠7
The angle alternate to ∠15 is ∠5
Question: 5
In Figure, line l ∥ m and n is transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.
line l ∥ m and n is transversal
Solution:
Given that,
∠1 = 40°
∠1 and ∠2 is a linear pair
= ∠1 + ∠2 = 180°
= ∠2 = 180° – 40°
= ∠2 = 140°
∠2 and ∠6 is a corresponding angle pair
So, ∠6 = 140°
∠6 and ∠5 is a linear pair
= ∠6 + ∠5 = 180°
= ∠5 = 180° – 140°
= ∠5 = 40°
∠3 and ∠5 are alternative interior angles
So, ∠5 = ∠3 = 40°
∠3 and ∠4 is a linear pair
= ∠3 + ∠4 = 180°
= ∠4 = 180° – 40°
= ∠4 = 140°
∠4 and ∠6 are a pair interior angles
So, ∠4 = ∠6 = 140°
∠3 and ∠7 are pair of corresponding angles
So, ∠3 = ∠7 = 40°
Therefore, ∠7 = 40°
∠4 and ∠8 are a pair corresponding angles
So, ∠4 = ∠8 = 140°
Therefore, ∠8 = 140°
So, ∠1 = 40°, ∠2 = 140°, ∠3 = 40°, ∠4 = 140°, ∠5 = 40°, ∠6 = 140°, ∠7 = 40°, ∠8 = 140°
Question: 6
In Figure, line l ∥ m and a transversal n cuts them P and Q respectively. If ∠1 = 75°, find all other angles.
line l ∥ m and a transversal n cuts them P and Q
Solution:
Given that, l ∥ m and ∠1 = 75∘
We know that,
∠1 + ∠2 = 180° → (linear pair)
= ∠2 = 180° – 75°
= ∠2 = 105°
here, ∠1 = ∠5 = 75° are corresponding angles
∠5 = ∠7 = 75° are vertically opposite angles.
∠2 = ∠6 = 105° are corresponding angles
∠6 = ∠8 = 105° are vertically opposite angles
∠2 = ∠4 = 105° are vertically opposite angles
So, ∠1 = 75°, ∠2 = 105°, ∠3 = 75°, ∠4 = 105°, ∠5 = 75°, ∠6 = 105°, ∠7 = 75°, ∠8 = 105°
Question: 7
In Figure, AB ∥ CD and a transversal PQ cuts at L and M respectively. If ∠QMD = 100°, find all the other angles