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:
(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.
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.
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.
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.
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.
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.
Solution:
Given that, AB ∥ CD and ∠QMD = 100°
We know that,
Linear pair,
∠QMD + ∠QMC = 180°
= ∠QMC = 180° – ∠QMD
= ∠QMC = 180° – 100°
= ∠QMC = 80°
Corresponding angles,
∠DMQ = ∠BLM = 100°
∠CMQ = ∠ALM = 80°
Vertically Opposite angles,
∠DMQ = ∠CML = 100°
∠BLM = ∠PLA = 100°
∠CMQ = ∠DML = 80°
∠ALM = ∠PLB = 80°
Question: 8
In Figure, l ∥ m and p || q. Find the values of x, y, z, t.
Solution:
Give that, angle is 80°
∠z and 80° are vertically opposite angles
= ∠z = 80°
∠z and ∠t are corresponding angles
= ∠z = ∠t
Therefore, ∠t = 80°
∠z and ∠y are corresponding angles
= ∠z = ∠y
Therefore, ∠y = 80°
∠x and ∠y are corresponding angles
= ∠y = ∠x
Therefore, ∠x = 80°
Question: 9
In Figure, line l ∥ m, ∠1 = 120° and ∠2 = 100°, find out ∠3 and ∠4.
Solution:
Given that, ∠1 = 120° and ∠2 = 100°
∠1 and ∠5 a linear pair
= ∠1 + ∠5 = 180°
= ∠5 = 180° – 120°
= ∠5 = 60°
Therefore, ∠5 = 60°
∠2 and ∠6 are corresponding angles
= ∠2 = ∠6 = 100°
Therefore, ∠6 = 100°
∠6 and ∠3 a linear pair
= ∠6 + ∠3 = 180°
= ∠3 = 180° – 100°
= ∠3 = 80°
Therefore, ∠3 = 80°
By, angles of sum property
= ∠3 + ∠5 + ∠4 = 180°
= ∠4 = 180° – 80° – 60°
= ∠4 = 40°
Therefore, ∠4 = 40°
Question: 10
In Figure, l ∥ m. Find the values of a, b, c, d. Give reasons.
Solution:
Given that, l ∥ m
Vertically opposite angles,
∠a = 110°
Corresponding angles,
∠a = ∠b
Therefore, ∠b = 110°
Vertically opposite angle,
∠d = 85°
Corresponding angles,
∠d = ∠c
Therefore, ∠c = 85°
Hence, ∠a = 110°, ∠b = 110°, ∠c = 85°, ∠d = 85°
Question: 11
In Figure, AB ∥ CD and ∠1 and ∠2 are in the ratio of 3: 2. Determine all angles from 1 to 8.
Solution:
Given that,
∠1 and ∠2 are 3: 2
Let us take the angles as 3x, 2x
∠1 and ∠2 are linear pair
= 3x + 2x = 180°
= 5x = 180°
= x = 180°/5
= x = 36°
Therefore, ∠1 = 3x = 3(36) = 108°
∠2 = 2x = 2(36) = 72°
∠1 and ∠5 are corresponding angles
= ∠1 = ∠5
Therefore, ∠5 = 108°
∠2 and ∠6 are corresponding angles
= ∠2 = ∠6
Therefore, ∠6 = 72°
∠4 and ∠6 are alternate pair of angles
= ∠4 = ∠6 = 72°
Therefore, ∠4 = 72°
∠3 and ∠5 are alternate pair of angles
= ∠3 = ∠5 = 108°
Therefore, ∠5 = 108°
∠2 and ∠8 are alternate exterior of angles
= ∠2 = ∠8 = 72°
Therefore, ∠8 = 72°
∠1 and ∠7 are alternate exterior of angles
= ∠1 = ∠7 = 108°
Therefore, ∠7 = 108°
Hence, ∠1 = 108°, ∠2 = 72°, ∠3 = 108°, ∠4 = 72°, ∠5 = 108°, ∠6 = 72°, ∠7 = 108°, ∠8 = 72°
Question: 12
In Figure, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠1, ∠2 and ∠3.
Solution:
Linear pair,
= ∠4 + 60° = 180°
= ∠4 = 180° – 60∘
= ∠4 = 120°
∠4 and ∠1 are corresponding angles
= ∠4 = ∠1
Therefore, ∠1 = 120°
∠1 and ∠2 are corresponding angles
= ∠2 = ∠1
Therefore, ∠2 = 120°
∠2 and ∠3 are vertically opposite angles
= ∠2 = ∠3
Therefore, ∠3 = 120°
Question: 13
In Figure, if l ∥ m ∥ n and ∠1 = 60°, find ∠2
Solution:
Given that,
Corresponding angles:
∠1 = ∠3
= ∠1 = 60°
Therefore, ∠3 = 60°
∠3 and ∠4 are linear pair
= ∠3 + ∠4 = 180°
= ∠4 = 180° – 60°
= ∠4 = 120°
∠3 and ∠4 are alternative interior angles
= ∠4 = ∠2
Therefore, ∠2 = 120°
Question: 14
In Figure, if AB ∥ CD and CD ∥ EF, find ∠ACE
Solution:
Given that,
Sum of the interior angles,
= ∠CEF + ∠ECD = 180°
= 130° + ∠ECD = 180°
= ∠ECD = 180° – 130°
= ∠ECD = 50°
We know that alternate angles are equal
= ∠BAC = ∠ACD
= ∠BAC = ∠ECD + ∠ACE
= ∠ACE = 70° – 50°
= ∠ACE = 20°
Therefore, ∠ACE = 20°
Question: 15
In Figure, if l ∥ m, n ∥ p and ∠1 = 85°, find ∠2.
Solution:
Given that, ∠1 = 85°
∠1 and ∠3 are corresponding angles
So, ∠1 = ∠3
= ∠3 = 85°
Sum of the interior angles
= ∠3 + ∠2 = 180°
= ∠2 = 180° – 85°
= ∠2 = 95°
Question: 16
In Figure, a transversal n cuts two lines l and m. If ∠1 = 70° and ∠7 = 80°, is l ∥ m?
Solution:
We know that if the alternate exterior angles of the two lines are equal, then the lines are parallel.
Here, ∠1 and ∠7 are alternate exterior angles, but they are not equal
= ∠1 ≠ ∠7 ≠ 80°
Question: 17
In Figure, a transversal n cuts two lines l and m such that ∠2 = 65° and ∠8 = 65°. Are the lines parallel?
Solution:
vertically opposite angels,
∠2 = ∠3 = 65°
∠8 = ∠6 = 65°
Therefore, ∠3 = ∠6
Hence, l ∥ m
Question: 18
In Figure, Show that AB ∥ EF.
Solution:
We know that,
∠ACD = ∠ACE + ∠ECD
= ∠ACD = 35° + 22°
= ∠ACD = 57° = ∠BAC
Thus, lines BA and CD are intersected by the line AC such that, ∠ACD = ∠BAC
So, the alternate angles are equal
Therefore, AB ∥ CD — 1
Now,
∠ECD + ∠CEF = 35° + 45° = 180°
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
Question: 19
In Figure, AB ∥ CD. Find the values of x, y, z.
Solution:
Linear pair,
= ∠x + 125° = 180°
= ∠x = 180° – 125°
= ∠x = 55°
Corresponding angles
= ∠z = 125°
Adjacent interior angles
= ∠x + ∠z = 180°
= ∠x + 125° = 180°
= ∠x = 180° – 125°
= ∠x = 55°
Adjacent interior angles
= ∠x + ∠y = 180°
= ∠y + 55° = 180°
= ∠y = 180° – 55°
= ∠y = 125°
Question: 20
In Figure, find out ∠PXR, if PQ ∥ RS.
Solution:
We need to find ∠PXR
∠XRS = 50°
∠XPR = 70°
Given, that PQ ∥ RS
∠PXR = ∠XRS + ∠XPR
∠PXR = 50° + 70°
∠PXR = 120°
Therefore, ∠PXR = 120°
Question: 21
In Figure, we have
(i) ∠MLY = 2∠LMQ
(ii) ∠XLM = (2x - 10)° and ∠LMQ = (x + 30) °, find x.
(iii) ∠XLM = ∠PML, find ∠ALY
(iv) ∠ALY = (2x - 15) °, ∠LMQ = (x + 40) °, find x.
Solution:
(i) ∠MLY and ∠LMQ are interior angles
= ∠MLY + ∠LMQ = 180°
= 2∠LMQ + ∠LMQ = 180°
= 3∠LMQ = 180°
= ∠LMQ = 180°/3
= ∠LMQ = 60°
(ii) ∠XLM = (2x - 10)° and ∠LMQ = (x + 30) °, find x.
∠XLM = (2x - 10) ° and ∠LMQ = (x + 30) °
∠XLM and ∠LMQ are alternate interior angles
= ∠XLM = ∠LMQ
= (2x - 10) ° = (x + 30) °
= 2x – x = 30° + 10°
= x = 40°
Therefore, x = 40°
(iii) ∠XLM = ∠PML, find ∠ALY
∠XLM = ∠PML
Sum of interior angles is 180 degrees
= ∠XLM + ∠PML = 180°
= ∠XLM + ∠XLM = 180°
= 2∠XLM = 180°
= ∠XLM = 180°/2
= ∠XLM = 90°
∠XLM and ∠ALY are vertically opposite angles
Therefore, ∠ALY = 90°
(iv) ∠ALY = (2x - 15) °, ∠LMQ = (x + 40) °, find x.
∠ALY and ∠LMQ are corresponding angles
= ∠ALY = ∠LMQ
= (2x - 15) °= (x + 40) °
= 2x – x = 40° + 15°
=x = 55°
Therefore, x = 55°
Question: 22
In Figure, DE ∥ BC. Find the values of x and y.
Solution:
We know that,
ABC, DAB are alternate interior angles
∠ABC = ∠DAB
So, x = 40°
And ACB, EAC are alternate interior angles
∠ACB = ∠EAC
So, y = 40°
Question: 23
In Figure, line AC ∥ line DE and ∠ABD = 32°, Find out the angles x and y if ∠E = 122°.
Solution:
∠BDE = ∠ABD = 32° – Alternate interior angles
= ∠BDE + y = 180°– linear pair
= 32° + y = 180°
= y = 180° – 32°
= y = 148°
∠ABE = ∠E = 32° – Alternate interior angles
= ∠ABD + ∠DBE = 122°
= 32° + x = 122°
= x = 122° – 32°
= x = 90°
Question: 24
In Figure, side BC of ΔABC has been produced to D and CE ∥ BA. If ∠ABC = 65°, ∠BAC = 55°, find ∠ACE, ∠ECD, ∠ACD.
Solution:
Corresponding angles,
∠ABC = ∠ECD = 55°
Alternate interior angles,
∠BAC = ∠ACE = 65°
Now, ∠ACD = ∠ACE + ∠ECD
= ∠ACD = 55° + 65°
= 120°
Question: 25
In Figure, line CA ⊥ AB ∥ line CR and line PR ∥ line BD. Find ∠x, ∠y, ∠z.
Solution:
Given that, CA ⊥ AB
= ∠CAB = 90°
= ∠AQP = 20°
By, angle of sum property
In ΔAPD
= ∠CAB + ∠AQP + ∠APQ = 180∘
= ∠APQ = 180° – 90° – 20°
= ∠APQ = 70°
y and ∠APQ are corresponding angles
= y = ∠APQ = 70°
∠APQ and ∠z are interior angles
= ∠APQ + ∠z = 180°
= ∠z = 180° – 70°
= ∠z = 110°
Question: 26
In Figure, PQ ∥ RS. Find the value of x.
Solution:
Given,
Linear pair,
∠RCD + ∠RCB = 180°
= ∠RCB = 180° – 130°
= 50°
In ΔABC,
∠BAC + ∠ABC + ∠BCA = 180°
By, angle sum property
= ∠BAC = 180° – 55° – 50°
= ∠BAC = 75°
Question: 27
In Figure, AB ∥ CD and AE ∥ CF, ∠FCG = 90° and ∠BAC = 120°. Find the value of x, y and z.
Solution:
Alternate interior angle
∠BAC = ∠ACG = 120°
= ∠ACF + ∠FCG = 120°
So, ∠ACF = 120° – 90°
= 30°
Linear pair,
∠DCA + ∠ACG = 180°
= ∠x = 180° – 120°
= 60°
∠BAC + ∠BAE + ∠EAC = 360°
∠CAE = 360° – 120° – (60° + 30°)
= 150°
Question: 28
In Figure, AB ∥ CD and AC ∥ BD. Find the values of x, y, z.
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Solution:
(i) Since, AC ∥ BD and CD ∥ AB, ABCD is a parallelogram
Adjacent angles of parallelogram,
∠CAD + ∠ACD = 180°
= ∠ACD = 180° – 65°
= 115°
Opposite angles of parallelogram,
= ∠CAD = ∠CDB = 65°
= ∠ACD = ∠DBA = 115°
(ii) Here,
AC ∥ BD and CD ∥ AB
Alternate interior angles,
∠DCA = x = 40°
∠DAB = y = 35°
Question: 29
In Figure, state which lines are parallel and why?
Solution:
Let, F be the point of intersection of the line CD and the line passing through point E.
Here, ∠ACD and ∠CDE are alternate and equal angles.
So, ∠ACD = ∠CDE = 100°
Therefore, AC ∥ EF
Question: 30
In Figure, the corresponding arms of ∠ABC and ∠DEF are parallel. If ∠ABC = 75°, find ∠DEF.
Solution:
Let, G be the point of intersection of the lines BC and DE
Since, AB ∥ DE and BC ∥ EF
The corresponding angles,
= ∠ABC = ∠DGC = ∠DEF = 100°