Lines and Angles - 2 - Class 10 The Complete SAT Course Free MCQ Test

Test: Lines & Angles - 2 - Question 1

In the given figure AB || CD, ∠A = 128°, ∠E = 144°. Then, ∠FCD is equal to :

A.
72°
B.
64°
C.
136°
D.
92°

Detailed Solution: Question 1

As per the given figure,
Through E draw EE’ || AB || CD.
Then, ∠AEE’ = 180° ‒ ∠BAE = (180° ‒ 128°) = 52°.
(Interior angles on the same side of the transversal are supplementary.)
Now, ∠E’EC = (144° ‒ 52°) = 92°.
∠FCD = ∠E’EC = 92° (Corr. ∠s).
Hence, option D is correct.

Test: Lines & Angles - 2 - Question 2

In the trapezium PQRS, QR || PS, ∠Q = 90°, PQ = QR and ∠PRS = 20°. If ∠TSR = θ, then the value of θ is:

A.
75°
B.
55°
C.
65°
D.
45°

Detailed Solution: Question 2

Since QR || PS and PQ is transversal, the consecutive angles must be supplementary.
- ∠PQR + ∠QPS = 180°.
- But ∠PQR = 90° (given).
- So, ∠QPS = 90°.
Now, consider triangle PQR:
- As before, ∠QPR = ∠QRP = 45°.
Now, look at triangle PRS:
- We know ∠PRS = 20°.
- We need to find ∠PSR = θ.
- But we need more information.

First, find ∠SRQ:

Points Q, R, S are not colinear.
The angle between QR and RS is ∠QRS.
From triangle PQR, ∠QRP = 45°.
The angle between PR and RS is ∠PRS = 20°.
Therefore, ∠QRS = ∠QRP + ∠PRS = 45° + 20° = 65°.

Now, since QR || PS and RS is a transversal:

∠QRS and ∠PSR are alternate angles.
Therefore, ∠PSR = ∠QRS = 65°.

But if θ is ∠PSR, then θ = 65°.

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Test: Lines & Angles - 2 - Question 3

In the adjoining figure, ∠ABC = 100°, ∠EDC = 120° and AB || DE. Then, ∠BCD is equal to:

A.
80°
B.
60°
C.
40°
D.
20°

Detailed Solution: Question 3

In the given figure,
Produce AB to meet CD at F.
∠BFD = ∠EDF = 120° (alt. ∠s)
∠BFC = (180° ‒ 120°) = 60°.
∠CBF = (180° ‒ 100°) = 80°.
∴ ∠BCF = 180° ‒ (60° + 80°) = 40°.
Hence, option C is correct.

Test: Lines & Angles - 2 - Question 4

In the given figure, AB || CD, ∠ABO = 40° and ∠CDO = 30°. If ∠DOB = x°, then the value of x is:

A.
35°
B.
110°
C.
70°
D.
140°

Detailed Solution: Question 4

In the given figure,

Through O draw EOF parallel to AB & so to CD.
∴ ∠BOF = ∠ABO = 40° (alt. ∠s)
Similarly, ∠FOD = ∠CDO = 30° (alt. ∠s)
∴ ∠BOD = (40° + 30°) = 70°.
So, x = 70°.

Hence, option C is correct.

Test: Lines & Angles - 2 - Question 5

In the given figure, AB || CD, m∠ABF = 45° and m∠CFC = 110°. Then, m∠FDC is:

A.
25°
B.
45°
C.
35°
D.
30°

Detailed Solution: Question 5

As in the given figure,
∠FCD = ∠FBA = 45° (alt. ∠s)
∴ ∠FDC = 180° ‒ (110° + 45°) = 25°.
Hence, option A is correct.

Test: Lines & Angles - 2 - Question 6

In the given figure, line CE is drawn parallel to DB. If ∠BAD = 110°, ∠ABD = 30°, ∠ADC = 75° and ∠BCD = 60°, then the value of x° is:

A.
45°
B.
75°
C.
85°
D.
120°

Detailed Solution: Question 6

As in the given figure,
∠ADB = 180° ‒ (110° + 30°) = 40°.
So, ∠BDC = (75° ‒ 40°) = 35°.

∴ ∠DBC = 180° ‒ (60° + 35°) = 85°.

∴ ∠BCE = ∠DBC = 85° (alt. ∠s).

So, x = 85°.
Hence, option C is correct.

Test: Lines & Angles - 2 - Question 7

If two supplementary angles differ by 44°, then one of the angle is:

A.
72°
B.
102°
C.
65°
D.
68°

Detailed Solution: Question 7

Let the two angles are x and y. Therefore, as per the given information,
x ‒ y = 44° and
x + y = 180°
[ As the total of supplementary angles is 180°]
On solving these two linear equations we get,
2x = 224,
x = 112°.
Therefore the other angle y = 180° ‒ 112° = 68°
Hence, option D is correct.

Test: Lines & Angles - 2 - Question 8

Consider the following statements
If two straight lines intersect, then
I. vertically opposite angles are equal.
II. vertically opposite angles are supplementary.
III. adjacent angles are complementary.
Which of the statements given above is/are correct?

A.
Only III
B.
Only I
C.
II and III
D.
II and II

Detailed Solution: Question 8

Here, AB and CD are two lines.

If two straight lines intersect, then opposite vertically angles are equal.
Hence, option B is correct.

Test: Lines & Angles - 2 - Question 9

In the figure given above LOM is a straight line. What is the value of x°?

A.
45°
B.
60°
C.
70°
D.
80°

Detailed Solution: Question 9

From the given figure,
∠LOQ + ∠QOP + ∠POM = 180°
(straight line)
∴ (x° + 20°) + 50° + (x° – 10°) = 180°
⇒ 2x° + 60° = 180° ⇒ 2x° = 120°
∴ x° = 60°
Hence, optuion B is correct.

Test: Lines & Angles - 2 - Question 10

In the figure given above, EC is parallel to AB, ∠ECD = 70° and ∠BDO = 20°. What is the value of ∠OBD?

A.
20°
B.
30°
C.
40°
D.
50°

Detailed Solution: Question 10

Given that, EC || AB
∴ ∠ECO + ∠AOC = 180°

⇒ ∠AOC = 180° – 70° = 110°

∴ ∠BOD = ∠AOC = 110°

(alternate angle)

Now, in ΔOBD

∠BOD + ∠ODB + ∠DBO = 180°

∴ 110° + 20° + x° = 180° ⇒ x° = 50°.

Hence, option D is correct.

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Lines & Angles - 2 - Free MCQ Practice Test with solutions, Class 10

MCQ Practice Test & Solutions: Test: Lines & Angles - 2 (10 Questions)

In the given figure AB || CD, ∠A = 128°, ∠E = 144°. Then, ∠FCD is equal to :

In the trapezium PQRS, QR || PS, ∠Q = 90°, PQ = QR and ∠PRS = 20°. If ∠TSR = θ, then the value of θ is:

In the adjoining figure, ∠ABC = 100°, ∠EDC = 120° and AB || DE. Then, ∠BCD is equal to:

In the given figure, AB || CD, ∠ABO = 40° and ∠CDO = 30°. If ∠DOB = x°, then the value of x is:

In the given figure, AB || CD, m∠ABF = 45° and m∠CFC = 110°. Then, m∠FDC is:

In the given figure, line CE is drawn parallel to DB. If ∠BAD = 110°, ∠ABD = 30°, ∠ADC = 75° and ∠BCD = 60°, then the value of x° is:

If two supplementary angles differ by 44°, then one of the angle is:

Consider the following statements
If two straight lines intersect, then
I. vertically opposite angles are equal.
II. vertically opposite angles are supplementary.
III. adjacent angles are complementary.
Which of the statements given above is/are correct?

In the figure given above LOM is a straight line. What is the value of x°?

In the figure given above, EC is parallel to AB, ∠ECD = 70° and ∠BDO = 20°. What is the value of ∠OBD?

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Lines & Angles - 2 - Free MCQ Practice Test with solutions, Class 10

MCQ Practice Test & Solutions: Test: Lines & Angles - 2 (10 Questions)

In the given figure AB || CD, ∠A = 128°, ∠E = 144°. Then, ∠FCD is equal to :

In the trapezium PQRS, QR || PS, ∠Q = 90°, PQ = QR and ∠PRS = 20°. If ∠TSR = θ, then the value of θ is:

In the adjoining figure, ∠ABC = 100°, ∠EDC = 120° and AB || DE. Then, ∠BCD is equal to:

In the given figure, AB || CD, ∠ABO = 40° and ∠CDO = 30°. If ∠DOB = x°, then the value of x is:

In the given figure, AB || CD, m∠ABF = 45° and m∠CFC = 110°. Then, m∠FDC is:

In the given figure, line CE is drawn parallel to DB. If ∠BAD = 110°, ∠ABD = 30°, ∠ADC = 75° and ∠BCD = 60°, then the value of x° is:

If two supplementary angles differ by 44°, then one of the angle is:

Consider the following statementsIf two straight lines intersect, then I. vertically opposite angles are equal. II. vertically opposite angles are supplementary. III. adjacent angles are complementary.Which of the statements given above is/are correct?

In the figure given above LOM is a straight line. What is the value of x°?

In the figure given above, EC is parallel to AB, ∠ECD = 70° and ∠BDO = 20°. What is the value of ∠OBD?

The Complete SAT Course

The Complete SAT Course

Top Courses for Class 10

About this Class 10 Practice Test

Explore more Class 10 Study Resources

Consider the following statements
If two straight lines intersect, then
I. vertically opposite angles are equal.
II. vertically opposite angles are supplementary.
III. adjacent angles are complementary.
Which of the statements given above is/are correct?