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CBSE Class 10  >  Class 10 Test  >  The Complete SAT Course  >  Test: Lines & Angles - 3 - Class 10 MCQ

Lines & Angles - 3 - Free MCQ Practice Test with solutions, Class 10


MCQ Practice Test & Solutions: Test: Lines & Angles - 3 (10 Questions)

You can prepare effectively for Class 10 The Complete SAT Course with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Lines & Angles - 3". These 10 questions have been designed by the experts with the latest curriculum of Class 10 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 20 minutes
  • - Number of Questions: 10

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Test: Lines & Angles - 3 - Question 1
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An angle which is greater than 180° but less than 360° is called :

Detailed Solution: Question 1

An angle which is greater than 180° but less than 360° is called a reflex angle.

Test: Lines & Angles - 3 - Question 2
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The complement of 72° 40’ is :

Detailed Solution: Question 2

Complement of 72° 40’ is 90° ‒ (72° 40’)
= (89° 60’) ‒ (72° 40’) {since 1° = 60’}
= 17° 20’

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Test: Lines & Angles - 3 - Question 3
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The supplement of 154° 30’ is :

Detailed Solution: Question 3

Supplement of 154° 30’ is 180° ‒ (154° 30’)

= (179° 30’) ‒ (154° 30’) {since 1° = 60’}

= 25° 30’.

Test: Lines & Angles - 3 - Question 4
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Two straight lines AB and CD cut each other at O. If ∠BOD = 63°, then ∠BOC is:

Detailed Solution: Question 4

As given ∠BOD = 63° 
Since COD is a straight line, we have:

∠BOC + ∠BOD = 180°. So, ∠BOC = (180° ‒ 63°) = 117°.

Test: Lines & Angles - 3 - Question 5
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The straight lines AD and BC intersect one another at the point O. If ∠AOB + ∠BOD + ∠DOC = 274°, then ∠AOC is :

Detailed Solution: Question 5

As we know that the sum of all the angles around a point is 360°.

(∠AOB + ∠BOD + ∠DOC) + ∠AOC = 360°
∴ 274° + ∠AOC = 360° or ∠AOC = 86°.
Hence, option A is correct.

Test: Lines & Angles - 3 - Question 6
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In the given figure, AOB is a straight line. If ∠AOC + ∠BOD = 85°, then ∠COD is:

Detailed Solution: Question 6

Clearly,
∠AOC + ∠COD + ∠BOD = 180°
∴ 85° + ∠COD = 180°. So, ∠COD = (180° ‒ 85°) = 95°.
Hence, option C is correct.

Test: Lines & Angles - 3 - Question 7
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In the given figure, if AOB is a straight line, then the value of x is:

Detailed Solution: Question 7

As, (x + 30°) + 45° + (x + 15°) = 180°
⇒ x = 45°.

Test: Lines & Angles - 3 - Question 8
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In the given figure, the value of x, that would make POQ a straight line, is :

Detailed Solution: Question 8

POQ will be a straight line,
If 80° + 66° + x = 180°, i.e. x = 34°.

Test: Lines & Angles - 3 - Question 9
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If two angles are complementary of each other, then each angle is :

Detailed Solution: Question 9

If two angles are complementary, then clearly each angle is less than 90° and is therefore an acute angle.

Test: Lines & Angles - 3 - Question 10
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In the given figure, if AB || CD, then ∠FXE is equal to:

Detailed Solution: Question 10

As per the given figure,
∠BFE = ∠CEF = 110° (alt. ∠s).
So, ∠XFE = ∠BFE ‒ ∠BFX = (110° ‒ 50°) = 60°.
And on straight line CD,
110° + ∠FEX + 30° = 180° ⇒ ∠FEX = 40°.
Now, ∠XFE + ∠FEX + ∠FXE = 180° ⇒ 60° + 40° + ∠FXE = 180°.
∴ ∠FXE = 80°.
Hence, option D is correct.

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About this Class 10 Practice Test

This test contains 10 MCQs on Lines & Angles - 3 with detailed solutions for each question. It is part of the The Complete SAT Course course on EduRev, designed to align with the current Class 10 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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