MCQ: Complementary angles - 1 - SSC CGL MCQ

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.

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

POQ will be a straight line,
If 80° + 66° + x = 180°, i.e. x = 34°.
Hence, option C is correct.

As, (x + 30°) + 45° + (x + 15°) = 180°
⇒ x = 45°.
Hence, option B is correct.

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

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.

As given ∠BOD = 63°
Since COD is a straight line, we have:
∠BOC + ∠BOD = 180°. So, ∠BOC = (180° ‒ 63°) = 117°.
Hence, option B is correct.

Supplement of 154° 30’ is 180° ‒ (154° 30’)
= (179° 30’) ‒ (154° 30’) {since 1° = 60’}
= 25° 30’.
Hence, option A is correct.

Complement of 72° 40’ is 90° ‒ (72° 40’)
= (89° 60’) ‒ (72° 40’) {since 1° = 60’}
= 17° 20’
Hence, option C is correct.

An angle which is greater than 180° but less than 360° is called a reflex angle.
Hence, option D is correct.

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. 

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

Here, AB and CD are two lines.

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

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 C is correct.

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.