Test: Lines & Angles - 3 - SAT MCQ

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

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

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

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

= 25° 30’.

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

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

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.

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

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

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

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

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.