MCQ: Complementary angles - 2 - SSC CGL MCQ

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

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.

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.

In the given figure,
PQ = QR and ∠PQR = 90°
⇒ ∠QPR = ∠QRP = 45°.
∴ ∠QRS = (45° + 20°) = 65°.
∴ θ = ∠QRS = 65° (alt. ∠s)
Hence, option C is correct.

As per the given figure,
Through E draw EE’ || AB || C D.
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.

As, AP is a straight line and rays GO and GF stands on it.
∴ ∠AGO + ∠OGF + ∠PGF = 180°
⇒ (∠AGO + ∠PGE) + ∠OGF = 180°
⇒ 70° + ∠OGF = 180°
⇒ ∠OGF = 180° – 70°
⇒ ∠OGF = 110°
As, OQ is a straight line, rays GF and GP stands on it.
∠OGF + ∠PGF + ∠PGQ = 180°
Putting value of ∠OGF & ∠PGQ
110° + ∠PGF + 40° = 180°
∠PGF = 180° – 150° = 30°
Hence, option C is correct.

∠ALC = ∠LCD = 60° [∵ Alternate angles]
EC is the bisector of ∠LCD

∠CEF + ∠ECD = 180° [∵ Pair of interior angles]
∠CEF + 30° = 180°
∠CEF = 180° – 30° = 150°
Hence, option C is correct. 

Complementary angles: Complementary angles are angle pairs whose measures sum to one right angle (90°).
So, the required angle will be 10°
180° = π radian


Hence, option C is correct.

Given that,

⇒ 3(∠A + ∠B) + 2∠C = 480° ....(i)
Also, in ΔABC,
∠A + ∠B + ∠C = 180°
On multiplying both sides L.H.S. & R.H.S. by 3, we get
3(∠A + ∠B) + 3∠C = 540° ..... (ii)
On subtracting Eq. (i) from Eq. (ii), we get
∠C = 60°.
Hence, option C is correct.

If the arms of one angle are respectively parallel to the arms of another angle, then the two angles are not equal but supplementary.
Ex.

If l1 || l2 ⇒ ∠1 + ∠2 = 180° (Supplementary)
Hence, option B is correct.

In 1 min = 60 s distance travelled by the wheel
= 12 × Its circumference
= 12 × 2πr
∴ In 1 s distance travelled by the wheel

Which is the required angle.
Hence, option B is correct.

Statements I and II are both true, because the locus of points which are equidistant from two parallel lines is a line parallel to both of them and draw mid way between them.
Also, it is true that the perpendicular distances of any point on this locus line from two original parallel lines are equal. Further, no point outside this locus line has this property.

Hence, option C is correct.

∠PCT + ∠PCB = π (Linear pair)
∠PCB = π – (π – b°) = b° ..... (i) 

In ΔBPC,
∠PCB + ∠BPC + ∠PBC = π
∠PBC = π – ∠PCB – ∠BPC = π – b° – a° ..... (ii)
∵ ∠ABE + ∠EBC = π (∵ ∠PBC = ∠EBC) (linear pair)
∠ABE = π – ∠PBC = π – (π – b° – a°) = a° + b° ...(iii)
Now, in ΔABE
Sum of two interior angles = Exterior angle
∠EAB + ∠ABE = ∠BES ⇒ c° + b° + a° = x°
∴ x° = a° + b° + c°.
Hence, option C is correct.

Given that,
∠ABC = 65° and ∠CDE = 15°
Here, ∠ABC + ∠TCB = 180° (∵ AB || CD)
∠TCB = 180° – ∠ABC
∴ ∠TCB = 180° – 65° = 115°
∵ ∠TCB + ∠DCB = 180° (Linear pair)
∴ ∠DCB = 65°
Now, in ΔCDE
∠CED = 180° – (∠ECD + ∠EDC) (∵ ∠ECD = ∠BCD)
= 180° – (– 65° + 15° ) = 100°

∵ ∠DEC + ∠FEC = 180°
⇒ ∠FEC = 180° – 100° = 80°
Given that, AB = AE. i.e. ΔABE an isosceles triangle.
∴ ∠ABE = ∠AEB = 65°
∵ ∠AEB + ∠AEF + ∠FEC = 180° (straight line)
⇒ 65° + x° + 80° = 180°
∴ x° = 180° – 145° = 35°.
Hence, option B is correct.

An angle which is greater than 180° but less than 360° is called a reflex
Given that, BC || AE
∠CBA + ∠EAB = 180°
⇒ ∠EAB = 180° – 65° = 115°
∵ BC = AC
Hence, ΔABC is an isosceles triangle. 

⇒ ∠CBA = ∠CAB = 65°
Now, ∠EAB = ∠EAC + ∠CAB
⇒ 115° = x + 65° ⇒ x = 50°.
Hence, option D is correct.