MCQ: Complementary angles - 2 - SSC CGL MCQ

# MCQ: Complementary angles - 2 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Complementary angles - 2

MCQ: Complementary angles - 2 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Complementary angles - 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Complementary angles - 2 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Complementary angles - 2 below.
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MCQ: Complementary angles - 2 - Question 1

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

Detailed Solution for MCQ: Complementary angles - 2 - Question 1

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

MCQ: Complementary angles - 2 - Question 2

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

Detailed Solution for MCQ: Complementary angles - 2 - Question 2

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.

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MCQ: Complementary angles - 2 - Question 3

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

Detailed Solution for MCQ: Complementary angles - 2 - 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.

MCQ: Complementary angles - 2 - Question 4

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

Detailed Solution for MCQ: Complementary angles - 2 - Question 4

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.

MCQ: Complementary angles - 2 - Question 5

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

Detailed Solution for MCQ: Complementary angles - 2 - Question 5

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.

MCQ: Complementary angles - 2 - Question 6

In the given figure lines AP and OQ intersect at G If ∠AGO + ∠PGF = 70° and ∠PGQ = 40°. Find the angle value of ∠PGF.

Detailed Solution for MCQ: Complementary angles - 2 - Question 6

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.

MCQ: Complementary angles - 2 - Question 7

In the given figure AB || CD, ∠ALC = 60° and EC is the bisector of ∠LCD. If EF || AB then the value of ∠CEF is

Detailed Solution for MCQ: Complementary angles - 2 - Question 7

∠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.

MCQ: Complementary angles - 2 - Question 8

The complement angle of 80° is

Detailed Solution for MCQ: Complementary angles - 2 - Question 8

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

Hence, option C is correct.

MCQ: Complementary angles - 2 - Question 9

In a Δ ABC, then what is the value of ∠C?

Detailed Solution for MCQ: Complementary angles - 2 - Question 9

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.

MCQ: Complementary angles - 2 - Question 10

If the arms of one angle are respectively parallel to the arms of another angle, then the two angles are

Detailed Solution for MCQ: Complementary angles - 2 - Question 10

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.

MCQ: Complementary angles - 2 - Question 11

A wheel makes 12 revolutions per min. The angle in radian described by a spoke of the wheel in 1 s is:

Detailed Solution for MCQ: Complementary angles - 2 - Question 11

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.

MCQ: Complementary angles - 2 - Question 12

Consider the following statements
I. The locus of points which are equidistant from two parallel lines is a line parallel to both of them and drawn mid-way between them.
II. The perpendicular distance of any point on this locus line from two original parallel lines are equal. Further, no point outside this locus line has this property.
Which of the above statements is/are correct?

Detailed Solution for MCQ: Complementary angles - 2 - Question 12

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.

MCQ: Complementary angles - 2 - Question 13

The angles x°, a°, c° and (π – b)° are indicated in the figure given below. Which one of the following is correct?

Detailed Solution for MCQ: Complementary angles - 2 - Question 13

∠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.

MCQ: Complementary angles - 2 - Question 14

In the figure given below, AB is parallel to CD. ∠ABC = 65°, ∠CDE = 15° and AB = AE. What is the value of ∠AEF?

Detailed Solution for MCQ: Complementary angles - 2 - Question 14

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.

MCQ: Complementary angles - 2 - Question 15

In the figure given below, ABC is a triangle. BC is parallel to AE. If BC = AC, then what is the value of ∠CAE?

Detailed Solution for MCQ: Complementary angles - 2 - Question 15

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.

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