Explore topic-wise InterviewSolutions in Current Affairs.

The correct option is (A) 150°.

An angle is 60° then its complementary angle will be 30°.

∠e + 125° = 180°, (Linear pair angle)

⇒ ∠e = 180° – 125° = 55°,

∠e = ∠f = 55°, (Vertically opposite angles)

In figure, p ॥ q and t is a transversal

∠a = ∠f, (Alternate angle)

∠a = 55°, (∵ ∠f = 55°)

∠d = 125°, (Corresponding angels)

∠c = ∠a = 55°, (Vertically opposite angles)

and ∠b = ∠d = 125°, (Vertically opposite angles)

Thus, ∠a = 55°, ∠b = 125°, ∠c = 55°, ∠d = 125°, ∠e = 55° and ∠f= 55°

Wind mill, Comer of walls.

Sum of internal angles made by a transversal on its one side will be 180°.

Sum of two complementary angles will be 90°.

Supplementary angle of 87° is 93°.

(i) ∵∠x+ 125°= 180°

∠x = 180°- 125° = 55°

Two lines intersect each other

∴ ∠y = 125° (vertically opposite angles)

(ii) ∠x + 20° = 180°

[ ∠x = 180° – 20°= 160°]

(iii) Two lines intersect each other

∠x = 55° (vertically opposite angles)

∵ ∠x + ∠z = 180° (x, z is a linear pair whose sum is 180°)

55° + ∠z = 180°

⇒ ∠z = 180°- 55°= 125°

Clearly ∠y + ∠x = 90°

∠y + 55° = 90°

∠y = 90° – 55° = 35°

(i) True
(ii) False
(iii) True
(iv) True

? = 70° 

[ ∵ from the figure, these two angles are exterior angles on the same side of the transversal] 

EF || GH

115° + (x + 32°)= 180°

(The sum of the interior angles on the same side of a transversal is supplementary)

⇒ x + 147° = 180°

⇒ x = 180° – 147° 

= 33°

11x + 2 = 75° 

11x = 75 – 2 = 73

∴ x = 73/11

(∴ pair of corresponding angles are equal).

A pair of angles in which one arm of both the angles is on the same side of the transversal and their other arms are directed in the same sense is called a pair of corresponding angles.

Therefore corresponding angles are

∠ALM = ∠CMQ = 60° [given]

Vertically opposite angles are

∠LMD = ∠CMQ = 60° [given]

Vertically opposite angles are

∠ALM = ∠PLB = 60°

Here, ∠CMQ + ∠QMD = 180° are the linear pair

On rearranging we get

= ∠QMD = 180° – 60°

= 120°

Corresponding angles are

∠QMD = ∠MLB = 120°

Vertically opposite angles

∠QMD = ∠CML = 120°

Vertically opposite angles

∠MLB = ∠ALP = 120°

We are given that AB || CD and CD || EF

∴ ∠CQR = ∠QRF = z …(i)

(Alternate angles)

and ∠APQ = ∠CQR = x …(ii)

(Corresponding angles)

From (i) and (ii), we get

x = z …(iii)

Also ∠CQP + ∠CQR= 180°

(Linear pair)

⇒ y + z = 180°

But y : z = 3 : 7

⇒ y = 3k and z = 7k

⇒ 3k + 7k = 180°

10k = 180°

k = 18°

∴ y = 3k = 3 × 18° = 54°

and z = 7k = 7 × 18° = 126°

Hence. x = z = 126° [using (iii)]

x = 110° (alternate exterior angles)

Let the angle measured be x

Given that,

The angles measured will be differ by 10°

x° – (90° – x) = 10°

2x = 100°

x = 50°

∴ n║p and m is transversal 

∴ ∠1 = ∠3 = 85⁰ [Corresponding angles] 

 Also m║I & p is transversal 

∴ ∠2 + ∠3 = 180⁰ [∵Consecutive interior angles] 

⇒ ∠2 + 85⁰ = 180⁰ 

⇒ ∠2 + 180⁰ - 85⁰ 

⇒ ∠2 = 95⁰  

let angles is x⁰ then its complement is 90 - x⁰ . 

Now given x⁰ - (90 - x⁰ ) = 10 

⇒ x⁰ - 90⁰ + x⁰ = 10 

⇒ 2x⁰ = 10 + 90 = 100 

⇒ x⁰ = 100⁰/2 = 50⁰

∴ Required angle is 50⁰ . 

Given ∠AOE = 40⁰ & ∠BOD = 35⁰ 

Clearly ∠AOC = ∠BOD [Vertically opposite angles] 

⇒ ∠AOC = 35⁰ Ans. 

∠BOF = ∠AOE [Vertically opposite angles] 

⇒ ∠BOF = 40⁰ Ans. 

Now, ∠AOB = 180⁰ [Straight angles] 

⇒ ∠AOC + ∠COF + ∠BOF = 180⁰ [Angles sum property] 

⇒ 35⁰ + ∠COF + 40⁰ = 180⁰ 

⇒ ∠COF = 180⁰ - 75⁰ = 105⁰ 

Now, ∠DOE = ∠COF [Vertically opposite angles] 

∴ ∠DOE = 105⁰

∴ OP bisects ∠BOC 

∴ ∠POC = 1/2 ∠BOC ...(i) 

 Also OQ bisects ∠AOC 

∴ ∠COQ = 1/2 ∠AOC ...(ii) 

∴ OC stands on AB 

∴ ∠AOC + ∠BOC = 180⁰ [Linear pair] 

⇒ 1/2 ∠AOC + 1/2 ∠BOC = 1/2 × 180⁰ 

⇒ ∠COQ + ∠POC = 90⁰ [Using (i) & (ii)] 

⇒ ∠POQ = 90⁰ [By angle sum property] 

(c) 198

Because, we know that the numbers end with 2, 3, 7 and 8 are not perfect squares.

(d) 59²

Because, the square of a number having 1 or 9 at the units place ends in 1.

It is given that

Radius of the cone = 35cm

Height of the cone = 12cm

We know that

Volume of the cone = 1/3 πr²h

By substituting the values

Volume of the cone = 1/3 × (22/7) × 35² × 12

On further calculation

Volume of the cone = 15400 cm³

We know that

Slant height l = √(r² + h²)

By substituting the values

l = √ (35² + 12²)

On further calculation

l = √ 1369

So we get

l = 37 cm

We know that

Curved surface area of a cone = πrl

By substituting the values

Curved surface area of a cone = (22/7) × 35 × 37

So we get

Curved surface area of a cone = 4070 cm²

We know that

Total surface area of cone = πr (l + r)

By substituting the values

Total surface area of cone = (22/7) × 35 × (37 + 35)

On further calculation

Total surface area of cone = 22 × 5 × 72

So we get

Total surface area of cone = 7920 cm²

(i) 52 

Unit digit of (52)² = unit digit of (2)² = 4 

(ii) 977 

Unit digit of (977)² = unit digit of (7)² = 9 

(iii) 4583 

Unit digit of (4583)² = unit digit of (3)² = 9 

(iv) 78367 

Unit digit of (78367)² = unit digit of (7)² = 9 

(v) 52698 

Unit digit of (52698)² = unit digit of (8)² = 4 

(vi) 99880 

Unit digit of (99880)² = unit digit of (0)² = 0 

(vii) 12796 

Unit digit of (12796)² = unit digit of (6)² = 6 

(viii) 55555 

Unit digit of (55555)² = unit digit of (5)² = 5 

(ix) 53924 

Unit digit of (53924)² = unit digit of (4)² = 6

(b) 10

The natural numbers lie between 5² and 6² i.e. 25 and 36 are,

26, 27, 28, 29, 30, 31, 32, 33, 34 and 35

The square of the following numbers will have 1 at their units place as (1)² = 1, (9)² = 81 

81 and 139 i.e., (i) and (iii)

(b) 62²

Because, the square of a number having 2 or 8 at the units place ends in 4.

62 × 62 = 3844

(c) 1

Because, the square of a number having 1 or 9 at the units place ends in 1.

Example: – 11 × 11 = 121

We know that the ending digit (the digit at units place) of the square of a number is 0, 1, 4, 5, 6, or 9

So, the following numbers can be squares : 3051, 5684, and 50699 i.e., (i), (iii), and (v)

The square of the following numbers will not have 1 at their units place : as only (1)² = 1, (9)² = 81 have 1 at then units place 

322, 572, 2652 i.e., (i), (ii) and (v)

The squares of the following numbers, Will not have 6 at their units place as only (4)² = 16, (6)² = 36 has but its units place 35, 23 and 98 i.e., (i), (ii), and (v)

(i) Given as cosec^-1 (-√2)

Let y = cosec^-1 (-√2)

cosec y = -√2

– cosec y = √2

– cosec (π/4) = √2

– cosec (π/4) = cosec (-π/4) [since – cosec θ = cosec (-θ)]

So, the range of principal value of cosec^-1 [-π/2, π/2] – {0} and cosec (-π/4) = – √2

cosec (-π/4) = – √2

So, the principal value of cosec^-1 (-√2) is – π/4

(ii) Given as cosec^-1 (-2)

Let y = cosec^-1 (-2)

cosec y = -2

– cosec y = 2

– cosec (π/6) = 2

– cosec (π/6) = cosec (-π/6) [since – cosec θ = cosec (-θ)]

So, the range of principal value of cosec^-1 [-π/2, π/2] – {0} and cosec (-π/6) = – 2

cosec (-π/6) = – 2

So, the principal value of cosec^-1 (-2) is – π/6

(iii) Given as cosec^-1 (2/√3)

Let y = cosec^-1 (2/√3)

cosec y = (2/√3)

cosec (π/3) = (2/√3)

So, the range of principal value of cosec^-1 is [-π/2, π/2] – {0} and cosec (π/3) = (2/√3)

Hence, the principal value of cosec^-1 (2/√3) is π/3

(iv) Given as cosec^-1 (2 cos(2π/3))

As we know that cos (2π/3) = – ½

So, 2 cos (2π/3) = 2 × – ½

2 cos (2π/3) = -1

Substitute these values in cosec^-1 (2 cos(2π/3)) we get,

cosec^-1 (-1)

Let y = cosec^-1 (-1)

– cosec y = 1

– cosec (π/2) = cosec (-π/2) [since –cosec θ = cosec (-θ)]

So, the range of principal value of cosec^-1 [-π/2, π/2] – {0} and cosec (-π/2) = – 1

cosec (-π/2) = – 1

So, the principal value of cosec^-1 (2 cos(2π/3)) is – π/2

Given tan^-1 {tan (15π/4)}

= tan^-1 {tan (4π - π/4)}

We know that tan (2π – θ) = -tan θ

= tan^-1 (-tan π/4)

= tan^-1 (-1)

= -π/4

∴ tan^-1 {tan (15π/4)} = -π/4

Given

LHS = sin^-1 (2x - √ (1 – x²))

Let x = sin θ

= sin^-1 (2sin θ √ (1 – sin² θ))

We know that 1 – sin² θ = cos² θ

= sin^-1 (2 sin θ cos θ)

= sin^-1 (sin² θ)

= 2θ

= 2 sin^-1 x

= RHS 

∴ sin^-1 (2x - √ (1 – x²)) = 2 sin^-1 x

(a) [0, 1]

Since, cos-1 x is defined for x ∈ [-1, 1]

So, f(x) = cos-1 (2x – 1) is defined if

-1 ≤ 2x – 1 ≤ 1

0 ≤ 2x ≤ 2

Hence,

0 ≤ x ≤ 1

Given sin^-1 (sin 3π/5)

= sin^-1 [sin (π – 2π/5)]

= sin^-1 (sin 2π/5)

= 2π/5

∴ sin^-1 (sin 3π/5) = 2π/5

Given sin^-1 (sin 3π/5)

We know that sin^-1 (sin θ) = π – θ, if θ ∈ [π/2, 3π/2]

= π – 3π/5

= 2π/5

∴ sin^-1 (sin 3π/5) = 2π/5

Domain of sec^-1x is (-∞, -1] ⋃ [1, ∞)

Domain of tan^-1x is R

Union of (1) and (2) will be domain of given function

(–∞,–1]⋃[1,∞) ⋃ R

⇒ (–∞,–1]⋃[1,∞)

\(\therefore\) The domain of given function is (–∞,–1]⋃[1,∞).

Now the domain of cot x is R

While the domain of cot^–1x is [0,π ]

∴ The union of these two will give the domain of f(x)

⇒ R ⋃ [0,π]

= [0,π]

∴ The domain of f(x) is [0,π]

Range of sin^-1x is \([-\frac{\pi}2,\frac{\pi}2].\)

Given that

(sin^-1x)² + (sin^-1y)² + (sin^-1z)² = \(\frac 34\)π²

Each of sin^-1x, sin^-1y and sin^-1z takes the value of \(\frac{\pi}2.\)

x = 1, y = 1, and z = 1.

Hence,

= x² + y² + z²

= 1 + 1 + 1

= 3.

Domain of cos^-1x lies in the interval [-1, 1].

Therefore, the domain of cos^–1(x² – 4) lies in the interval [–1, 1].

-1 ≤ x² - 4 ≤ 1

3 ≤ x² ≤ 5

±√3 ≤ x ≤ ± √5

-√5 ≤ x ≤ - √3 and √3 ≤ x ≤ √5

Domain of cos^-1(x² - 4) is [-√5, -√3] ⋃ [√3, √5].

Domain of cos^-1x lies in the interval [–1, 1].

Domain of cos x lies in the interval [0, π] = [0, 3.14]

\(\therefore\) Domain of cos^-1x + cos x lies in the interval [–1, 1].

Domain of cos^-1x lies in the interval [-1, 1].

Therefore, the domain of cos^-1(2x) lies in the interval [-1, 1].

-1 ≤ 2x ≤ 1

\(\frac{-1}2\) ≤ x ≤ \(\frac 12\)

Domain of cos^-1(2x) is \([\frac{-1}2,\frac 12].\)

Domain of sin^-1(2x) + sin^-1x lies in the interval \([\frac{-1}2,\frac 12].\)

Domain of sin^-1 lies in the interval [-1, 1].

-1≤ x ≤ 1.

The domain of sin x lies in the interval\([-\frac{\pi}2,\frac{\pi}2]\)

\(-\frac{\pi}2\leq x\leq\frac{\pi}2\)

-1.57 ≤ x ≤ 1.57

From the above we can see that the domain of sin^–1x + sin x is the intersection of the domains of sin^–1x and sin x.

So domain of sin^–1x + sin x is [–1, 1].

From the given figure, we can write as

∠BAD = ∠ADC = 52° are alternate angles

We know that the sum of all the angles of a triangle is 180°.

Therefore, consider △DEC, we have

x + 40° + 52° = 180°

x = 88°

In the given figure, we have a quadrilateral and also we know that sum of all angles is quadrilateral is 360°.

Thus,

35° + 45° + 50° + reflex ∠ADC = 360°

On rearranging we get,

Reflex ∠ADC = 230°

230° + x = 360° (A complete angle)

x = 130°