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Check the options to see which satisfies both conditions: sum of digits is 8 and adding 18 reverses the number.

For option 35: 3 + 5 = 8 (Condition 1 satisfied).

Adding 18 to the number: 35 + 18 = 53.

The digits 3 and 5 are reversed to 5 and 3 (Condition 2 satisfied).

∴ The correct answer is 35.

Alternate Method

Given:

Sum of digits of a two-digit number = 8

Number + 18 = Reversed number

Formula Used:

A two-digit number with tens digit x and units digit y is represented as: 10x + y.

Calculations:

Let the tens digit be x and the units digit be y.

⇒ x + y = 8 → (Equation 1)

The original number is 10x + y and the reversed number is 10y + x.

⇒ (10x + y) + 18 = 10y + x

⇒ 10x − x + y − 10y = −18

⇒ 9x − 9y = −18

⇒ x − y = −2

⇒ y − x = 2 → (Equation 2)

Adding (Equation 1) and (Equation 2):

⇒ (x + y) + (y − x) = 8 + 2

⇒ 2y = 10

⇒ y = 5

Substituting y = 5 in Equation 1:

⇒ x + 5 = 8

⇒ x = 3

⇒ The number is 10(3) + 5 = 35.

∴ The correct answer is 35.

Additional Information

Difference between Number and its Reverse

The difference between a two-digit number (10x + y) and its reverse (10y + x) is always equal to 9 × |x − y|. In this case, 18 ÷ 9 = 2, which is the difference between the digits.

Sum of a Number and its Reverse

The sum of a two-digit number and its reverse is always a multiple of 11, specifically 11 × (x + y).

Algebraic Representation

Always represent a two-digit number as 10T + U, where T is the tens digit and U is the units digit, to solve digit-reversal problems effectively.

Concept used:

sin 30° = \(\frac{1}{{2}}\)

Explanation:

According to the question 

Let us suppose sin 30° in the place of sin 31° + sin 32° 

⇒ (\(\frac{1}{{2}}\) + \(\frac{1}{{2}}\)) = \(\frac{2}{{2}}\) = 1

Here, we can clearly see that if we put the value of sin 30° we get its value be 1.

So, if we find the value of sin 31° + sin 32°, we have to put the sin value greater than 1.

∴ The required answer is Its value is greater than 1.

Given:

1 + tan θ = √2

Concept used:

tanθ = 1/cotθ 

Calculation: 

1 + tan θ = √2

⇒ tanθ = (√2 - 1)

⇒ cotθ = 1/(√2 - 1)

Multiplying the numerator and denominator with(√2 + 1), we get

⇒ (\(\frac{1}{{√2 - 1}}\) × \(\frac{√2+1}{{√2 + 1}}\))

⇒ \(\frac{√2+1}{{2 - 1}}\) = √2 + 1

⇒ cotθ = √2 + 1

⇒ cotθ  - 1 = √2

∴ The required value is √2.

Given:

tanθ = \(\frac{3}{4}\)

Formula used:

tanθ = Perpendicular/Base

sinθ = Perpendicular/Hypotenuse

By Pythagoras' theorem,

H² = P² + B²

Calculation:

tan θ = Perpendicular/Base = \(\frac{3}{4}\)

So, by Pythagoras' theorem,

H2 = P2 + B²

⇒ H² = 3² + 4²

⇒ H² = (9 + 16)

⇒ H² = 25

⇒ H = 5 cm

⇒ sin θ = Perpendicular/Hypotenuse = \(\frac{3}{5}\)

∴ The required value is \(\frac{3}{5}\).

Given:

sin 25° sin 35° sec 65° sec 55° 

Concept used:

sec θ = 1/cos θ 

cos(90 - θ) = sinθ 

Calculation:

According to the question

sin 25° sin 35° sec 65° sec 55° 

⇒ sin 25°. sin 35°. \(\frac{1}{{cos65° }}\). \(\frac{1}{{cos55° }}\)

⇒ sin 25°. sin 35°. \(\frac{1}{{cos(90^{\circ}-25° )}}\). \(\frac{1}{{cos(90^{\circ}-35°) }}\)

⇒ sin 25°. sin 35°. \(\frac{1}{{sin25° }}\). \(\frac{1}{{sin35° }}\)

⇒ 1 

∴ The required answer is 1.

Concept used:

Following condition is must in any triangle

(sum of two sides) > Third side

Calculation:

According to the given options

Option 1:

(3, 4, 5)

(sum of two sides) > Third side

⇒ (3 + 4) > 5

⇒ 7 > 5

It represent the sides of a triangle.

Option 2:

(4, 7, 10)

(sum of two sides) > Third side

⇒ (4 + 7) > 10

⇒ 11 > 10

It represent the sides of a triangle.

Option 3:

(3, 6, 8)

(sum of two sides) > Third side

⇒ (3 + 6) > 8

⇒ 9 > 8

It represent the sides of a triangle.

Option 4:

(2, 3, 6)

(sum of two sides) > Third side

⇒ (2 + 3) > 6

⇒ 5 > 6

It does not represent the sides of a triangle.

∴ The required option 4 is correct.

Given:

A round balloon of unit radius subtends an angle of 90° at the eye of an observer standing at a point, say A.

Concept used:

cos θ = Base / hypotenuse

cos 45° = \(\frac{1}{{√ 2}}\)

Calculation:

According to the above figure,

In ΔOAB,

sin 45° = \(\frac{OB}{{OA}}\)

⇒ \(\frac{1}{{√ 2}}\) = \(\frac{1}{{OA}}\)

⇒ OA = √2 cm

∴ The distance of the center of the balloon from point A is√2.

Given:

The shadow of a pole is √3 times the length of the pole.

Concept used:

tanθ = Perpendicular/base

tan 30° = \(\frac{1}{{\sqrt 3}}\)

Calculation:

The height of the tower be h.

So, the height of the shadow be √3h

Now, In ΔABC,

⇒ tan ∠ACB = \(\frac{h}{{\sqrt 3h}}\)

⇒ tan ∠ACB = \(\frac{1}{{\sqrt 3}}\)

⇒ tan ∠ACB = tan 30° 

⇒ ∠ACB = 30° 

∴ The angle of elevation is 30°

Concept used:

Right angle = 90° 

We know that the sum of all the sides of the triangle is 180°

tan 60° = √3

Calculation:

According to the above concept,

The sum of all the sides of the triangle is 180°

α  + β + 90° = 180 

⇒ α = 180° - (90° + β)

⇒ α = 180° - (90° + 30°)

⇒ α = (180° - 120°)

⇒ α = 60° 

So, the value of tan α = tan 60° = √3

∴ The required value of tan α is √3.

Given:

The top of a building 90 m high.

The angle of depression from the top and the bottom of a tree are 30° and 45° respectively. 

Concept used:

tan 45° = 1

tan 30° = \(\frac{1}{{√ 3}}\)

Calculation:

Let the height of the tree = h m.

According to the above figure,

In ΔADB,

tan 45° = \(\frac{AB}{{DB}}\)

⇒ 1 = \(\frac{90}{{DB}}\)

⇒ DB = 90 m

So, CE = DB = 90 m

In ΔCAE,

tan 30° = \(\frac{AE}{{CE}}\)

⇒ \(\frac{1}{{√ 3}}\) = \(\frac{AE}{{90}}\)

⇒ AE = \(\frac{90}{{√ 3}}\) = 30√3 m.

So, h = 90 - AE = (90 - 30√3) m

∴ The required height of the tree is 90 - 30√3 m.