Trigonometry MCQ Quiz - Objective Question with Answer for Trigonometry - Download Free PDF

Given:

sin A = 3/4

Formula used:

cos²A = 1 - sin²A

cos A = √(1 - sin²A)

Calculation:

cos²A = 1 - sin²A

⇒ cos²A = 1 - (3/4)²

⇒ cos²A = 1 - 9/16

⇒ cos²A = (16/16) - (9/16)

⇒ cos²A = 7/16

⇒ cos A = √(7/16)

⇒ cos A = √7 / 4

(2 × cos A) / √7:

⇒ (2 × (√7 / 4)) / √7

⇒ (2√7 / 4) / √7

⇒ (2√7) / (4√7)

⇒ 2 / 4

⇒ 1 / 2

∴ The correct answer is option (3).

Given:

Height of first pole = 12 m

Height of second pole = 20 m

Distance between the bottom of the poles = 15 m

Formula Used:

Distance between the tops of the poles = √((Horizontal distance)² + (Difference in heights)²)

Calculation:

Horizontal distance between poles = 15 m

Difference in heights = 20 - 12 = 8 m

Distance between the tops = √((15)² + (8)²)

⇒ Distance between the tops = √(225 + 64)

⇒ Distance between the tops = √289

⇒ Distance between the tops = 17 m

The distance between the tops of the poles is 17 m.

Given:

Expression: 3 + tan²θ + cot²θ − sec²θ × cosec²θ

Formula used:

tan²θ = sec²θ − 1

cot²θ = cosec²θ − 1

Calculation:

⇒ 3 + (sec²θ − 1) + (cosec²θ − 1) − sec²θ × cosec²θ

⇒ 3 + sec²θ + cosec²θ − 2 − sec²θ × cosec²θ

⇒ (1 + sec²θ + cosec²θ − sec²θ × cosec²θ)

Now let’s simplify by assuming θ = 45° (a valid identity angle where all values are defined)

⇒ sec 45° = √2, so sec²θ = 2

⇒ cosec 45° = √2, so cosec²θ = 2

⇒ Expression = 1 + 2 + 2 − (2 × 2)

⇒ 1 + 4 − 4 = 1

∴ The value of the expression is 1.

Given:

cot θ = 3/4

Formula used:

sin 3θ = 3sin θ - 4sin³ θ

Calculation:

cot θ = 3/4

⇒ tan θ = 4/3

⇒ sin θ = 4/5

⇒ sin³ θ = (4/5)³ = 64/125

⇒ sin 3θ = 3 × (4/5) - 4 × (64/125)

⇒ sin 3θ = 12/5 - 256/125

⇒ sin 3θ = 300/125 - 256/125

⇒ sin 3θ = 44/125

∴ The correct answer is option (1).

Given:

Height of the tower (h) = 250 m

Distance of the point from the foot of the tower (b) = 250 m

Formula used:

In a right-angled triangle, tan =

Here, Opposite side = Height of the tower

Adjacent side = Distance from the foot of the tower

Calculations:

Let be the angle of elevation.

tan =

tan =

⇒ tan =

We know that tan(60°) =

⇒ = 60°

∴ The angle of elevation is 60°.

GIVEN:

BC = 18 m

CONCEPT:

FORMULAE USED:

Tanθ = Perpendicular/Base

Cosθ = Base/Hypotenuse

CALCULATION:

Height of the tree = AB + AC

Tan 30° = AB/18

⇒ (1/√3) = AB/18

⇒ AB = (18/√3)

Cos 30° = BC/AC = 18/AC

⇒ √3/2 = 18/AC

⇒ AC = 36/√3

Hence, AB + AC = 18/√3 + 36/√3 = 54 / √3

⇒ 54/√3 × √3 /√3  (rationalizing to remove root from denominator)

⇒ 54√3 / 3 = 18√3

∴ Height of the tree = 18√3.

Mistake Point: Here, total height of tree is (AB + AC).

The above Question is previous year Question taken directly from NCERT class 10th. Correct answer will be 18√3

We can find the angle of elevation by using the following steps:

Calculation:

Label the height difference between the two points on the ground as "h" and the horizontal distance between the two points as "d".

Use the tangent function to find the angle of elevation:

tan(θ) = .

Solve for the angle of elevation:

In this case, h = 20 m and d = 20√3 m, so:

θ = 30°

So the angle of elevation is 30°.

Given:

tan 53° = 4/3

Formula Used:

tan(x – y) = (tanx – tany)/(1 + tanxtany)

Calculation:

We know, 8° = 53° - 45°

Tan8° = tan(53° - 45°)

⇒ tan8° = (tan53° - tan45°)/(1 + tan53° tan45°)

⇒ tan8° = (4/3 – 1)/(1 + 4/3 × 1)

⇒ tan8° = (1/3)/(7/3)

⇒ tan8° = 1/7

Concept used: 

sec²(x) = 1 + tan²(x)

Calculation:

⇒ sec²θ + tan²θ = 5/3

⇒ 1 + tan²θ + tan²θ = 5/3

⇒ 2tan²θ = 2/3

⇒ tanθ = 1/√3

⇒ θ = 30

Given:

tanθ + cotθ = √3

Formula used:

(a + b)³ = a³ + b³ + 3ab(a + b)

a² + b² = (a + b)² - 2(a × b)

tanθ × cotθ = 1

Calculation:

tanθ + cotθ = √3

Taking cube on both sides, we get

(tanθ + cotθ)³ = (√3)³

⇒ tan³θ + cot³θ + 3 × tanθ × cotθ × (tanθ + cotθ) = 3√3

⇒ tan3θ + cot3θ + 3√3  = 3√3

⇒ tan3θ + cot3θ = 0  

Taking square on the both sides

(tan3θ + cot3θ)² = 0

⇒ tan⁶θ + cot⁶θ + 2 × tan3θ × cot3θ = 0

⇒ tan6θ + cot6θ + 2 = 0    

⇒ tan6θ + cot6θ = - 2

∴ The value of tan6θ + cot6θ is - 2.

As,

⇒ sec²θ = 1 + tan²θ

We have,

⇒ (sec²θ)² – sec²θ = 3

⇒ (1 + tan²θ)² – (1 + tan²θ) = 3

⇒ (1 + tan⁴θ + 2tan²θ) – (1 + tan²θ) = 3

⇒ 1 + tan⁴θ + 2tan²θ – 1 – tan²θ = 3

⇒ tan⁴θ + tan²θ = 3

Formula Used:

1/Cosec Ø = Sin Ø 

Sin²Ø + Cos²Ø = 1

Calculation:

Cos2Ø + 1/Cosec2Ø + 17 = x

⇒ Cos2Ø + Sin2Ø + 17 = x

⇒ 1 + 17 = x

⇒ x = 18

⇒ x² = 324

∴ The value of x² is 324.

Given:

A woman is standing 30 m away from her house. An angle of elevation from the top of her is 30° towards the top of house and angle of elevation from her foot is 60° towards the top of the house

Calculation:

In ΔABC,

⇒ tan30° = AB/BC

⇒ 1/√3 = AB/30 

⇒ AB = 30/√3

⇒ AB = 30√3/(√3 × √3) 

⇒ AB = 10√3 m

In ΔAED,

⇒ tan60° = AE/ED

⇒ √3 = (AB + BE)/30

⇒ AB + BE = 30√3

⇒ BE = 30√3 – 10√3

⇒ BE = 20√3 m

Total height of the house = 10√3 + 20√3 = 30√3

Height of the women = CD = BE = 20√3

Total height of the house and women = 30√3 + 20√3 = 50√3

∴ Total height of the house and women is 50√3 

Given:

secθ - cosθ = 14 and 14 secθ = x

Concept used:

Calculations:

According to the question,

⇒ 

⇒ 

⇒ 

⇒       ----()

∴ The value of x is .

Given:

The given expression = 

Formula used:

cos 67° = sin (90° - 67°) = sin 23°

sin 67° = cos (90° - 67°) = cos 23°

sin² θ + cos² θ = 1

sec² θ - tan² θ = 1

Calculation:

= 

= 

= sin2 23° + 1 + cos2 23° + 1

= 1 + 1 + 1

= 3

∴ The value of the given expression is 3