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

Given:

Expression = sinθ/cosθ

Calculations:

The ratio of the sine of an angle to the cosine of the same angle is defined as the tangent of that angle.

⇒ \(\dfrac{\sin\theta}{\cos\theta}\) = tanθ

∴ The correct  answer is option 1.

Given:

In a right-angled triangle, perpendicular = base

Formula used:

sinθ = opposite / hypotenuse

cosθ = adjacent / hypotenuse

tanθ = opposite / adjacent

Calculations:

Since perpendicular = base, the triangle is isosceles right-angled.

Let each side = 1 unit.

Hypotenuse = √(1² + 1²) = √2

sinθ = 1 / √2 = 1 / 1.414 = 0.707 ≈ 1/√2

cosθ = 1 / √2 = 1/√2 (not 1/2)

tanθ = 1 / 1 = 1 (not √3)

sinθ ≠ 1/2

∴ The correct statement is: sinθ = 1/√2.

Given:

Height of the tree (opposite side) = 1 meter

Length of the shadow (adjacent side) = \(\frac{1}{\sqrt{3}}\) meters

Formula Used:

In a right-angled triangle, the tangent of the angle of elevation is given by:

tan(θ) = Opposite side / Adjacent side

Calculation:

Let θ be the angle of elevation of the sun.

tan(θ) = Height of tree / Length of shadow

tan(θ) = \(\frac{1}{1/\sqrt{3}}\)

tan(θ) = \(1 \times \sqrt{3}\)

tan(θ) = \(\sqrt{3}\)

We know that tan(60°) = \(\sqrt{3}\).

Therefore, θ = 60°

∴ The angle of elevation of the sun is 60°.

Given:

When seen from a light-house 50 m above sea-level, the angle of depression of a boat is 30°. 

Concept used:

tanθ = Height ÷ Base

\(\sqrt3 \approx 1.732\)

Calculation:

Here,

AB = Height of the light-house from the sea level

BC = Distance between the foot of the light-house and the boat

Angle of depression = ∠DAC = 30°

Since, AD is parallel to BC, ∠DAC = 30° = ∠ACB

ΔABC is a right-angled triangle at B.

So,

tan 30° = AB/BC

⇒ \(\frac {1}{\sqrt3}\) = \(\frac {50}{BC}\)

⇒ BC = \(50\sqrt3\)

⇒ BC = 50 × 1.732

⇒ BC ≈ 86.6

∴ The boat is 86.6 m and 50√3 m away from the lighthouse.

Given:

 A man of height 1.86 m is standing on top of a 13.14-m-high building. He observes a tower and the angles of elevation and depression of the top and bottom of the tower are 30º and 60º, respectively.

Concept used:

1. tan 60° = \({\sqrt3}\)

2. tan 30° = \(\frac{1}{\sqrt3}\)

3. Tangent of an acute angle of a right-angled triangle = \(\frac{Perpendicular}{Base}\)

Calculation:

Let's suppose, AB is the 1.86m tall man standing on a 13.6m long tower, BC. As the man observes another tower DG, the angles of elevation i.e. ∠EAD, and depression i.e. ∠EAG of the top and bottom of the tower, DG are 30º and 60º, respectively.

So, AB = 1.86, BC = 13.14, ∠EAD = 30º &  ∠EAG = 60º

Since ∠EAG & ∠AGC are alternate angles, ∠EAG = 60º = ∠AGC

AE is drawn parallel to CG.

Hence, AE = CG and AC = EG

Both ΔAED & ΔACG are right-angled triangles at E and C, respectively.

According to the concept,

In ΔACG,

tan ∠AGC = \(\frac{AC}{CG}\)

⇒ tan 60º = \(\frac {AB + BC}{CG}\)

⇒ \(\sqrt3 = \frac{1.86 + 13.14}{CG}\)

⇒ CG = \(\frac {15}{\sqrt3}\)

⇒ CG = \(5\sqrt3\)

⇒ AE = \(5\sqrt3\)    ....(1)

Now, in ΔAED,

tan ∠EAD = \(\frac{DE}{AE}\)

⇒ tan 30º = \(\frac {DE}{5\sqrt3}\) (From 1)

⇒ \(\frac{1}{\sqrt3}\) = \(\frac {DE}{5\sqrt3}\)

⇒ DE = 5

Now, the length of DG

⇒ EG + DE

⇒ AC + DE

⇒ AB + BC + DE

⇒ 13.14 + 1.86 + 5

⇒ 20

∴ The height of the tower is 20m.

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) = \(\frac{h}{d}\).

Solve for the angle of elevation:

\(θ = tan^-1(\frac{h}{d}).\)

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

\(tan(θ) = \frac{20 }{ (20√3)}\)

\(tan(θ) = \frac{1 }{ √3}\)

\(θ = tan^-1(\frac{1}{ √3})\)
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)

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

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:

The given expression = \(\frac{{\sin 23^\circ \cos 67^\circ + \sec52^\circ \sin38^\circ + \cos 23^\circ \sin 67^\circ + \rm cosec52^\circ \cos 38^\circ }}{{\rm cose{c^2}20^\circ - {{\tan }^2}70^\circ }}\)

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:

\(\frac{{\sin 23^\circ \cos 67^\circ + \sec52^\circ \sin38^\circ + \cos 23^\circ \sin 67^\circ + \rm cosec52^\circ \cos 38^\circ }}{{\rm cose{c^2}20^\circ - {{\tan }^2}70^\circ }}\)

= \(\frac{{\sin 23^\circ \sin 23^\circ +\frac{1}{cos\,52^\circ} \sin38^\circ + cos 23^\circ cos 23^\circ + \frac{1}{sin\,52^\circ} \cos 38^\circ }}{sec^2\,70^\circ-tan^2\,70^\circ}\)

= \(\frac{sin^2\,23^\circ+\frac{sin\,38^\circ}{sin\,38^\circ}+cos^2\,23^\circ+\frac{cos\,38^\circ}{cos\,38^\circ}}{1}\)

= sin2 23° + 1 + cos2 23° + 1

= 1 + 1 + 1

= 3

∴ The value of the given expression is 3

Given:

secθ - cosθ = 14 and 14 secθ = x

Concept used:

\(Sec\theta =\frac{1}{Cos\theta}\)

Calculations:

According to the question,

⇒ \(sec\theta - cos\theta= 14\)

⇒ \(\sec\theta-\frac{1}{sec\theta}=14\)

⇒ \( sec²\theta-1=14sec\theta\)

⇒ \(\tan^2\theta=14sec\theta\)      ----(\(sec²\theta-1=tan^2\theta\))

\(\ tan²\theta=x\)

∴ The value of x is \(tan²\theta\).

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