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

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:

tan2A - 6tanA + 9 = 0

Concept used:

tanθ = P/B

cotθ = 1/tanθ

cosθ = B/H

Here,

P = Perpendicular

B = Base

H = Hypotenuse 

Pythagoras theorem = H2 = P2 + B2

Calculation:

tan2A - 6tanA + 9 = 0

⇒ tan2A - 2 × 3 × tanA + 9 = 0

⇒ (tanA - 3)² = 0

⇒ (tanA - 3) = 0

⇒ tanA  = 3

So, P/B = 3/1

Now,

H² = 3² + 1²

⇒ H2 = 9 + 1

⇒ H2 = 10

⇒ H = √10

So, cosA = 1/√10

Now,

6 cot A + 8\(√{10} \) cos A = 6 × (1/3) + 8\(√{10} \) × (1/√10)

⇒ 2 + 8

⇒ 10

∴ Required answer is 10.

Shortcut Trick

tan2A - 6tanA + 9 = 0

⇒ (tanA - 3)2 = 0

⇒ (tanA - 3) = 0

⇒ tanA  = 3

⇒ cotA = 1/3

cosA = 1 ÷  \(\sqrt{sec^2A}\)

⇒ 1 ÷\(\sqrt{1 + tan^2A}\)

⇒ 1 ÷\(\sqrt{1 + 9}\) = \(\frac {1}{\sqrt{10}}\)

Now,

6 cot A + \(8\sqrt{10}\) cos A

⇒ 6 × (1/3) + \(8\sqrt{10}\) × \(\frac {1}{\sqrt{10}}\)

⇒ 2 + 8 = 10

∴ The value of 6 cot A + 8\(\sqrt{10} \) cos A is 10.

Given:

cos (x - y) = √3/2

sin (x + y) = 1/2

Formula:

cos30° = √3/2

sin30° = 1/2

Calculation:

cos(x - y) = √3/2

⇒ cos(x - y) = cos30°

(x - y) = 30°     ---- (1)

sin(x + y) = 1/2

⇒ sin(x + y) = sin30°

(x + y) = 30    ---- (2)

Add equation (1) and equation (2), we get

2x = 60°

∴ x = 30°

Given:

If sec 4A = cosec (3A - 50°), where 4A and 3A are acute angles, find the value of cosec (A + 25º).

Formula Used:

We know that sec θ = cosec (90° - θ).

Calculation:

Given sec 4A = cosec (3A - 50°)

 ⇒ 4A = 90° - (3A - 50°)

⇒ 4A = 90° - 3A + 50°

⇒ 4A + 3A = 140°

⇒ 7A = 140°

⇒ A = 20°

Now, we need to find A + 25:

⇒ A + 25 = 20° + 25°

⇒ A + 25 = 45°

cosec (A + 25º) = cosec 45° = √2

∴ The correct answer is √2.

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

Calculation:

cot⁴θ + cot²θ = 3

⇒ cos⁴θ/sin⁴θ + cos ²θ/sin ²θ 

⇒ cos 2θ(cos 2θ+ sin 2θ )/sin4θ = 3 (Taking LCM)

⇒ cos 2θ/sin⁴θ = 3

⇒ cot²θ . cosec²θ = 3

Now, cosec⁴θ – cosec²θ

⇒ cosec²θ(cosec²θ – 1)

⇒ cosec²θcot²θ = 3

∴ cosec4θ – cosec2θ = 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 \(tan²\theta\).