Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 2

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
Two men on either side of a temple of 30 metres high observe its top at the angles of elevation α and β respectively. (as shown in the figure above). The distance between the two men is 40√3 metres and the distance between the first person A and the temple is 30√3 meters.
∠CAB = α =

A
A. sin^-1 (2/√3)

B
B. sin^-1 (1/2)

C
C. sin⁻¹ (2)

D
D. sin⁻¹ (√3/2)

Solution

In Δ BDA
sin α = BD/AB
AB² = AD² + BD²
= (30√3)² + 30²
= 60²
AB = 60 m
Now, sinα = 30/60
Sinα = ½
I.e. ∠CAB = α = sin^-1 (½)
Question 2
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
Two men on either side of a temple of 30 metres high observe its top at the angles of elevation α and β respectively. (as shown in the figure above). The distance between the two men is 40√3 metres and the distance between the first person A and the temple is 30√3 meters.
Domain and Range of cos⁻¹ x =

A
A. (−1, 1), (0, π)

B
B. [−1, 1], (0, π)

C
C. [−1, 1], [0, π]

D
D. (−1, 1), [-(π/2), (π/2)]

Solution

Since cos x is defined at x = 0, and x = π
Domain of cos⁻¹ x includes -1 and 1
Range of cos⁻¹ x also includes 0 and π
Question 3
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan^-1 (½) and tan^-1 (⅓) respectively.
The value of sin A is _______.

A
A. ½

B
B. ⅓

C
C. 1/√5

D
D. 2/√5

Solution

A = tan^-1 (½)
⇒ tan A = ½
∴ sin A = 1/√5
Question 4
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan^-1 (½) and tan^-1 (⅓) respectively.
The third angle, ∠C = _______.

A
A. π/4

B
B. π/2

C
C. π/3

D
D. 3π/4

Solution

∠C = π – (A + B)
= π - (π/4)
= 3π/4
Question 5
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.
Principal value of sin^–1 (½) is

A
A. π/6

B
B. π/3

C
C. π/4

D
D. π/2

Solution

sin^-1 (½) = y
sin y = ½
Principal value branch of sin^–1 is (-π/2, π/2)
and sin(π/6) = ½
⇒ Principal value of sin^-1 (½) is π/6.
Question 6
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.
Principal value of 2cos^–1 (1) + 5tan^–1 (1) is:

A
A. 3π/4

B
B. π/4

C
C. π/2

D
D. 5π/4

Solution

2cos^-1 (1) + 5tan^-1 (1)
= 2 x 0 + 5 x π/4
= 5π/4
Question 7
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
Two men on either side of a temple of 30 metres high observe its top at the angles of elevation α and β respectively. (as shown in the figure above). The distance between the two men is 40√3 metres and the distance between the first person A and the temple is 30√3 meters.
∠CAB = α =

A
A. cos^-1 (⅕)

B
B. cos⁻¹ (⅖)

C
C. cos⁻¹ (√3/2)

D
D. cos⁻¹ (⅘)

Solution

In Δ BDA
Cosα = AD/AB
Cosα = 30√3/60
α = cos^-1 (√3/2)
∴∠CAB = α = cos^-1 (√3/2)
Question 8
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan^-1 (½) and tan^-1 (⅓) respectively.
cos(A + B + C) = _______.

A
A. 1

B
B. 0

C
C. -1

D
D. ½

Solution

Since ABC is a triangle,
∴ A + B + C = 180°
cos (A + B + C) = cos 180°
= -1
Question 9
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.
Principal value of tan^–1 (1)

A
A. π/4

B
B. π/2

C
C. π

D
D. π/3

Solution

tan^-1 (1) =tan^-1 (tan(π/4))
= π/4
Question 10
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
Two men on either side of a temple of 30 metres high observe its top at the angles of elevation α and β respectively. (as shown in the figure above). The distance between the two men is 40√3 metres and the distance between the first person A and the temple is 30√3 meters.
∠BCA = β =

A
A. tan⁻¹ (½)

B
B. tan⁻¹ (2)

C
C. tan⁻¹ (1/√3)

D
D. tan⁻¹ (√3)

Solution

DC = AC - AD
= 40√3 - 30√3
= 10√3 m
In ΔBDC
Question 11
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan^-1 (½) and tan^-1 (⅓) respectively.
If B = cos^–1 x, then x = _______.

A
A. 1/√5

B
B. 3/√10

C
C. 1/√10

D
D. 2/√5

Solution

Given B = tan^-1 (⅓)
⇒ tan B = ⅓
∴ cos B = 3/√10
B = cos^-1 (3/√10)
⇒ x = 3/√10
Question 12
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.
Principal value of cot^-1 (√3) is :

A
A. π/3

B
B. π

C
C. π/6

D
D. π/2

Solution

cot ^-1 (√3) =cot-1 (cot (π/6) = π/6
Question 13
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
Two men on either side of a temple of 30 metres high observe its top at the angles of elevation α and β respectively. (as shown in the figure above). The distance between the two men is 40√3 metres and the distance between the first person A and the temple is 30√3 meters.
∠ABC =

A
A. π/4

B
B. π/6

C
C. π/2

D
D. π/3

Solution

Since,
Sin α = ½
i.e., sin α = sin 30⁰ [∵ sin 30⁰ = ½]
∴ α = 30⁰
We have tan β = √3
tan β = tan 60⁰
∴ β = 60⁰
Now In ΔABC
∠ABC + ∠BCA + ∠CAB = 180⁰
∠ABC + 60⁰+ 30⁰= 180⁰
∠ABC = 90⁰
∴ ∠ABC = π/2
Question 14
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan^-1 (½) and tan^-1 (⅓) respectively.
If A = sin^–1 x; then the value of x is:

A
A. 1/√5

B
B. 2/√5

C
C. 1/√10

D
D. 3/√10

Solution

A = tan^-1 (½)
⇒ tan A = ½
∴ Sin A = 1/√5
A = sin^-1 (1/√5)
⇒ x = 1/√5
Question 15
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:
The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.
Principal value of sin ^-1 (1) + sin^-1 (1/√2) is

A
A. 2π

B
B. π

C
C. 3π/4

D
D. π/3

Solution

sin^-1 (1) + sin-1(1/√2) = π//2 + π/4
= 3π/4