Trigonometry Test

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Trigonometry Test - 1

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Question 1
3 / -1

A student is standing with a banner at the top of a 100 m high college building. From a point on the ground, the angle of elevation of the top of the student is 60° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the student.

A
35 m

B
73.2 m

C
50 m

D
75m

Solution

Let BC be the height of the tower and DC be the height of the student.
In rt. ΔABC,
AB = BC cot 45° = 100 m
In rt. ΔABD, AB = BD cot 60° = (BC + CD) cot 60° = (10 + CD) * (1 / √3)
∵ AB = 100 m
⇒ (10 + CD) * 1 / √3 = 100
⇒ (10 + CD) = 100√3
⇒ CD = 100√3 - 100 = 100 (1.732 - 1) = 100 x 0.732 = 73.2 m
Question 2
3 / -1

If sin (A + B) = √3 / 2 and tan (A – B) = 1. What are the values of A and B?

A
37, 54

B
50, 10

C
45, 15

D
52.5, 7.5

Solution
Question 3
3 / -1

If Cos x – Sin x = √2 Sin x, find the value of Cos x + Sin x:

A
√2 Cos x

B
√2 Cosec x

C
√2 Sec x

D
√2 Sin x Cos x

Solution

Cos x – Sin x = √2 Sin x
⇒ Cos x = Sin x + √2 Sin x

⇒ Sin x = (√2 - 1) Cos x
⇒ Sin x = √2 Cos x - Cos x
⇒ Sin x + Cos x = √2 Cos x
Question 4
3 / -1

If tanØ + sinØ = m, tanØ - sinØ = n, find the value of m² - n².

A
2√mn

B
4√mn

C
m – n

D
2mn

Solution

Adding the two equations, tanØ = (m + n) / 2
Subtracting the two equations, sinØ = (m - n) / 2
Since, there are no available direct formula for relation between sinØ tanØ.
But we know that: cosec²Ø - cot²Ø = 1
Question 5
3 / -1

If cos A + cos² A = 1 and a sin¹² A + b sin¹⁰ A + c sin⁸ A + d sin⁶ A - 1 = 0. Find the value of a+b / c+d

A
4

B
1

C
6

D
3

Solution

Cos A = 1 - Cos²A
⇒ Cos A = Sin²A
⇒ Cos²A = Sin⁴A
⇒ 1 – Sin²A = Sin⁴A
⇒ 1 = Sin⁴A + Sin²A
⇒ 1³ = (Sin⁴A + Sin²A)³
⇒ 1 = Sin¹²A + Sin⁶A + 3 Sin⁸A + 3 Sin¹⁰A
⇒ Sin¹²A + Sin⁶A + 3 Sin⁸A + 3 Sin¹⁰A – 1 = 0
On comparing,
a = 1, b = 3 , c = 3 , d = 1
⇒ (a+b)/(c+d) = 1
Hence, the answer is 1
Question 6
3 / -1

3sinx + 4cosx + r is always greater than or equal to 0. What is the smallest value ‘r’ can to take?

A
5

B
-5

C
4

D
3

Solution

Therefore, the answer is Option A.
Question 7
3 / -1

A right angled triangle has a height ‘p’, base ‘b’ and hypotenuse ‘h’. Which of the following value can h² not take, given that p and b are positive integers?

A
74

B
52

C
13

D
23

Solution

We know that,
h² = p² + b² Given, p and b are positive integer, so h2 will be sum of two perfect squares.

We see
a) 7² + 5² = 74
b) 6² + 4² = 52
c) 3² + 2² = 13
d) Can’t be expressed as a sum of two perfect squares
Therefore the answer is Option D.
Question 8
3 / -1

If Cos x – Sin x = √2 Sin x, find the value of Cos x + Sin x:

A
√2 Cos x

B
√2 Cosec x

C
√2 Sec x

D
√2 Sin x Cos x

Solution

Cos x – Sin x = √2 Sin x

=> Cos x = Sin x + √2 Sin x
=> Cos x = Sin x + √2 Sin x
=> Sin x = Cosx/(√2+1) * Cos x
=> Sin x = (√2−1)/(√2−1) * 1/(√2+1) * Cos x
=> Sin x = (√2−1)/((√2)²−(1)²)* Cos x
=> Sin x = (√2 - 1) Cos x
=> Sin x = √2 Cos x – Cos x
=> Sin x + Cos x = √2 Cos x
Hence, the correct answer is Option A.
Question 9
3 / -1

You are standing on the corner of a square whose side length is 25 feet. Standing on the opposite corner from you is a tall tree. The angle of elevation from your position to the top of the tree is exactly 60°. How tall is the tree?

A
25√ 2

B
25√ 3

C
25√ 6

D
50√ 3

Solution

First find the distance of the diagonal d along the ground from corner to corner. Using Pythagorean theorem with sides 25 and 25, we get:
25² + 25² = d²
2 × 25² = d²
d = 25√ 2 .
Then to obtain the height h of the tree, use the tangent ratio with angle 60°.
tan 60° = x / (25√ 2 )
√ 3 = x / (25√ 2 )
x = 25√ 2 × √ 3 = 25√ 6
Question 10
3 / -1

Anil looked up at the top of a lighthouse from his boat and found the angle of elevation to be 30 degrees. After sailing in a straight line 50 m towards the lighthouse, he found that the angle of elevation changed to 45 degrees. Find the height of the lighthouse.

A
25

B
25√3

C
25(√3-1)

D
25(√3+1)

Solution

If we look at the above image, A is the previous position of the boat. The angle of elevation from this point to the top of the lighthouse is 30 degrees.
After sailing for 50 m, Anil reaches point D from where the angle of elevation is 45 degrees. C is the top of the lighthouse.
Let BD = x
Now, we know tan 30 degrees = 1/ √3 = BC/AB
Tan 45 degrees = 1
=> BC = BD = x
Thus, 1/ √3 = BC/AB = BC / (AD+DB) = x / (50 + x)
Thus x (√3 -1) = 50 or x= 25(√3 +1) m
The answer is Option D.