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QUESTION: 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.

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

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

Solution:

QUESTION: 3

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

If tanØ + sinØ = m, tanØ - sinØ = n, find the value of **m ^{2} - n^{2}.**

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:__ cosec^{2}Ø - cot^{2}Ø = 1

QUESTION: 5

If cos A + cos^{2} A = 1 and a sin^{12} A + b sin^{10} A + c sin^{8} A + d sin^{6} A - 1 = 0. Find the value of a+b / c+d

Solution:

Cos A = 1 - Cos^{2}A

⇒ Cos A = Sin^{2}A

⇒ Cos^{2}A = Sin^{4}A

⇒ 1 – Sin^{2}A = Sin^{4}A

⇒ 1 = Sin^{4}A + Sin^{2}A

⇒ 1^{3} = (Sin^{4}A + Sin^{2}A)^{3}

⇒ 1 = Sin^{12}A + Sin^{6}A + 3 Sin^{8}A + 3 Sin^{10}A

⇒ Sin^{12}A + Sin^{6}A + 3 Sin^{8}A + 3 Sin^{10}A – 1 = 0

On comparing,

a = 1, b = 3 , c = 3 , d = 1

⇒ (a+b)/(c+d) = 1

Hence, the answer is 1

QUESTION: 6

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

Solution:

Therefore, the answer is **Option A.**

QUESTION: 7

A right angled triangle has a height ‘p’, base ‘b’ and hypotenuse ‘h’. Which of the following value can h^{2} not take, given that p and b are positive integers?

Solution:

We know that,

h^{2} = p^{2} + b^{2} Given, p and b are positive integer, so h2 will be sum of two perfect squares.

We see

a) 7^{2} + 5^{2} = 74

b) 6^{2} + 4^{2} = 52

c) 3^{2} + 2^{2} = 13

d) Can’t be expressed as a sum of two perfect squares

Therefore the answer is **Option D.**

QUESTION: 8

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

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?

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:

25^{2} + 25^{2} = *d*^{2}

2 × 25^{2} = *d*^{2}

*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

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.

Solution:

If we look at the above image, A is the previous position of the boat. Angle of elevation from this point to the top of the light house is 30 degrees.

After sailing for 50 m, Anil reaches point D from where angle of elevation is 45 degrees. C is the top of the light house.

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.**

### Maths test :-Some applications of trigonometry.

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