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QUESTION: 1

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

Solution:

QUESTION: 2

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: 3

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: 4

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: 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

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

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