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# Test: Trigonometry- 1

## 5 Questions MCQ Test Quantitative Aptitude (Quant) | Test: Trigonometry- 1

Description
This mock test of Test: Trigonometry- 1 for SSC helps you for every SSC entrance exam. This contains 5 Multiple Choice Questions for SSC Test: Trigonometry- 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Trigonometry- 1 quiz give you a good mix of easy questions and tough questions. SSC students definitely take this Test: Trigonometry- 1 exercise for a better result in the exam. You can find other Test: Trigonometry- 1 extra questions, long questions & short questions for SSC on EduRev as well by searching above.
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 m2 - n2.

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: 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 + 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

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