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**Introduction**

Trigonometry is the branch of mathematics that deals with the study of relationships between sides and angles of a triangle. It is derived from the Greek word, **Tri** meaning three and **Gon** means Angle and **Metron** means Measure.

**Trigonometry**

- The ratio of the lengths of two sides of a right-angled triangle is called a
**trigonometric ratio**. There are six trigonometric ratios. __Consider a right-angled triangle ABC as shown below with:__

AC = Hypotenuse of triangle

AB = Side adjacent to angle A

BC = Side opposite to angle A__Trigonometric ratios of right-angled triangle ABC with ∠B = 90°:__**(i)**sin A = Sine of ∠A = (Side opposite to ∠A / hypotenuse) = (BC/AC)**(ii)**cos A = Cosine of ∠A = (Side adjacent to ∠A / hypotenuse) = (AB/AC)**(iii)**tan A = Tangent of ∠A = (Side opposite to ∠A / Side adjacent to ∠A) = (BC/AB)**(iv)**cosec A = Cosecant of ∠A = 1/sin A**(v)**sec A = Secant of ∠A = 1/cos A**(vi)**cot A = Cotangent of ∠A = 1/tan A

**➢ **

- The trigonometric ratios of some special angles, i.e. 0°, 30°,45°,60°,90° follow a pattern and are easy to remember. Identifying and remembering these patterns helps in solving problems involving these angles.

**Table: Trigonometric Ratios of Specific Angles**

- Two angles are said to be complementary if their sum is 90°. Thus ϴ and (90 – ϴ) are complementary angles.
- Trigonometric ratios of complementary angles help in simplifying the problems.
- Representing complementary angles in terms of these standard angles helps in solving a complex problem involving trigonometric ratios.

**➢ **

- Sin (90° – A) = Cos A
- Cos (90° – A) = Sin A
- Tan (90° – A) = Cot A
- Cot (90° – A) = Tan A
- Sec (90° – A) = Cosec A
- Cosec (90° – A) = Sec A

**➢ **

- sin
^{2}θ + cos^{2}θ = 1 - tan
^{2}θ + 1 = sec^{2}θ - cot
^{2}θ + 1 = cosec^{2}θ

Question 2:The value of sin θ and cos (90° – θ)

**➢ **

- There are two special triangles we need to know
**45°-45°-90° and 30°-60°-90° triangles**. They are depicted in the figures below.

- The figures show how to find the side lengths of those types of these special triangles.
- In a 45-45-90 triangle ABC shown above, ratio of side AC:BC:AB = 1:1:1/√2
- In a 60-30-90 triangle ABC shown above, ratio of sides AC:BC:AB = 1: 1/√3: 2

__Let us look at some examples to understand the concept better:__

Question 3:The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

**Example 1: ****Anil looked up at the top of a lighthouse from his boat and found the angle of elevation to be 30º. After sailing in a straight line 50 m towards the lighthouse, he found that the angle of elevation changed to 45º. Find the height of the lighthouse.****a)** 25**b)** 25√3**c)** 25(√3-1)**d) **25(√3+1)

**Ans**: d

**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/A

► 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.

Question 4:From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:

Solution:

Let C and D be the two aeroplanes, and A be the point of observation. Then,

Therefore, the distance between the two planes is 6250 m.

**Ans**: a**Solution:**

Let the higher plane fly such that at point A the angle of elevation from point D is 60º.

Let the height of the plane flying at a lower level be CD = y (since tan 45º= 1)

► tan 60º = 3000/y

► √3 = 3000/y

► y = 3000/√3 = 1000√3

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