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

**Fig: 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 âˆ B = 90Â° :

- Sin A = sine of âˆ A = (side opposite to âˆ A)/hypotenuse = (BC/AC)
- cos A = cosine of âˆ A = (side adjacent to âˆ A)/hypotenuse = (AB/AC)
- tan A = tangent of âˆ A = (side opposite to âˆ A)/(side adjacent to âˆ A) = (BC/AB)
- cosec A = cosecant of âˆ A = 1/sin A
- sec A = secant of âˆ A = 1/cos A
- cot AA = cotangent of âˆ A = 1/tan A

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

Two angles are said to be complementary if their sum is 90Â°. Thus Ï´ and (90 â€“ Ï´) 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.

**Trigonometric ratios of complementary angles:**

- Sin (90 â€“ A) = cos A
- Cos (90 â€“ A) = sin A
- Tan (90 â€“ A) = cot A
- Cot (90 â€“ A) = tan A
- Sec (90 â€“ A) = cosec A
- Cosec (90 â€“ A) = sec A

**Trigonometric Identities**

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

**Two Special triangles**

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/âˆš2
- In a 60-30-90 triangle ABC shown above, ratio of sides AC:BC:AB = 1: 1/âˆš3: 2

Let us look at some examples to understand the concept better.

**Example 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)** 25âˆš3**c)** 25(âˆš3-1)**d) **25(âˆš3+1)

**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/âˆš3 = BC/A

Tan 45 degrees = 1

â‡’ BC = BD = x

Thus, 1/âˆš3 = BC/AB = BC/(AD+DB) = x/(50 + x)

Thus x (**âˆš**3 -1) = 50 or x= 25(**âˆš**3 +1) m.**Example 2: ****An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30 ^{0} and 60^{0}, respectively. Find the distance between the two planes at that instant.**

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.

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 45^{o}= 1)

Tan 60 degrees = 3000/y

â‡’ âˆš3 = 3000/y

â‡’ y = 3000/âˆš3 = 1000 âˆš3

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