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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.

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° :

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

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.

**Trigonometric ratios of complementary angles:**- 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

**Trigonometric Identities**- sin
^{2}Ꝋ + cos^{2}Ꝋ = 1 - tan
^{2}Ꝋ + 1 = sec^{2}Ꝋ - cot
^{2}Ꝋ + 1 = cosec^{2}Ꝋ

- sin
**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/√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.

**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) 25√3

c) 25(√3-1)

d) 25(√3+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/√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.**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.**Example 3: **An airplane flying at 3000 m above the ground passes vertically above another plane at an instant when the angle of elevation of the two planes from the same point on the ground are 60^{0} and 45^{0} respectively. The height of the lower plane from the ground is

a) 1000√3 m

b) 1000/√3 m

c) 500 m

d) 1500(√3+1)**Solution:**

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

Let the height of the plane flying at a lower level be “h” QR=h (since tan 45o= 1)

Tan 60 degrees = 3000/h

⇒ √3 = 3000/h

⇒ h = 3000/√3 = 1000 √3

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