**Trigonometry Study Material PDF**

**Trigonometric ratios:**

The most important task of trigonometry is to find the remaining side and angle of a triangle when some of its side and angles are given. This problem is solved by using some ratio of sides of a triangle with respect to its acute angle. These ratio of acute angle are called trigonometric ratio of angle. Let us now define various trigonometric ratio.

Consider an acute angle ∠YAX = ? with initial side AX and terminal side AY. Draw PM perpendicular from P on AX to get right angle triangle AMP. In right angle triangle AMP.

Base = AM = x

Perpendicular = PM = y and

Hypotenuse = AP = r.

r=√(x^2+y^2 )

**We define the following six trigonometric Ratios:**

- sinθ=Perpendicular/Hypotenuse=y/r
- cosθ=Base/Hypotenuse=x/r
- tanθ=Perpendicular/Base=y/x
- cosecθ=Hypotenuse/Perpendicular=r/y
- secθ=Hypotenuse/Base=r/x
- cotθ=Base/Perpendicular=x/y

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**Important formula:-**

- sin^2θ+cos^2θ=1
- cosec^2θ-cot^2θ=1
- sec^2θ=tan^2θ=1
- sin(90°-θ)=cosθ
- cos(90°-θ)=sinθ
- tan(90°-θ)=cotθ⇒ cot〖(90°-θ)=tanθ 〗
- cosec(90°-θ)=secθ
- sec(90°-θ)=cosecθ

**Example 1: In a ΔABC right angled at B if AB = 12, and BC = 5 find sin A and tan A, cos C and cot C**

**Solution: **

AC=√((AB)^2+(BC)^2 )

=√(〖12〗^2+5^2 )

=√(144+25)

=√169=13

When we consider t-ratios of∠A we have

Base AB = 12

Perpendicular = BC = 5

Hypotenuse = AC = 13

sinA=Perpendicular/Hypotenuse=5/13

tanA=Perpendicular/Base=5/12

When we consider t-ratios of ∠C, we have

Base = BC = 5

Perpendicular = AB = 12

Hypotenuse = AC = 13

cosC=Base/Hypotenuse=5/13

cotC=Base/Perpendicular=5/12

**Example 2: In a right triangle ABC right angle at B the six trigonometric ratios of ∠C**

**Solution:**

sinA=Perpendicular/Hypotenuse=3/5

Base=√((Hypotenuse)^2-(Perpendicualr)^2 )

=√(5^2-3^2 )

=√(25-9)=√16=4

Now

sinC=BC/AC=4/5,cosecC=5/4

cosC=3/5=AB/AC,secC=5/3

tanC=AB/AC=4/3,cotC=3/4

**Example 3: Find the value of 2 sin2 30° tan 60° – 3 cos2 60° sec2 30°**

**Solution:**

2(1/2)^2×√3-3(1/2)^2×(2/√3)^2

=2×1/4×√3-3×1/4×4/3=√3/2-1=(√3-2)/2

**Example 4:bFind the value θ sin2θ=√3**

**Solution:**

sin2θ= √3/2

2θ = 60

θ = 30°

**Example 5: Find the value of x. Tan 3x = sin 45° cos 45° + sin 30°**

**Solution:**

tan3x=1/√2×1/√2+1/2

=1/2+1/2=1

⇒tan3x=1 ⇒tan3x=tan〖45°〗

3x = 45°

X = 15°

**HEIGHT AND DISTANCE**

Sometimes, we have to find the height of a tower, building, tree, distance of a ship, width of a river, etc. Though we cannot measure them easily, we can determine these by using trigonometric ratios.

**Line of Sight**

The line of sight or the line of vision is a straight line to the object we are viewing. If the object is above the horizontal from the eye, we have to lift up our head to view the object. In this process, our eye move, through an angle. This angle is called the angle of elevation of the object.

If the object is below the horizontal from the eye, then we have no turn our head downwards no view the object. In this process, our eye move through an angle. This angle is called the angle of depression of the object.

**Example: A 25 m long ladder is placed against a vertical wall of a building. The foot of the ladder is 7m from base of the building. If the top of the ladder slips 4m, then the foot of the Ladder will slide by how much distance.**

**Sol:** Let the height of the wall be h.

Now, h = √(〖25〗^2-7^2 )

= √(576 ) = 24m

QS = √(625-400)

= √(225 )=15m

Required distance, X = (15-7) = 8m

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