FIND TRIGONOMETRIC RATIOS USING RIGHT TRIANGLES

To find all trigonometric ratios from the given right triangle, first we have to name the sides as hypotenuse side, opposite side and adjacent side. 

Hypotenuse Side :

The side which is opposite to 90 degree is known as hypotenuse side.

Opposite Side :

The side which is opposite to θ is known as opposite side

Adjacent Side :

Apart from hypotenuse and opposite side, the remaining third side of the triangle is known as adjacent side.

Generally we have 6 trigonometric ratios, those are sin θ, cos θ, tan θ, csc θ, sec θ and cot θ.

Formulas to find the values of the above six trigonometric ratios. 

sin θ  =  Opposite side/hypotenuse side

cos θ  =  Adjacent side/hypotenuse side

tan θ  =  Opposite side/Adjacent side

csc θ  =  Hypotenuse side/Opposite side

sec θ  =  Hypotenuse side/Adjacent side

cot θ  =  Adjacent side/Opposite side

From the above formulas, we ca get the following results.

sin θ and csc θ are reciprocal to each other 

cos θ and sec θ are reciprocal to each other

tan θ and cot θ are reciprocal to each other   

Examples

Example 1 :

In the right triangle shown below, find the six trigonometric ratios of the angle θ. 

Solution :

From the above triangle, we come to know that the right angled at B.

AC - hypotenuse side  =  10

AB - opposite side  =  6 

BC - Adjacent side  =  8

Finding the vale of sin θ :  

sin θ  =  Opposite side/Hypotenuse side

sin θ  =  AB/AC

sin θ  =  6/10

sin θ  =  3/5

Finding the vale of cos θ :  

cos θ  =  Adjacent side/Hypotenuse side 

cos θ  =  BC/AC

cos θ  =  8/10

cos θ  =  4/5

Finding the vale of tan θ :  

tan θ  =  Opposite side/Adjacent side 

tan θ  =  AB/BC

tan θ  =  6/8

tan θ  =  3/4

Finding the vale of csc θ :  

csc θ = Hypotenuse side/Opposite side

csc θ  =  AC/AB

csc θ  =  10/6

csc θ  =  5/3

Finding the vale of sec θ :  

sec θ = Hypotenuse side/Adjacent side

sec θ  =  AC/BC

sec θ  =  10/8

sec θ  =  5/4

Finding the vale of cot θ :  

cot θ = Adjacent side/opposite side

cot θ  =  BC/AB

cot θ  =  8/6

cot θ  =  4/3

Example 2 :

In the right triangle shown below, find the six trigonometric ratios of the angle θ. 

Solution :

In the triangle above, right angle is at C.

AB - hypotenuse side = 25

AC - opposite side = 7 

BC - Adjacent side = 24

Finding the vale of sin θ :  

sin θ  =  Opposite side/Hypotenuse side 

sin θ  =  AC/AB

sin θ  =  7/25

Finding the vale of cos θ :  

cos θ  =  Adjacent side/Hypotenuse side 

cos θ  =  BC/AB

cos θ  =  24/25

Finding the vale of tan θ :  

tan θ  =  Opposite side/Adjacent side 

tan θ  =  AC/BC

tan θ  =  7/24 

Finding the vale of csc θ :  

csc θ  =  Hypotenuse side/Opposite side

csc θ  =  AB/AC

csc θ  =  25/7 

Finding the vale of sec θ :  

sec θ  =  Hypotenuse side/Adjacent side

 sec θ  =  AB/BC

sec θ   =  25/24

Finding the vale of cot θ :  

cot θ  =  Adjacent side/opposite side

cot θ  =  BC/AC

cot θ  =  24/7

Example 3 :

In the right triangle shown below, find the six trigonometric ratios of the angle θ. 

Solution :

From the above triangle right angled at C.

AB - hypotenuse side = 37

AC - opposite side = 35 

BC - Adjacent side = 12

Finding the vale of sin θ : 

sin θ  =  Opposite side/Hypotenuse side 

sin θ  =  AC/AB

sin θ  = 35/37

Finding the vale of cos θ : 

cos θ  =  Adjacent side/Hypotenuse side 

cos θ  =  BC/AB

cos θ  =  12/37

Finding the vale of tan θ :

tan θ  =  Opposite side/Adjacent side

tan θ  =  AC/BC

tan θ  =  35/12 

Finding the value of csc θ : 

csc θ  =  Hypotenuse side/Opposite side

csc θ  =  AB/AC

csc θ  =  37/35 

Finding the value of sec θ : 

sec θ  =  Hypotenuse side/Adjacent side

sec θ  =  AB/BC

sec θ  =  37/12

Finding the value of cot θ : 

cot θ  =  Adjacent side/opposite side

cot θ  =  BC/AC

cot θ  =  12/35

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