Trigonometric ratios for 60 degrees


 
 
Concept Explanation
 

Trigonometric ratios for 60 degrees

Trigonometric Ratios for 60 Degrees: Value for the Trigonometrical Ratios for certain angle such as 30^0,60^090^0 and 45^0 are commonly called standard angles and the trigonometrical ratios of these angles are frequently used to solve particular angles. In this section we will derive the value of trigonometric ratios for 60^0.

Consider an equilateral  Delta ; ABC with each side = 2a units

We know that  each angle of Delta ABC; is ; 60^o. Moreover in an equilateral triangle the altitude and median coincide. From A, draw AD ; perp ; BC.

 As AD is the median also therefore D is the mid - point of BC,

As BC = 2a units

BD = DC = a units

Also,   angle ADB=90^o [ AD is perpendicular BC ]

and  angle ABD = 60^0     [ Each angle in equilateral triangle is 60^0 ]

 Therefore in triangle ABD      angle BAD=30^o    [Angle Sum Property]

Now Delta ADB, is a right angled triangle so by pythagorous theorem

AB^2 = AD^2 + BD^2

AD =sqrt {AB^2-BD^2}            

Substituting the value of AB and BD

          =sqrt {(2a)^2-a^2}=sqrt {3a^2}=sqrt {3}a; units

In right Delta ADB, we have angle BAD;=30^0

Adjacent; angle A= AD=sqrt {3}a, ; Opposite ;angle A=BD=a ;

; and ;Hypoteneous =AB=2a

Thus, we have

sin ; 60^o=frac{Opposite}{Hypoteneous}=frac {AD}{AB}=frac {sqrt {3}a}{2a}=frac {sqrt {3}}{2}       

cos ; 60^o=frac{Adjacent}{Hypoteneous}=frac {BD}{AB}=frac {a}{2a}=frac {1}{2} 

tan ; 60^o=frac{Opposite}{Adjacent}=frac {AD}{BD}=frac {sqrt {3}a}{a}=sqrt {3}  

cosec ; 60^o=frac {1}{sin ; 60^o}=frac {1}{sqrt {3}}

sec ; 60^o=frac {1}{cos ; 60^o}=2

cot ; 60^o=frac {1}{tan ; 60^o}=frac {1}{sqrt {3}}

large angle Asin Acos Atan Acosec Asec Acot A
large 60^{circ}large frac{sqrt{3}}{2}large frac{1}{2}large sqrt{3}large frac{2}{sqrt{3}}2

large frac{1}{sqrt{3}}

 

Illustration: Simplify the given expression (tan ;60^0 +1) times (1- cot; 60^0)

Solution: To solve this expression we will substitute the value of the ratios at 60^0

(tan ;60^0 +1) times (1- cot; 60^0)

=(sqrt3 +1);(1-frac{1}{sqrt3})

=(sqrt3 +1);times (frac{sqrt3 -1}{sqrt3})

=frac{(sqrt3 +1);times (sqrt3 -1)}{sqrt3}

=frac{3 -1}{sqrt3}

=frac{2}{sqrt3}

Sample Questions
(More Questions for each concept available in Login)
Question : 1

frac{2;tan; 30^0}{1+tan^230^0}  is equal to ___________________.

Right Option : C
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Explanation
Question : 2

Solve the following expression :

(tan;60^0+2)+(1+sin;60^0+cos;60^0)

Right Option : B
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Explanation
Question : 3

If frac{cos; theta }{1-sin; theta }+frac{cos; theta }{1+sin; theta }=4 ,then

Right Option : D
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Explanation
 
 
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