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Trigonometric ratios for 60 degrees

Introduction to Trigonometry I - Concepts

Identifying Hypotenuses In A Right Triangle

Identifying Adjacent Side To An Angle

Identifying Opposite Side To An Angle

Trigonometric Ratios

Reciprocal Trigonometric Ratios

Trigonometric ratios for 30 degrees

Trigonometric ratios for 45 degrees

Trigonometric ratios for 60 degrees

Trigonometric ratios for 90 degrees

Simplifying Trigonometric Expressions

Class - 10th CBSE Subjects

Concept Explanation

Trigonometric ratios for 60 degrees

Trigonometric Ratios for 60 Degrees: Value for the Trigonometrical Ratios for certain angle such as $30^0$ , $60^0$ , $90^0$ 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

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

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

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

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

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

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

$large angle A$	sin A	cos A	tan A	cosec A	sec A	cot 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

The value for the given expression is ___________ $frac{sqrt{13};cos ;60^0+2;sin; 60^0}{sqrt{13};cos; 60^0 -2;sin;60^0}$
A. $sqrt{13}-2sqrt3$ $sqrt{13}-2sqrt3$	B. $sqrt{13}+2sqrt3$	C. $(sqrt{13}-2sqrt3)^2$	D. $(sqrt{13}+2sqrt3)^2$
Right Option : D
View Explanation

Question : 2

Simplify the given expression: $sin^2;60^0;cos^2;60^0+4;tan^2;60^0+frac{1}{2};sin^2;60^0-2;cos^2;60^0+frac{1}{24};cos^2;60^0$
A. 401.23	B. 12.052	C. 15.021	D. 130.025
Right Option : B
View Explanation

Question : 3

Solve the following expression : $sec;60^0+tan;60^0 +(1+cot;60^0;sin;60^0)$
A. $frac{3+2sqrt5}{2}$	B. $frac{7+2sqrt3}{5}$	C. $frac{7+2sqrt3}{2}$	D. $frac{6+2sqrt3}{2}$
Right Option : C
View Explanation

Chapters

Polynomial I

Polynomials II

Pair of Linear Equations I

Pair of Linear Equations II

Real Numbers I

Real Numbers II

Coordinate Geometry I

Coordinate Geometry II

Quadratic Equations I

Quadratic Equations II

Arithmetic Progression I

Arithmetic Progression II

Introduction to Trigonometry I

Introduction to Trigonometry II

Surface Area and Volume I

Surface Area and Volume II

Areas Related to Circles

Some Applications of Trigonometry

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