कुछ विशिष्ट कोणों के त्रिकोणमितीय अनुपात: Difference between revisions
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In this section, we will find the values of the trigonometric ratios for angles of <math>0^\circ ,30^\circ , 45^\circ, 60^\circ , 90^\circ | |||
</math>. | |||
== Trigonometric Ratios of 45° == | |||
[[File:Right angle triangle.jpg|alt=Fig.1 Triangle|thumb|Fig.1 Triangle]] | |||
In <math>\bigtriangleup ABC</math> right angled at <math>B</math> , If <math>\angle A =45^\circ</math>, <math>\angle C =45^\circ</math> | |||
<math>BC=AB=a</math> | |||
Using Pythagoras Theorem | |||
<math>AB^2+BC^2=AC^2</math> | |||
<math>a^2+a^2=2a^2</math> | |||
<math>AC= a\sqrt{2} </math> | |||
<math>sin \ 45^\circ = \frac{side \ opposite \ to \ angle \ 45^\circ}{hypotenuse}= \frac{BC} {AC} =\frac{a} {a\sqrt{2}}=\frac{1} {\sqrt{2}}</math> | |||
<math>cos \ 45^\circ = \frac{side \ adjacent \ to \ angle \ 45^\circ}{hypotenuse}= \frac{AB} {AC} =\frac{a} {a\sqrt{2}}=\frac{1} {\sqrt{2}}</math> | |||
<math>tan \ 45^\circ = \frac{side \ opposite \ to \ angle \ 45^\circ}{side \ adjacent \ to \ angle \ 45^\circ}= \frac{BC}{AB} =\frac{a}{a}=1</math> | |||
<math>cosec \ 45^\circ = \frac{1}{sin \ 45^\circ}=\sqrt{2}</math> , <math>sec \ 45^\circ = \frac{1}{cos \ 45^\circ}=\sqrt{2}</math> , <math>cot \ 45^\circ = \frac{1}{tan \ 45^\circ}=1</math> | |||
== Trigonometric Ratios of 30° and 60° == | |||
[[File:Triangle -1.jpg|alt=Fig. 2 - Triangle|thumb|Fig. 2 Triangle]] | |||
Consider an equilateral <math>\bigtriangleup ABC</math>. Each angle in an equilateral triangle is <math>60^\circ</math>, therefore,<math>\angle A = \angle B =\angle C =60^\circ</math> . | |||
Draw a perpendicular <math>AD</math> from <math>A</math> to the side <math>BC</math> (see Fig. 2). | |||
Now <math>\bigtriangleup ABD=\bigtriangleup ACD</math> | |||
Therefore, <math>BD=DC</math> and <math>\angle BAD = \angle CAD</math> (Corresponding Parts of Congruent Triangles) | |||
<math>\bigtriangleup ABD</math> is a right angled triangle , right angled at <math>D</math> with <math>\angle BAD = 30^\circ</math> and <math>\angle ABD = 60^\circ</math> | |||
Let <math>AB=2a</math> , Hence <math>BC=AC=AB=2a</math> | |||
<math>BD=\frac{1}{2}BC=\frac{1}{2} \times 2a=a</math> | |||
<math>AD^2=AB^2-BD^2=(2a)^2-a^2=3a^2</math> | |||
<math>AD=a\sqrt{3}</math> | |||
<math>sin \ 30^\circ = \ \frac{BD} {AB} =\frac{a} {2a}=\frac{1} {2}</math> , <math>cos \ 30^\circ = \ \frac{AD} {AB} =\frac{a\sqrt{3}} {2a}=\frac{\sqrt{3}} {2}</math> , <math>tan \ 30^\circ = \ \frac{BD} {AD} =\frac{a} {a\sqrt{3}}=\frac{1}{\sqrt{3}} </math> | |||
<math>cosec \ 30^\circ = \frac{1}{sin \ 30^\circ}=2</math> , <math>sec \ 30^\circ = \frac{1}{cos \ 30^\circ}=\frac{2}{\sqrt{3}}</math> , <math>cot \ 30^\circ = \frac{1}{tan \ 30^\circ}=\sqrt{3}</math> | |||
Similarly | |||
<math>sin \ 60^\circ = \ \frac{AD} {AB} =\frac{a\sqrt{3}} {2a}=\frac{\sqrt{3}} {2}</math> , <math>cos \ 60^\circ = \ \frac{BD} {AB} =\frac{a} {2a}=\frac{1} {2}</math> , <math>tan \ 60^\circ = \ \frac{AD} {BD} =\frac{a\sqrt{3}} {a}=\sqrt{3} </math> | |||
<math>cosec \ 60^\circ = \frac{2}{\sqrt{3}}</math> , <math>sec \ 60^\circ = \frac{1}{cos \ 60^\circ}=2</math> , <math>cot \ 60^\circ = \frac{1}{tan \ 60^\circ}=\frac{1}{\sqrt{3}}</math> | |||
{| class="wikitable" | |||
|+Trigonometric ratios of 0°, 30°, 45°, 60° and 90° | |||
!<math>\angle A</math> | |||
!<math>0^\circ</math> | |||
!<math>30^\circ</math> | |||
!<math>45^\circ</math> | |||
!<math>60^\circ</math> | |||
!<math>90^\circ</math> | |||
|- | |||
|<math>sin \ A</math> | |||
|<math>0</math> | |||
|<math>\frac{1} {2}</math> | |||
|<math>\frac{1}{\sqrt{2}}</math> | |||
|<math>\frac{\sqrt{3}} {2}</math> | |||
|<math>1</math> | |||
|- | |||
|<math>cos \ A</math> | |||
|<math>1</math> | |||
|<math>\frac{\sqrt{3}} {2}</math> | |||
|<math>\frac{1}{\sqrt{2}}</math> | |||
|<math>\frac{1} {2}</math> | |||
|<math>0</math> | |||
|- | |||
|<math>tan \ A</math> | |||
|<math>0</math> | |||
|<math>\frac{1}{\sqrt{3}}</math> | |||
|<math>1</math> | |||
|<math>\sqrt{3} </math> | |||
|Not Defined | |||
|- | |||
|<math>cosec \ A</math> | |||
|Not Defined | |||
|<math>2</math> | |||
|<math>\sqrt{2} </math> | |||
|<math>\frac {2}{\sqrt{3}}</math> | |||
|<math>1</math> | |||
|- | |||
|<math>sec \ A</math> | |||
|<math>1</math> | |||
|<math>\frac {2}{\sqrt{3}}</math> | |||
|<math>\sqrt{2} </math> | |||
|<math>2</math> | |||
|Not Defined | |||
|- | |||
|<math>cot \ A</math> | |||
|Not Defined | |||
|<math>\sqrt{3} </math> | |||
|<math>1</math> | |||
|<math>\frac{1}{\sqrt{3}}</math> | |||
|<math>0</math> | |||
|} | |||
[[Category:गणित]] | [[Category:गणित]] | ||
[[Category:कक्षा-10]] | [[Category:कक्षा-10]] | ||
[[Category:त्रिकोणमिति के कुछ अनुप्रयोग]] | [[Category:त्रिकोणमिति के कुछ अनुप्रयोग]] | ||
Revision as of 12:26, 7 June 2024
In this section, we will find the values of the trigonometric ratios for angles of .
Trigonometric Ratios of 45°
In right angled at , If ,
Using Pythagoras Theorem
, ,
Trigonometric Ratios of 30° and 60°
Consider an equilateral . Each angle in an equilateral triangle is , therefore, .
Draw a perpendicular from to the side (see Fig. 2).
Now
Therefore, and (Corresponding Parts of Congruent Triangles)
is a right angled triangle , right angled at with and
Let , Hence
, ,
, ,
Similarly
, ,
, ,
Not Defined | |||||
Not Defined | |||||
Not Defined | |||||
Not Defined |