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If sin a =3/5, find cos a |
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Answer» Sin a= 3/5=p/h and b=✓h×h-p×p =✓25-9=4 So, cos a= b/h =4/5 {tex}sin \\ a =\xa0{3\\over 5}{/tex}we know,\xa0{tex}sin ^2\\theta + cos^2\\theta = 1 {/tex}So,\xa0{tex}cos \\ a = \\sqrt {1-sin^2a}{/tex}{tex}cos \\ a = \\sqrt {1- ({3\\over 5})^2}{/tex}{tex}=> cos \\ a = \\sqrt {1- {9\\over 25}}{/tex}{tex}=> cos \\ a = \\sqrt {16\\over 25}{/tex}{tex}=> cos \\ a = {4\\over 5}{/tex} Sin a =3/5 P/H Cos a = B/H =H(SQUARE) =B(SQUARE) +P(SQUARE) =5(SQUARE)=B(SQUARE) +3(SQUARE) =25=B(SQUARE) +9=25-9= 16 B(SQUARE)B=4 COS a =b/h = 4/5. |
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