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If sin A =9/41,comput3 cos A and tan A |
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Answer» SinA=1/cos =41/9 Given,\xa0{tex}\\angle{/tex}B =\xa090°.Let BC = 9 and AC = 41sin A{tex}= \\frac { \\text { Perpendicular } } { \\text { Hypotenuse } } = \\frac { \\mathrm { BC } } { \\mathrm { AC } } = \\frac { 9 } { 41 }{/tex}By Pythagoras\' theorem, we haveAC2\xa0= AB2 + BC2{tex}\\Rightarrow{/tex}AB2= AC2\xa0- BC2= 412\xa0- 92\xa0= 1681 - 81\xa0= 1600{tex}\\Rightarrow AB = 40{/tex}{tex}\\cos A = \\frac { \\text { Base } } { \\text { Hypotenuse } } = \\frac { A B } { A C } = \\frac { 40 } { 41 }{/tex}{tex}\\tan A = \\frac { \\text { Perpendicular } } { \\text { Base } } = \\frac { B C } { A B } = \\frac { 9 } { 40 }{/tex} |
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