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Given seco=13/12, calculate all trigonometrc ratios. |
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Answer» Consider a triangle ABC in which {tex}\\angle A = \\theta {/tex}\xa0and {tex}\\angle B = {90^o}{/tex}Let AB = 12k and AC = 13kThen, using Pythagoras theorem,\xa0{tex}BC=\\sqrt { ( \\mathrm { AC } ) ^ { 2 } - ( \\mathrm { AB } ) ^ { 2 } } = \\sqrt { ( 13 k ) ^ { 2 } - ( 12 k ) ^ { 2 } }{/tex}{tex}= \\sqrt { 169 k ^ { 2 } - 144 k ^ { 2 } } = \\sqrt { 25 k ^ { 2 } } = 5 k{/tex}{tex}\\therefore {/tex}\xa0{tex}\\sin \\theta = \\frac { B C } { A C } = \\frac { 5 k } { 13 k } = \\frac { 5 } { 13 }{/tex}{tex}\\cos \\theta = \\frac { A B } { A C } = \\frac { 12 k } { 13 k } = \\frac { 12 } { 13 } \\tan \\theta = \\frac { B C } { A B } = \\frac { 5 k } { 12 k } = \\frac { 5 } { 12 }{/tex}{tex}\\cot \\theta = \\frac { A B } { B C } = \\frac { 12 k } { 5 k } = \\frac { 12 } { 5 } \\cos e c \\theta = \\frac { A C } { B C } = \\frac { 13 k } { 5 k } = \\frac { 13 } { 5 }{/tex} Sec theata = hypotenuse/ baseSo, base =12 Hypotenuse=13Prependicular =√( hypotenuse)2 - ( base)2 √(13)2-(12)2=√169-144=√25 =5 Sin theata =P/H =5/13Cos theata=B/H =12/13Tan theata=P/B =5/12Cot theata=B/P =12/5Cosec theata= H/P = 13 /5 SecA=H/B,Here secA=13/12So, H=13 and B=12 We know that, P2=H2-B2P2=164-144P2=25P=5Now, All trignometric ratiosSinA=P/H=5/13CosA=B/H=12/13TanA=P/B=5/12CosecA=H/P=13/5CotA=B/P=12/5 |
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