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Perhatikan segitiga siku-siku dibawah ini! Tu...

Perhatikan segitiga siku-siku dibawah ini!



Tunjukkan bahwa left parenthesis scs space straight A right parenthesis squared minus left parenthesis cot space straight A right parenthesis squared equals 1

Jawaban:

Kita asumsikan bahwa pada gambar tersebut, segitiga tersebut mempunyai titik straight Astraight B, dan straight C, sehingga segitiga tersebut diberi nama segitiga ABC.

Kita asumsikan juga bahwa yang yang diminta di dalam soal adalah tunjukkan bahwa left parenthesis csc space straight A right parenthesis squared minus left parenthesis cot space straight A right parenthesis squared equals 1


table attributes columnalign right center left columnspacing 0px end attributes row cell csc space straight A end cell equals cell fraction numerator 1 over denominator sin space straight A end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style bevelled straight a over straight c end style end fraction end cell row cell csc space straight A end cell equals cell straight c over straight a end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row cell cot space straight A end cell equals cell fraction numerator 1 over denominator tan space straight A end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style bevelled straight a over straight b end style end fraction end cell row cell cot space straight A end cell equals cell straight b over straight a end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis csc space straight A right parenthesis squared minus left parenthesis cot space straight A right parenthesis squared end cell equals 1 row cell open parentheses straight c over straight a close parentheses squared minus open parentheses straight b over straight a close parentheses squared end cell equals 1 row cell fraction numerator straight c squared minus straight b squared over denominator straight a squared end fraction end cell equals 1 end table


Dalam segitiga siku-siku berlaku hubungan straight a squared plus straight b squared equals straight c squared sesuai Teorema Pythagoras sehingga straight c squared minus straight b squared equals straight a squared. Maka


table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator straight c squared minus straight b squared over denominator straight a squared end fraction end cell equals 1 row cell straight a squared over straight a squared end cell equals 1 row 1 equals cell 1 space space space space left parenthesis Terbukti right parenthesis end cell end table


Jadi terbukti bahwa left parenthesis csc space straight A right parenthesis squared minus left parenthesis cot space straight A right parenthesis squared equals 1.

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