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Pertanyaan

Pada gambar berikut, diketahui AB equals 10 space cm comma space BC equals 5 space cm comma space angle BAP equals straight x degree, dan BC tegak lurus AB.

Proyeksi titik straight B dan straight C ke garis AP berturut-turut adalah titik straight D dan straight E.

a. Tunjukkanlah

    table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses straight i close parentheses space EC end cell equals cell 10 space sin space x degree plus 5 space cos space x degree end cell row cell open parentheses ii close parentheses space AE end cell equals cell 10 space cos space x degree minus 5 space sin space x degree end cell end table

b. Nyatakanlah :

    table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses straight i close parentheses space EC space dalam space bentuk space R space sin space open parentheses x plus alpha close parentheses degree end cell row blank blank cell open parentheses ii close parentheses space AE space dalam space bentuk space R space sin space open parentheses x minus alpha close parentheses degree end cell end table

c. Hitunglah nilai maksimum dan minimum dari :

    table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses straight i close parentheses space EC end cell row blank blank cell open parentheses ii close parentheses AE end cell end table 

A. Hadiannur

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Pembahasan

Ingat konsep :

  • sin space open parentheses A plus B close parentheses equals sin space A space cos space B plus cos space A space sin space B 
  • sin space open parentheses A minus B close parentheses equals sin space A space cos space B minus cos space A space sin space B
  • table attributes columnalign right center left columnspacing 0px end attributes row cell R space sin space open parentheses x plus-or-minus alpha close parentheses end cell equals cell a space sin space x plus-or-minus b space cos space x comma space dengan end cell row R equals cell square root of a squared plus b squared end root space dan space alpha equals tan to the power of negative 1 end exponent open parentheses b over a close parentheses end cell end table
  • sin squared x plus cos squared x equals 1 
  • table attributes columnalign right center left columnspacing 0px end attributes row y equals cell a space cos space x plus b space sin space x comma space space 0 degree less or equal than x less than 360 degree end cell row cell y subscript m i n end subscript end cell equals cell R equals square root of a squared plus b squared end root comma space pada space x equals 180 degree plus alpha end cell row cell y subscript m a k s end subscript end cell equals cell R equals square root of a squared plus b squared end root comma space pada space x equals alpha end cell row blank blank blank end table
  • Perhatikan segitiga berikut :

sin space theta equals B over Z cos space theta equals A over Z tan space theta equals B over A

Dari soal diketahui :

angle E G A equals 90 degree minus x angle B G C equals 90 degree minus x angle B C F equals 180 degree minus open parentheses 90 degree plus 90 degree minus x close parentheses equals x

a. Berdasarkan konsep di atas maka diperoleh penyelesaian :

table attributes columnalign right center left columnspacing 0px end attributes row cell E F end cell equals cell B D end cell row cell fraction numerator B D over denominator A B end fraction end cell equals cell sin space x degree space rightwards double arrow space fraction numerator B D over denominator 10 end fraction equals sin space x degree space rightwards double arrow space B D equals 10 space sin space x degree end cell row cell fraction numerator C F over denominator B C end fraction end cell equals cell sin space x degree space rightwards double arrow space fraction numerator C F over denominator 5 end fraction equals sin space x degree space rightwards double arrow C F equals 5 space sin space x degree end cell row cell E C end cell equals cell E F plus C F equals B D plus C F equals 10 space sin space x degree plus 5 space sin space x degree end cell row cell fraction numerator A D over denominator A B end fraction end cell equals cell cos space x degree space rightwards double arrow space A D equals A B space cos space x degree space equals 10 space cos space x degree end cell row cell fraction numerator B F over denominator B C end fraction end cell equals cell sin space x degree space rightwards double arrow space B F equals B C space sin space x degree space equals 5 space sin space x degree end cell row cell A E end cell equals cell A D minus D E equals 10 space cos space x degree minus 5 space sin space x degree end cell row blank blank blank end table

Dengan demikian, terbukti  EC equals 10 space sin space x degree plus 5 space cos space x degree dan AE equals 10 space cos space x degree minus 5 space sin space x degree.

b. Berdasarkan konsep penjumlahan sinus dua sudut maka diperoleh :

 table attributes columnalign right center left columnspacing 0px end attributes row cell sin space open parentheses A plus B close parentheses end cell equals cell sin space A space cos space B plus cos space A space sin space B end cell row cell E C end cell equals cell 10 space sin space x degree plus 5 space cos space x degree equals R space sin space open parentheses x plus alpha close parentheses degree end cell row blank equals cell R space sin space x degree space cos space alpha plus R space cos space x degree space sin space alpha end cell row cell R space sin space x degree space cos space alpha end cell equals cell 10 space sin space x degree space rightwards double arrow space R space cos space alpha equals 10 space rightwards double arrow cos space alpha equals 10 over R end cell row cell R space cos space x degree space sin space alpha end cell equals cell 5 space cos space x degree space rightwards double arrow R space sin space alpha equals 5 space rightwards double arrow space sin space alpha equals 5 over R end cell row cell R squared space sin squared alpha plus R squared space cos squared alpha end cell equals cell 5 squared plus 10 squared equals 25 plus 100 equals 125 end cell row cell R squared open parentheses sin squared alpha plus cos squared alpha close parentheses end cell equals 125 row cell R squared end cell equals cell 125 space rightwards double arrow R equals square root of 125 equals 5 square root of 5 end cell end table

Karena sin space alpha greater than 0 space dan space cos space alpha greater than 0 maka alpha di kuadran pertama. Diperoleh :

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space alpha end cell equals cell fraction numerator sin space alpha over denominator cos space alpha end fraction equals fraction numerator begin display style 5 over 2 end style R over denominator begin display style 10 over R end style end fraction equals 1 half end cell row alpha equals cell tan to the power of negative 1 end exponent space open parentheses 1 half close parentheses equals 26 comma 56 degree end cell end table

Dengan demikian, diperoleh persamaan yaitu E C equals 5 square root of 5 space sin space open parentheses x plus 26 comma 56 close parentheses to the power of degree.

Berdasarkan konsep selisih sinus dua sudut maka diperoleh :

 table attributes columnalign right center left columnspacing 0px end attributes row cell sin space open parentheses A minus B close parentheses end cell equals cell sin space A space cos space B minus cos space A space sin space B end cell row cell A E end cell equals cell 10 space cos space x degree minus 5 space sin space x degree equals R space sin space open parentheses x minus alpha close parentheses degree end cell row blank equals cell R space sin space x degree space cos space alpha minus R space cos space x degree space sin space alpha end cell row cell R space sin space x degree space cos space alpha end cell equals cell negative 5 space sin space x degree space rightwards double arrow space R space space cos space alpha equals negative 5 space space rightwards double arrow cos space alpha equals negative 5 over R end cell row cell negative R space cos space x degree space sin space alpha end cell equals cell 10 space cos space x degree space rightwards double arrow R space sin space alpha equals negative 10 space rightwards double arrow space sin space alpha equals negative 10 over R end cell row cell R squared space sin squared alpha plus R squared space cos squared alpha end cell equals cell open parentheses negative 10 close parentheses squared plus open parentheses negative 5 close parentheses squared equals 100 plus 25 equals 125 end cell row cell R squared open parentheses sin squared alpha plus cos squared alpha close parentheses end cell equals 125 row cell R squared end cell equals cell 125 space rightwards double arrow R equals square root of 125 equals 5 square root of 5 end cell end table

Karena sin space alpha less than 0 space dan space cos space alpha less than 0 maka alpha di kuadran ketiga. Perhatikan gambar berikut :

dimana alpha equals straight pi plus straight theta

Karena alpha equals straight pi plus straight theta maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space alpha end cell equals cell fraction numerator sin space alpha over denominator cos space alpha end fraction equals fraction numerator begin display style bevelled fraction numerator begin display style negative 10 end style over denominator R end fraction end style over denominator begin display style bevelled fraction numerator begin display style negative 5 end style over denominator R end fraction end style end fraction equals 2 end cell row cell tan space theta end cell equals 2 row theta equals cell tan to the power of negative 1 end exponent space 2 equals 63 comma 43 degree end cell row alpha equals cell straight pi plus straight theta equals 243 comma 43 degree end cell end table

Dengan demikian, diperoleh persamaan yaitu A E equals 5 square root of 5 space sin space open parentheses x minus 243 comma 43 close parentheses to the power of degree.

c. Mencari nilai maksimum dan minimum

table attributes columnalign right center left columnspacing 0px end attributes row EC equals cell 10 space sin space x degree plus 5 space cos space x degree end cell row R equals cell square root of a squared plus b squared end root equals square root of 5 squared plus 10 squared end root equals square root of 25 plus 100 end root equals square root of 125 equals 5 square root of 5 end cell end table

Dengan demikian, E C subscript m i n end subscript equals negative 5 square root of 5 space dan space E C subscript m a k s end subscript equals 5 square root of 5

AE equals 10 space cos space x degree minus 5 space sin space x degree R equals square root of a squared plus b squared end root equals square root of 10 squared plus open parentheses negative 5 close parentheses squared end root equals square root of 100 plus 25 end root equals square root of 125 equals 5 square root of 5 

Dengan demikian, A E subscript m i n end subscript equals negative 5 square root of 5 space dan space A E subscript m a k s end subscript equals 5 square root of 5.

 

 

 

 

 

 

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