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ABCD adalah jajaran genjang, diketahui A(3,−3,2), B(5,−6,8), dan C(2,−1,0). Tentukan tan ∠ACD.

Pertanyaan

ABCD adalah jajaran genjang, diketahui straight A open parentheses 3 comma negative 3 comma 2 close parentheses comma space straight B open parentheses 5 comma negative 6 comma 8 close parentheses comma dan straight C open parentheses 2 comma negative 1 comma 0 close parentheses. Tentukan tan space angle ACD.

K. Putri

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Ganesha

Jawaban terverifikasi

Jawaban

 tan space angle ACD equals 1 over 20 atau tan space angle ACD equals negative 1 over 20.

Pembahasan

Pada jajaran genjang, pasangan sisi yang berhadapan selalu sejajar dan sama panjang. Berdasarkan soal maka dapat dikatakan bahwa  AB with rightwards arrow on top divided by divided by CD with rightwards arrow on top dan AC with rightwards arrow on top divided by divided by BD with rightwards arrow on top. Misalkan koordinat straight D adalah open parentheses straight a comma straight b comma straight c close parentheses. Jika AB with rightwards arrow on top divided by divided by CD with rightwards arrow on top, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell straight k CD with rightwards arrow on top end cell row cell straight B minus straight A end cell equals cell straight k open parentheses straight D minus straight C close parentheses end cell row cell open parentheses table row 5 row cell negative 6 end cell row 8 end table close parentheses minus open parentheses table row 3 row cell negative 3 end cell row 2 end table close parentheses end cell equals cell straight k open parentheses table row straight a row straight b row straight c end table minus open parentheses table row 2 row cell negative 1 end cell row 0 end table close parentheses close parentheses end cell row cell open parentheses table row 2 row cell negative 3 end cell row 6 end table close parentheses end cell equals cell open parentheses table row ka row kb row kc end table close parentheses minus open parentheses table row cell 2 straight k end cell row cell negative straight k end cell row 0 end table close parentheses end cell row cell open parentheses table row ka row kb row kc end table close parentheses end cell equals cell open parentheses table row cell 2 plus 2 straight k end cell row cell negative 3 minus straight k end cell row 6 end table close parentheses end cell end table  

table attributes columnalign right center left columnspacing 0px end attributes row blank rightwards double arrow cell ka equals 2 plus 2 straight k............. left parenthesis straight i right parenthesis end cell row blank rightwards double arrow cell kb equals negative 3 minus straight k........... left parenthesis ii right parenthesis end cell row blank rightwards double arrow cell kc equals 6................... left parenthesis iii right parenthesis end cell end table    

Jika AC with rightwards arrow on top divided by divided by BD with rightwards arrow on top, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell AC with rightwards arrow on top end cell equals cell straight k BD with rightwards arrow on top end cell row cell straight C minus straight A end cell equals cell straight k open parentheses straight D minus straight B close parentheses end cell row cell open parentheses table row 2 row cell negative 1 end cell row 0 end table close parentheses minus open parentheses table row 3 row cell negative 3 end cell row 2 end table close parentheses end cell equals cell straight k open parentheses table row straight a row straight b row straight c end table minus open parentheses table row 5 row cell negative 6 end cell row 8 end table close parentheses close parentheses end cell row cell open parentheses table row cell negative 1 end cell row 2 row cell negative 2 end cell end table close parentheses end cell equals cell open parentheses table row ka row kb row kc end table close parentheses minus open parentheses table row cell 5 straight k end cell row cell negative 6 straight k end cell row cell 8 straight k end cell end table close parentheses end cell row cell open parentheses table row ka row kb row kc end table close parentheses end cell equals cell open parentheses table row cell negative 1 plus 5 straight k end cell row cell 2 minus 6 straight k end cell row cell negative 2 plus 8 straight k end cell end table close parentheses end cell end table   

rightwards double arrow ka equals negative 1 plus 5 straight k........... left parenthesis iv right parenthesis rightwards double arrow kb equals 2 minus 6 straight k............... left parenthesis straight v right parenthesis rightwards double arrow kc equals negative 2 plus 8 straight k........... left parenthesis vi right parenthesis     

Tentukan nilai straight k dengan menggunakan salah satu pasangan kesamaan yaitu open parentheses iii close parentheses dan open parentheses vi close parentheses sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row kc equals kc row 6 equals cell negative 2 plus 8 straight k end cell row 8 equals cell 8 straight k end cell row 1 equals straight k end table  

Substitusi nilai straight k masing-masing ke salah satu persamaan untuk menentukan koordinat straight D sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell table row cell open parentheses straight i close parentheses rightwards double arrow end cell cell ka equals 2 plus 2 straight k end cell row blank cell open parentheses 1 close parentheses straight a equals 2 plus 2 open parentheses 1 close parentheses end cell row blank cell straight a equals 2 plus 2 end cell row blank cell straight a equals 4 end cell row cell open parentheses straight v close parentheses rightwards double arrow end cell cell kb equals 2 minus 6 straight k end cell row blank cell open parentheses 1 close parentheses straight b equals 2 minus 6 open parentheses 1 close parentheses end cell row blank cell straight b equals 2 minus 6 end cell row blank cell straight b equals negative 4 end cell row cell open parentheses iii close parentheses rightwards double arrow end cell cell kc equals 6 end cell row blank cell open parentheses 1 close parentheses straight c equals 6 end cell row blank cell straight c equals 6 end cell end table end cell end table   

Sehingga diperoleh koordinat straight D adalah open parentheses 4 comma negative 4 comma 6 close parentheses. Kemudian, pada angle ACD artinya besar sudut open parentheses theta close parentheses terletak di titik straight C yang merupakan perpotongan AC with rightwards arrow on top dan DC with rightwards arrow on top. Vektor AC with rightwards arrow on top dan DC with rightwards arrow on top berturut-turut dapat ditentukan sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell AC with rightwards arrow on top end cell equals cell straight C minus straight A end cell row cell AC with rightwards arrow on top end cell equals cell open parentheses table row 2 row cell negative 1 end cell row 0 end table close parentheses minus open parentheses table row 3 row cell negative 3 end cell row 2 end table close parentheses end cell row cell AC with rightwards arrow on top end cell equals cell open parentheses table row cell negative 1 end cell row 2 row cell negative 2 end cell end table close parentheses end cell row blank blank blank row cell DC with rightwards arrow on top end cell equals cell straight C minus straight D end cell row cell DC with rightwards arrow on top end cell equals cell open parentheses table row 2 row cell negative 1 end cell row 0 end table close parentheses minus open parentheses table row 4 row cell negative 4 end cell row 6 end table close parentheses end cell row cell DC with rightwards arrow on top end cell equals cell open parentheses table row cell negative 2 end cell row 3 row cell negative 6 end cell end table close parentheses end cell end table    

Setelah mengetahui vektor AC with rightwards arrow on top dan DC with rightwards arrow on top maka dapat ditentukan panjang vektor AC with rightwards arrow on top dan DC with rightwards arrow on top berturut-turut sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar AC with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 1 close parentheses squared plus open parentheses 2 close parentheses squared plus open parentheses negative 2 close parentheses squared end root end cell row cell open vertical bar AC with rightwards arrow on top close vertical bar end cell equals cell square root of 1 plus 4 plus 4 end root end cell row cell open vertical bar AC with rightwards arrow on top close vertical bar end cell equals cell square root of 9 end cell row cell open vertical bar AC with rightwards arrow on top close vertical bar end cell equals cell plus-or-minus 3 end cell row blank blank blank row cell open vertical bar DC with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 2 close parentheses squared plus open parentheses 3 close parentheses squared plus open parentheses negative 6 close parentheses squared end root end cell row cell open vertical bar DC with rightwards arrow on top close vertical bar end cell equals cell square root of 4 plus 9 plus 36 end root end cell row cell open vertical bar DC with rightwards arrow on top close vertical bar end cell equals cell square root of 49 end cell row cell open vertical bar DC with rightwards arrow on top close vertical bar end cell equals cell plus-or-minus 7 end cell end table 

panjang selalu bernilai positif maka open vertical bar AC with rightwards arrow on top close vertical bar equals 3 dan open vertical bar DC with rightwards arrow on top close vertical bar equals 7.

Sehingga, besar sudut open parentheses theta close parentheses antara vektor AC with rightwards arrow on top dan DC with rightwards arrow on top dapat ditentukan dengan rumus sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator stack A C with rightwards arrow on top times stack D C with rightwards arrow on top over denominator open vertical bar stack A C with rightwards arrow on top close vertical bar times open vertical bar stack D C with rightwards arrow on top close vertical bar end fraction end cell row cell cos space theta end cell equals cell fraction numerator open parentheses negative 1 comma 2 comma negative 2 close parentheses times open parentheses negative 2 comma 3 comma negative 6 close parentheses over denominator 3 times 7 end fraction end cell row cell cos space theta end cell equals cell fraction numerator open parentheses negative 1 close parentheses times open parentheses negative 2 close parentheses plus open parentheses 2 close parentheses times open parentheses 3 close parentheses plus open parentheses negative 2 close parentheses times open parentheses negative 6 close parentheses over denominator 21 end fraction end cell row cell cos space theta end cell equals cell fraction numerator 2 plus 6 plus 12 over denominator 21 end fraction end cell row cell cos space theta end cell equals cell 20 over 21 end cell end table 

Sesuai dengan rumus identitas trigonometri, maka:

 table attributes columnalign right center left columnspacing 0px end attributes row cell sin squared theta plus cos squared theta end cell equals 1 row cell sin squared theta end cell equals cell 1 minus cos squared theta end cell row cell sin space theta end cell equals cell square root of 1 minus cos squared theta end root end cell row cell sin space theta end cell equals cell square root of 1 minus open parentheses 20 over 21 close parentheses squared end root end cell row cell sin space theta end cell equals cell square root of 1 minus 400 over 441 end root end cell row cell sin space theta end cell equals cell square root of 1 over 441 end root end cell row cell sin space theta end cell equals cell plus-or-minus 1 over 21 end cell end table 

Diperoleh nilai sin space theta equals 1 over 21 atau sin space theta equals negative 1 over 21. Sesuai dengan rumus dasar trigonometri maka dapat ditentukan tan space theta.

Jika sin space theta equals 1 over 21 maka tan space theta sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space theta end cell equals cell fraction numerator sin space theta over denominator cos space theta end fraction end cell row cell tan space theta end cell equals cell fraction numerator open parentheses begin display style 1 over 21 end style close parentheses over denominator open parentheses begin display style 20 over 21 end style close parentheses end fraction end cell row cell tan space theta end cell equals cell 1 over 20 end cell end table 

Jika sin space theta equals negative 1 over 21 maka tan space theta sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space theta end cell equals cell fraction numerator sin space theta over denominator cos space theta end fraction end cell row cell tan space theta end cell equals cell fraction numerator open parentheses begin display style negative 1 over 21 end style close parentheses over denominator open parentheses begin display style 20 over 21 end style close parentheses end fraction end cell row cell tan space theta end cell equals cell negative 1 over 20 end cell end table 

Jadi, tan space angle ACD equals 1 over 20 atau tan space angle ACD equals negative 1 over 20.

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