Roboguru

Pertanyaan

Diketahui segitiga ABC dengan koordinat titik straight A open parentheses 1 comma space 2 close parentheses comma space straight B open parentheses 0 comma 0 close parentheses, dan straight C open parentheses 4 comma space 2 close parentheses. Nilai tan space angle ABC adalah ....

  1. 3 over 5 

  2. 3 over 4 

  3. 4 over 3 

  4. 9 over 16 

  5. 16 over 9 

W. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sriwijaya

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

Pembahasan

Diketahui segitiga ABC dengan koordinat titik straight A open parentheses 1 comma space 2 close parentheses comma space straight B open parentheses 0 comma 0 close parentheses, dan straight C open parentheses 4 comma space 2 close parentheses. Dapat digambarkan sebagai berikut:

Dapat dilihat bahwa angle ABC equals angle straight B berada diantara sisi BC dan BA, maka:

BC with rightwards arrow on top equals straight c with rightwards arrow on top minus straight b with rightwards arrow on top equals open parentheses table row 4 row 2 end table close parentheses minus open parentheses table row 0 row 0 end table close parentheses equals open parentheses table row 4 row 2 end table close parentheses rightwards arrow open vertical bar BC with rightwards arrow on top close vertical bar equals square root of 4 squared plus 2 squared end root equals square root of 16 plus 4 end root equals square root of 20 BA with rightwards arrow on top equals straight a with rightwards arrow on top minus straight b with rightwards arrow on top equals open parentheses table row 1 row 2 end table close parentheses minus open parentheses table row 0 row 0 end table close parentheses equals open parentheses table row 1 row 2 end table close parentheses rightwards arrow open vertical bar BA with rightwards arrow on top close vertical bar equals square root of 1 squared plus 2 squared end root equals square root of 1 plus 4 end root equals square root of 5 BC with rightwards arrow on top times BA with rightwards arrow on top equals open parentheses table row 4 row 2 end table close parentheses times open parentheses table row 1 row 2 end table close parentheses equals 4 plus 4 equals 8 

Sehingga:

cos space straight B equals fraction numerator BC with rightwards arrow on top times BA with rightwards arrow on top over denominator open vertical bar BC with rightwards arrow on top close vertical bar times open vertical bar BA with rightwards arrow on top close vertical bar end fraction cos space straight B equals fraction numerator 8 over denominator square root of 20 times square root of 5 end fraction cos space straight B equals fraction numerator 8 over denominator square root of 100 end fraction cos space straight B equals 8 over 10 cos space straight B equals 4 over 5 rightwards arrow cos equals fraction numerator sisi space samping over denominator sisi space miring end fraction 

Kemudian:

sisi space depan equals square root of 5 squared minus 4 squared end root sisi space depan equals square root of 25 minus 16 end root sisi space depan equals square root of 9 sisi space depan equals 3 

Oleh karena itu:

tan space angle ABC equals tan space straight B tan space angle ABC equals fraction numerator sisi space depan over denominator sisi space samping end fraction tan space angle ABC equals 3 over 4 

Jadi, jawaban yang benar adalah B.

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