Diketahui segitiga lancip ABC dengan sin C= 2/√13. jika tan A tan B = 13, nilai tan A+tan B=
maspur
Dalam sebuah segitiga jumlah ketiga sudutnya 180º. A + B + C = 180º C = 180 - (A + B) A + B = 180 - C
sin (A + B) = sin (180 - C) = sin C <-- nilai sinus di kuadran ke II positif
sin (A + B) = 2/√13
tan (A + B) = 2/3
....................tan A + tan B tan (A + B) = --------------------------.............k... silang ....................1 - (tan A) (tan B)
tan A + tan B = tan (A + B) [1 - tan A tan B] ....................= 2/3.............[1 - 13] ....................= 2/3.............[- 12] ....................= - 8..<-- jawaban
A + B + C = 180º
C = 180 - (A + B)
A + B = 180 - C
sin (A + B) = sin (180 - C) = sin C <-- nilai sinus di kuadran ke II positif
sin (A + B) = 2/√13
tan (A + B) = 2/3
....................tan A + tan B
tan (A + B) = --------------------------.............k... silang
....................1 - (tan A) (tan B)
tan A + tan B = tan (A + B) [1 - tan A tan B]
....................= 2/3.............[1 - 13]
....................= 2/3.............[- 12]
....................= - 8..<-- jawaban
Ok ?