jika alfa + beta = phi/ 3, alfa, beta sudut sudut lancip Dan tan alfa = 1/6 tan beta, maka sin alfa + sin beta =

Mapel : Matematika
Kelas : X SMA
Bab : Trigonometri

Pembahasan :
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Sin A + Sin B
= (√3)/(2√21) + (2√3)/(√21)
= (√3 + 4√3)/(2√21)
= (5√3)/(2√21)
= (15√7)/(42)
= 15/24 √7
= 5/8 √7

Bab Trigonometri
Matematika SMA Kelas X

alfa = a
beta = b
π/3 = 60°
a + b = 60°
b = 60° - a

tan a = (1/6) tan b
tan b = 6 . tan a

tan (60° - a) = tan b
(tan 60° - tan a) / (1 + tan 60° . tan a) = 6 . tan a
√3 - tan a = 6 tan a (1 + √3 . tan a)
6√3 tan² a + 6 tan a + tan a - √3 = 0
6√3 tan² a + 7 tan a - √3 = 0
6√3 p² + 7p - √3 = 0
a = 6√3; b = 7; c = -√3

p12 = (-7 +- √(7² - 4 . 6√3 . (-√3))/(2 . 6√3)
p12 = (-7 +- √(49 + 72)) / (12 √3)
p12 = (-7 +- √121) / (12√3)
p12 = (-7 +- 11) / (12√3)

p1 = (-7 + 11)/(12√3)
p1 = 4/12√3
p1 = 1/(3√3)
p1 = (1/9) √3

p2 = (-7 - 11)/12√3
p2 = -18/12√3
p2 = -3/2√3
p2 = (-1/2) √3

karena sudut lancip, maka yang memenuhi yang bernilai positif
tan a = (1/9) √3
tan a = (√3)/9 → y/x

r = √(x² + y²)
r = √(9² + √3²)
r = √84
r = 2√21

sin a = y/r
sin a = (√3) / (2.√21)
sin a = (√3) / (2 . √3 . √7)
sin a = 1/(2√7)
sin a = (1/14) √7

tan b = 6 . tan a
tan b = 6 . (1/9) √3
tan b = (2√3)/3 → y/x

r = √(x² + y²)
r = √(3²  + (2√3)²)
r = √21

sin b = y/r
sin b = 2√3/√21
sin b = 2√3 / (√7 . √3)
sin b = 2/√7
sin b = (2 √7) / 7

sin a + sin b = (1/14) √7 + (2/7) √7
sin a + sin b = (1/14) √7 + (4/14) √7
sin a + sin b = (5/14) √7