Jika sin 2a = 1/3 √5 untuk 0

Kelas 10 Matematika
Bab 6 - Trigonometri Dasar

sin 2a = (1/3) √5 → y/r

x = √(r² - y²)
x = √(3² - √5²)
x = √(9 - 5)
x = √4
x = 2

cos 2a = x/r
cos 2a = 2/3

tan 2a = sin 2a/cos 2a
tan 2a = (1/3 √5)/(2/3)
tan 2a = (1/2) √5

tan 2a = (1/2) √5
2 tan a / (1 - tan² a) = 1/2 √5
2 . (2/5) √5 . tan a = 1 - tan² a
tan² a + (4/5) √5 . tan a - 1 = 0

✩✩ dikali 5 ✩✩

5 tan² a + 4√5 . tan a - 5 = 0

tan a = p

5p² + 4p√5 - 5 = 0
a = 5; b = 4√5; c = -5

p12 = (-b +- √(b² - 4ac))/(2a)
p12 = (-4√5 +- √(4√5)² - 4 . 5 . (-5)))/(2 . 5)
p12 = (-4√5 +- √(80 + 100 ))/10
p12 = (-4√5 +- √180)/10
p12 = (-4√5 +- 6 √5)/10

p1 = (6√5 - 4√5)/10
p1 = (2√5)/10
tan a = (1/5) . √5

p2 = (-6√5 - 4√5)/10
tan a = (-10√5)/10
tan a = -√5

0 < a < 90° → kuadran I (positif)

tan a = (1/5) . √5