1.keliling segi -6 beraturan 12√3 Luas sego -6 tersebut adalah...cm3 a.12√2 b.12√3 c.24 d.18√2 e.18√3 2.

Trigonometri

1. K = 12√3 --> 6s = 12√3
s = 2√3
luas segi 6 = 6 x 1/2 x s² x sin 60
L = 3 (2√3)²(1/2 √3)
L = 3(12)(1/2) √3
L = 18√3 satuan

2.  pada ΔABC , <A terkecil
a, b, c = 2, 2√2, 2√3

cos A = {(2√2)² + (2√3)² -(2)²}/ ( 2 (2√2)(2√3))
cos A = (8+12 - 4)/ (8√6)
cos A = 16/(8√6)= 2/√6 = (√6)/(3) = x/r
y = √(3²-6) = √3
tan A = y/x = √3/√6 = ¹/₂  √2

3.
a = 2√5
b = 5√3
cari c = ....

tan C = (√6)/(3) = y/x
r² = x²+y² = 9+6 = 15
r = √15
cos C = x/r = (3)/(√15)

c² = a² +b² - 2ab cos C
c² = 20+ 75 - 2(2√5)(5√3)(3)/(√15)
c² = 95 -   60 √15/√15
c² = 95-60
c = √35