mohon bantuin secepatnya yaaa besok mau ujian

jarak P - AB = n
jarak P - CD = m
panjang AB = p

n = 3m
Luas CDXY = pm
Luas segitiga ABP = 1/2pn = 1/2p(3m)

pm : 1/2p(3m)
2pm : 3p²*m

16) (cos @)/a + (cos B)/b + (cos y)/c
= ((b² + c² - a²)/(2bc))/a + ((a² + c² - b²)/(2ac))/b + ((a² + b² - c²)/(2ab))/c
= (b² + c² - a²)/(2abc) + (a² + c² - b²)/(2abc) + (a² + b² - c²)/(2abc)
= (a² + b² + c²) / (2abc)

17) misal :
1) jarak P ke CD = x
P ke AB = 3 . (P ke CD) = 3x
2) AB = CD = y

a = luas CDXY = xy
b = luas ∆ABP = 1/2 . y . 3x = 3/2 xy

a : b = xy : 3/2 xy = 1 : 3/2 = 2/2 : 3/2 = 2 : 3