Pada segitiga ABC yang siku-siku di C, diketahui sin A. Sin B =2/5 dan sin (A-B) =5a, maka nilai a adalah

A+B+C = 180 krn C = 90 mk A+B = 90
cos (A+B) = cos A.cos B - sinA.sin B
0 = cos A.cos B - 2/5
cos A.cos B = 2/5

cos (A-B) = cos A.cos B + sin A.sin B = 2/5 + 2/5 = 4/5
sin (A-B) = √[1 - cos²(A-B)]
5a = √[1 - (4/5)²]
5a = 3/5
a = 3/25

Diketahui sin B x sin A = AC x BC/AB² = 2/5
AC = 2 ; AB = 5
BC = √25-4 = √21
sin A = BC/AB
sin(A-B) = sin A x cos B - cos A x sin B = 5a
BC²/AB² - 16/100 =5a
21/5 -16/100 = 5a
(420-16)/100 = 5a
404=500a
a = 808/1000 = 0.808