Question

Panjang BC adalah ... (lihat lampiran)

Answer 1

Mencari panjang AC
[AC/sin30] = [DA/sin 45]
AC = 10 ÷ (akar2/2) × 1/2
AC = 10/akar2

Mencari panjang BC
BC^2 = AC^2 + AB^2 - 2.AC.AB. Cos 60
BC^2 = (10/akar2)^2 + (10.akar2)^2 - 2. (10/akar2).(10.akar2). (1/2)
BC^2 = 50 + 200 - 100
BC^2 = 150
BC = akar 150
BC = 5. Akar 6

Answer 2

   AC / sin D = AD / sin C

AC / sin 30° = 10 / sin 45°

   AC / (1/2)  = 10 / (1/2√2)
               AC = 10/√2

BC² = AC² + AB² - 2 . AC .AB .cos A

        = (10/√2)² + (10√2)² - 2 . (10/√2) (10√2) cos 60°
        
        = 100 / 4 + 200 - 200 . 1/2

        = 25 + 200 - 100
 
        = 125
 
  BC = √125
       
  BC = 5√5