• Luas ΔABC = (50/3)√3 cm².
• Keliling ΔABC = [ 10 + 10√3 ] cm.

Penjelasan dengan langkah-langkah

Diketahui

• ΔABC siku-siku di B.
• BC = 10 cm.
• Besar ∠A = 60°.
• Besar ∠C = 30°.

Ditanyakan

• Luas dan keliling ΔABC

Penyelesaian

Luas ΔABC

L = ½·a·t
⇔ L = ½·AB·BC

• AB/BC = tan ∠C
  Maka: AB = BC tan ∠C.

⇔ L = ½·(BC tan ∠C)·BC
⇔ L = ½·BC²·tan ∠C
⇔ L = ½·10²·tan 30°
⇔ L = ½·100·(1/3)√3
⇔ L = (50/3)√3 cm²

Keliling ΔABC

K = AB + BC + AC

• AB = BC tan ∠C
• Karena BC/AC = sin ∠A, maka AC = BC / sin ∠A.

⇔ K = (BC tan ∠C) + BC + (BC / sin ∠A)
⇔ K = BC·[tan ∠C + 1 + (1/sin ∠A)]
⇔ K = 10·[tan 30° + 1 + (1/sin 60°)]
⇔ K = 10·[(1/3)√3 + 1 + (1 / ((½)√3))]
⇔ K = 10·[(1/3)√3 + 1 + (2/√3)]
⇔ K = 10·[(1/3)√3 + 1 + (2/3)√3]
⇔ K = 10·[1 + (1/3 + 2/3)√3]
⇔ K = 10·[1 + √3]
⇔ K = [ 10 + 10√3 ] cm