tentukan keliling segitiga ABC diatas! tolong beserta caranya yaa makasii:)

Mapel : Matematika
Materi : Segitiga - Sudut

Mencari AD

[tex]Sin\ (A) = \frac{AD}{AB} \\ \\Sin\ (30^o) = \frac{x}{8} \\ \\x=8\times Sin\ (30^o) \\ \\ x=8\times \frac{1}{2} \\ \\ x=4 [/tex]

Mencari BD

[tex]Cos\ (A) = \frac{BD}{AB} \\ \\Cos\ (30^o) = \frac{BD}{8} \\ \\BD=8\times Cos\ (30^o) \\ \\ BD=8\times \frac{1}{2} \sqrt{3} \\ \\ BD=4 \sqrt{3} [/tex]

Mencari AC

[tex] \frac{BD}{AD}= \frac{AB}{AC} \\ \\ \frac{4 \sqrt{3}}{4}= \frac{8}{AC} \\ \\ AC= \frac{8\times4}{4 \sqrt{3}} \\ \\ AC=8 \sqrt{3} [/tex]

Mencari BC

[tex]BC = \sqrt{AB^2+AC^2} \\ \\ BC= \sqrt{8^2+(8 \sqrt{3})}\\ \\ BC= \sqrt{64+(64\times3})}\\ \\ BC= \sqrt{64+192}\\ \\ BC= \sqrt{256}\\ \\ BC= 16 [/tex]

Keliling = AB + AC + BC
Keliling = 8 + 8√3 + 16
Keliling = 24 + 8√3

***Semoga Terbantu***