1.diketahui segitiga ABC dengan panjang AC = 5 akar 2 cm, sudut A = 105 derajat, dan sudut B = 30 derajat , panjang AB adalah 2. hasil dari akar 18 - 3 akar 7

Mapel : Matematika
Kelas : X SMA
Materi : Trigonometri dan bentuk akar

Pembahasan :
1¦¦

Aturan Sinus
AB/Sin C = AC/Sin B
AB/Sin 45° = (5√2)/Sin 30°
AB/(1/2 √2) = (5√2)/(1/2)
AB = 5/(1/2)
AB = 10 cm

2¦¦

√18 - 3√75 + √27
= √[9.2] - 3√[25.3] + √[9.3]
= 3√2 - 15√3 + 3√3
= 3√2 - 12√3
= 3(√2 - 4√3)

[tex]\displaystyle 1\\A+B+C=180^\circ\\105^\circ+30^\circ+C=180^\circ\\C=45^\circ\\\\\frac{AB}{\sin C}=\frac{AC}{\sin B}\\\frac{AB}{\sin45^\circ}=\frac{5\sqrt2}{\sin30^\circ}\\\frac{AB}{\frac12\sqrt2}=\frac{5\sqrt2}{\frac12}\\AB=5\sqrt2\cdot\sqrt2\\\boxed{\boxed{AB=10\,\text{cm}}}\\\\2\\\sqrt{18}-3\sqrt{75}+\sqrt{27}=3\sqrt{2}-3\cdot5\sqrt{3}+3\sqrt{3}\\\sqrt{18}-3\sqrt{75}+\sqrt{27}=3\sqrt{2}-15\sqrt{3}+3\sqrt{3}\\\boxed{\boxed{\sqrt{18}-3\sqrt{75}+\sqrt{27}=3\sqrt{2}-12\sqrt{3}}}[/tex]