a+b=10 sudut A=60 ,sudut B=45.Berapakah sisi-sisi segitiga tersebut ?

a+b = 10
b = 10-a
C = 180-105 = 75 djt
Aturan sin:
[tex]\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B} \\ \\ \frac{a}{\sin 60}=\frac{10-a}{\sin 45} \\ \frac{a}{\frac{1}{2}\sqrt{3}}=\frac{10-a}{\frac{1}{2}\sqrt{2}} \\ \\ \frac{a}{\sqrt{3}}=\frac{10-a}{\sqrt{2}} \\ \\ \sqrt{2}a=\sqrt{3}(10-a) \\ \sqrt{2}a=10\sqrt{3}-\sqrt{3}a \\ a(\sqrt{3}+\sqrt{2})=10\sqrt{3} \\ a=\frac{10\sqrt{3}}{\sqrt{3}+\sqrt{2}}=\frac{10\sqrt{3}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ \\ a=\frac{30-10\sqrt{6}}{3-2} \\ \\ a=30-10\sqrt{6}[/tex]
[tex]b=10-a \\ b=10-(30-10\sqrt{6}) \\ b=10\sqrt{6}-20 \\ \\ \displaystyle \frac{b}{\sin B}=\frac{c}{\sin C} \\ \\ \frac{10\sqrt{6}-20}{\frac{1}{2}\sqrt{2}}=\frac{c}{\frac{1}{4}(\sqrt{6}+\sqrt{2})} \\ 10\sqrt{6}-20=\frac{c}{\frac{1}{2}(\sqrt{3}+1)} \\ c=\frac{1}{2}(\sqrt{3}+1)(10\sqrt{6}-20) \\ c=(\sqrt{3}+1)(5\sqrt{6}-10) \\ c=5\sqrt{18}-10\sqrt{3}+5\sqrt{6}-10 \\ c=5\sqrt{6}-10\sqrt{3}+15\sqrt{2}-10[/tex]