(gk ush pake WA, jawab lgsg disini)
Jawab:
Penjelasan dengan langkah-langkah:
Penyelesaian Nomor 1.
[tex]\frac{30}{\sin 45} = \frac{x}{\sin 60}\\\sin30 \cdot 30 = x \cdot \sin 45\\\frac{1}{2} \times 30 = \frac{\sqrt2}{2}\times x\\15\sqrt{2} = x\\\text{Total sudut segitiga = 180, maka besar sudut B = 180 -105 = 75}\\\sin(30+45) = \sin30\cos45+\cos30\sin45\\\sin(30+45) = \frac{1}{2}\times \frac{\sqrt2}{2}+\frac{\sqrt3}{2}\times \frac{\sqrt2}{2}\\\sin(30+45) = \frac{1}{4}(\sqrt{6}+\sqrt{2})\\\\[/tex]
[tex]\frac{30}{\sin45}=\frac{y}{\sin 75}\\\sin75 \times 30 = \sin45 y\\\frac{1}{4}(\sqrt{6}+\sqrt{2})\times 30 = \frac{\sqrt2}{2}y\\y = 15\sqrt{3}+15\\\\\text{Keliling} = 30+x+y\\\text{Keliling} = 30+15\sqrt{2}+15\sqrt{3}+15\\\text{Keliling} = 45+15(\sqrt{2}+\sqrt{3})[/tex]
Penyelesaian Nomor 2.
[tex]\frac{4\sqrt{2}}{\sin A}=\frac{8}{\sin 45}\\\frac{4 \sqrt{2}}{\sin A}= \frac{8}{\frac{\sqrt{2}}{2}}\\\sin A = \frac{1}{2}\\A = \arcsin(0,5)\\A = 30, \\\\\tan \angle BAC = \tan A =\tan 30 = \frac{1}{3}\sqrt{3}\\[/tex]
Penyelesaian Nomor 3.
[tex](13)^2=(15)^2+(8)^2-2(15)(5)\cos A\\169=225+64-150 \cos A\\169 = 289- 150 \cos A\\120 = 150 \cos A\\\cos A = \frac{12}{15}\\\cos A = \frac{4}{5}\\\\\tan A =\frac{3}{4}[/tex]
Penyelesaian Nomor 4.
[tex]L_{\triangle} = \frac{1}{2}\times 8 \times 6 \times \sin 120\\L_{\triangle}=\frac{\sqrt3}{4} \times 48\\L_{\triangle} = 12\sqrt{3}[/tex]
Penyelesaian Nomor 5.
Pada KD.III, tan dan cot bernilai positif (+). Selainnya, akan bernilai negatif. Maka kita bisa memasukkan tan a kedalam segitiga siku siku dan mendapatkan perbandingan sebagai berikut :
Sisi miring segitiga = [tex]\sqrt{13}[/tex]
Sisi di depan sudut = 2
Sisi di samping sudut = 3
[tex]\frac{\cos \alpha + 6\sin \alpha}{3 \sin \alpha-\cos \alpha}\\= \frac{\frac{-3\sqrt{13}}{13}+6(-\frac{2\sqrt{13}}{13})}{3\frac{-2\sqrt{13}}{13}-(-\frac{3\sqrt{13}}{13}})\\\\=5[/tex]
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