Posisi 2 kapal laut dinyatakan dengan koordinat kutub yaitu

Jarak kedua kapal adalah \(2\sqrt{21}\).

Pembahasan

Untuk menghitung jarak antara dua kapal yang dinyatakan dalam koordinat kutub, kita dapat menggunakan rumus jarak dalam koordinat Kartesius setelah mengkonversi koordinat kutub ke koordinat Kartesius.

Koordinat kutub \((r, \theta)\) dapat dikonversi ke koordinat Kartesius \((x, y)\) menggunakan rumus:
* x = r \cos \theta
* y = r \sin \theta

**Kapal Pertama:**
Koordinat kutub: \((r_1, \theta_1) = (10, 15^{\circ})\)
Konversi ke koordinat Kartesius:
* \(x_1 = 10 \cos 15^{\circ}\)
* \(y_1 = 10 \sin 15^{\circ}\)

Kita tahu bahwa \(\cos 15^{\circ} = \cos(45^{\circ} - 30^{\circ}) = \cos 45^{\circ}\cos 30^{\circ} + \sin 45^{\circ}\sin 30^{\circ}
= (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2}) = \frac{\sqrt{6} + \sqrt{2}}{4}\)

Dan \(\sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) = \sin 45^{\circ}\cos 30^{\circ} - \cos 45^{\circ}\sin 30^{\circ}
= (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{2}}{2})(\frac{1}{2}) = \frac{\sqrt{6} - \sqrt{2}}{4}\)

Maka:
* \(x_1 = 10 \times \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{5}{2}(\sqrt{6} + \sqrt{2})\)
* \(y_1 = 10 \times \frac{\sqrt{6} - \sqrt{2}}{4} = \frac{5}{2}(\sqrt{6} - \sqrt{2})\)

**Kapal Kedua:**
Koordinat kutub: \((r_2, \theta_2) = (8, 75^{\circ})\)
Konversi ke koordinat Kartesius:
* \(x_2 = 8 \cos 75^{\circ}\)
* \(y_2 = 8 \sin 75^{\circ}\)

Kita tahu bahwa \(\cos 75^{\circ} = \cos(45^{\circ} + 30^{\circ}) = \cos 45^{\circ}\cos 30^{\circ} - \sin 45^{\circ}\sin 30^{\circ}
= (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{2}}{2})(\frac{1}{2}) = \frac{\sqrt{6} - \sqrt{2}}{4}\)

Dan \(\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \sin 45^{\circ}\cos 30^{\circ} + \cos 45^{\circ}\sin 30^{\circ}
= (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2}) = \frac{\sqrt{6} + \sqrt{2}}{4}\)

Maka:
* \(x_2 = 8 \times \frac{\sqrt{6} - \sqrt{2}}{4} = 2(\sqrt{6} - \sqrt{2})\)
* \(y_2 = 8 \times \frac{\sqrt{6} + \sqrt{2}}{4} = 2(\sqrt{6} + \sqrt{2})\)

**Menghitung Jarak:**
Jarak antara dua titik \((x_1, y_1)\) dan \((x_2, y_2)\) adalah:
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

\(x_2 - x_1 = 2(\sqrt{6} - \sqrt{2}) - \frac{5}{2}(\sqrt{6} + \sqrt{2})
= \frac{4(\sqrt{6} - \sqrt{2}) - 5(\sqrt{6} + \sqrt{2})}{2}
= \frac{4\sqrt{6} - 4\sqrt{2} - 5\sqrt{6} - 5\sqrt{2}}{2}
= \frac{- \sqrt{6} - 9\sqrt{2}}{2}\)

\(y_2 - y_1 = 2(\sqrt{6} + \sqrt{2}) - \frac{5}{2}(\sqrt{6} - \sqrt{2})
= \frac{4(\sqrt{6} + \sqrt{2}) - 5(\sqrt{6} - \sqrt{2})}{2}
= \frac{4\sqrt{6} + 4\sqrt{2} - 5\sqrt{6} + 5\sqrt{2}}{2}
= \frac{- \sqrt{6} + 9\sqrt{2}}{2}\)

\((x_2 - x_1)^2 = (\frac{- \sqrt{6} - 9\sqrt{2}}{2})^2 = \frac{(\sqrt{6} + 9\sqrt{2})^2}{4}
= \frac{6 + 18\sqrt{12} + 81 \times 2}{4} = \frac{6 + 18(2\sqrt{3}) + 162}{4} = \frac{168 + 36\sqrt{3}}{4} = 42 + 9\sqrt{3}\)

\((y_2 - y_1)^2 = (\frac{- \sqrt{6} + 9\sqrt{2}}{2})^2 = \frac{(\sqrt{6} - 9\sqrt{2})^2}{4}
= \frac{6 - 18\sqrt{12} + 81 \times 2}{4} = \frac{6 - 18(2\sqrt{3}) + 162}{4} = \frac{168 - 36\sqrt{3}}{4} = 42 - 9\sqrt{3}\)

\(d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 = (42 + 9\sqrt{3}) + (42 - 9\sqrt{3}) = 84\)

\(d = \sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21}\)

Jadi, jarak kedua kapal adalah \(2\sqrt{21}\).