Answer:
[tex]\underline{ \boxed{ \underline{part \: a}}}. \\the \: required \: distance \: a part \: be \to \: \boxed{\underline {\overline{p } =79.23 \: km} } \\ \\ \underline{ \boxed{ \underline{part \: b}}}. \\ the \: bearing \: of \: \boxed{ B } \: \: from \: \boxed{ \underline{A}} \: is \to \boxed{ \underline{324 }\degree }[/tex]
Step-by-step explanation:
[tex] \underline{ \boxed{ \underline{part \: a}}}. \\ l et \: the \: required \: distance \: apart \: be \to \: \boxed{ \overline {p}} \\ let \: \boxed{ \angle \: P} \: be \: the \: angle \: of \: seperation \: between \: both \: ships :given \: by \to \\ P = (244 - 180) - (196 - 180 )= (64 -16) = \boxed{48 \degree} \\ if \: ship \: A \: and \: B \: left \: the \: port\: at \: (10:30) \: \\ then \: at \: (14 :00) \: their \: time \: interval \: would \: be \to \\ (14 : 00) - (10 : 30) = \boxed{ 3.5 \: hrs} \\ ship \: \boxed{A }\: distance \: from \: the \: port =( v \times t) = (30 \times 3.5) = \boxed{105 \: km} \\ ship \: \boxed{B }\: distance \: from \: the \: port =( v \times t) = (24 \times 3.5) = \boxed{84\: km} \\n ow........we \: cant \: do \: much \: if \: their \: are \: no \: angles \: to \: work \: with \to \\ applying \:t he \: cosine \: rule : we \: have \to \\ \cos(P) = \frac{ {a}^{2} + {b}^{2} - \overline{{p}^{2} } }{2(ab)} \\ {a}^{2} + {b}^{2} - \overline{{p}^{2} } = 2(ab) \cos(P) \\ \overline{{p}^{2} } = {a}^{2} + {b}^{2} - \{2(ab) \cos(P) \} \\ \overline{{p}^{2} } = {84}^{2} + {105}^{2} - \{2(84)(105) \cos(48 \degree) \} \\\overline{{p}^{2} } = 18,081 - 11,803.463897 \\ \overline{{p}^{2} } = 6,277.536103 \\ \overline{p } = \sqrt{6,277.536103} \\ \boxed{\underline {\overline{p } =79.23 \: km} } \\ \\ \underline{ \boxed{ \underline{part \: b}}}. \\ to \: sove \: this : we \: first \: find \: angle \: \angle \: B : by \: applying \to \\ \frac{ \sin(B) }{b} = \frac{ \sin(P) }{p} \\ p \ast \sin(B) = b \ast \sin(P) \\ B = \sin {}^{ - 1} ( \frac{ b \ast \sin(P)}{p} ) \\ B = \sin {}^{ - 1} ( \frac{ 105 \times \sin(48 \degree)}{79.230903712} ) \\ \\ B = \sin {}^{ - 1} ( \frac{ 78.030206678}{79.230903712} ) \\ \\ B = \sin {}^{ - 1} (0.9848455971) \\ \boxed{B = 80 \degree} \\ hence \ : the \: bearing \: of \: \boxed{ B } \: \: from \: \boxed{ \underline{A}} \: is \to \\ 360 - \{180 - (80 + 48 + 16) \} \\ 360 -(180 - 144 )\\ 360 -36 = \boxed{324 \degree }[/tex]
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