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SVETLANKA909090 [29]
3 years ago
11

How to do this question plz ​

Mathematics
1 answer:
hjlf3 years ago
7 0

Answer:

\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 }

Step-by-step explanation:

\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 }

♨Rage♨

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