cshellsrock19
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Two ships leave whittier, Ak at 7 AM. The first ship sails towards Valdez on a 54° course at a constant rate of 36 mi/h. The second ship sails towards Knight island on a 144° course at a constant speed of 42 mi/h. Find the distance between the ships at 11 AM.



Answer :

[tex]11AM-7AM=4h\\\\a=36\ mi/h\cdot4h=36\cdot4\ miles=144\ miles\\\\b=42\ mi/h\cdot4h=42\cdot4\ miles=168\ miles\\\\\alpha=144^o-54^o=90^o\\\\x^2=144^2+168^2-2\cdot144\cdot168\cdot cos90^o\\\\x^2=20736+28224-2\cdot144\cdot168\cdot0\\\\x^2=48960\\\\x=\sqrt{48960}\\\\x\approx221.27\ (miles)[/tex]

[tex]or\ Pythagoras\ theorem:\\\\x^2=144^2+168^2\\\\x\approx\ 221.27\ (miles)\\\\\\Answer:221.27\ miles.[/tex]
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kate200468
[tex]t=11-7=4\ [h]\ \ \ \wedge\ \ \ S=V\cdot t\\ \\S_1=V_1\cdot 4=36 \cdot 4=144\ [mi]\ \ \ and\ \ \ S_2=V_2\cdot4=42 \cdot 4=168\ [mi]\\ \\x^2=(S_1)^2+(S_2)^2 \ \ \ \Rightarrow\ \ \ x^2=144^2+168^2\ \ \ \Rightarrow\ \ \ x^2=48960\\ \\x\approx221.3\ [mi]\\ \\Ans.\ The\ distance\ between\ the\ ships\ at\ 11\ AM\ is\ 221.3\ mi[/tex]

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