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An airport control tower measures the distance to one plane, that has
just taken off, to be \(500 \thinspace \rm{m}\), and the distance to a second plane, coming
in for landing, to be \(1\thinspace 050\thinspace \rm{m}\). If the angle between the lines of
observation on the radar screen is \(86^\circ\), how far apart are the two
planes? Round your answer to the nearest metre.
Write down what is given.
\(a = x\)
\(b = 1\thinspace 050 \rm{m}\)
\(c = 500 \rm{m}\)
\(\angle A = 86^\circ\)
\(\begin{align}
a^2 &= b^2 + c^2 - 2bc\cos A \\
x^2 &= 1\thinspace 050^2 + 500^2 - 2\left( 1\thinspace 050 \right)\left( {500} \right)\cos 86^\circ \\
x^2 &= 1\thinspace 279\thinspace 255.703... \\
x &= 1\thinspace 131.041... \\
x &\doteq 1\thinspace 131 \\
\end{align}\)
The two planes are approximately \(1\thinspace 131\rm{m}\) apart.
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Jeff is shooting pucks on an empty net during hockey practice.
He is standing to one side of the net. The goal post closest to him is
\(23\thinspace \rm{ft}\) away, and the goal post farthest from him is
\(26\thinspace \rm{ft}\) away. If the goal posts are \(6\thinspace
\rm{ft}\) apart, how many degrees of shooting variance does Jeff have
that will result in him getting a goal? Solve using the cosine law, and
round your answer to the nearest tenth of a degree.
Write down what is given.
\(a = 6\thinspace \rm{ft}\)
\(b = 23\thinspace \rm{ft}\)
\(c = 26\thinspace \rm{ft}\)
\(\angle A = \thinspace ?\)\[\begin{align}
\cos A &= \frac{{b^2 + c^2 - a^2 }}{{2bc}} \\
\cos A &= \frac{{\left( {23} \right)^2 + \left( {26} \right)^2 - \left( 6 \right)^2 }}{{2\left( {23} \right)\left( {26} \right)}} \\
\cos A &= \frac{{1\thinspace 169}}{{1\thinspace 196}} \\
\cos A &= 0.977... \\
\angle A &= \cos ^{ - 1} \left( {0.977...} \right) \\
\angle A &= 12.197...^\circ \\
\angle A &\doteq 12.2^\circ \\
\end{align}\]
Jeff must shoot the puck within a range of \(12.2^\circ\).
| For further information about the cosine law see pp. 116 to 119 of Pre-Calculus 11. |