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# An aeroplane leaving from Bismarck travels on a bearing of ${120^\circ}$, as shown in the figure. If the plane flew at an average rate of $400$miles per mile per hour, how many minutes (approx.) had it been in the air when it was $295$ miles from pierre?A. $51$B. $56$C. $29$D. $67$

Last updated date: 19th Sep 2024
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Hint: In this question we have found the plane flew from pierre. For that we are going to solve using trigonometric identities in angle. Plane flew south east direction from Bismarck. Next we went to find the direction using sine and the plane flew from pierre in the air when it was $295$ miles.

Formula used: $\sin {60^\circ} = \dfrac{{\sqrt 3 }}{2}$
To find the plane using trigonometric identities as shown in the figure, the plane flew south east direction that distance from Bismarck.

Here we chose sine for to find angle (blue curve). Angle (black coloured) mentioned to find cosine but we need only angle (coloured black curve)
$\sin {60^\circ}{\text{ = }}\dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}$

The angle that the plane makes with the line joining Bismarck and pierre is ${60^\circ}$
We thus have $\sin {60^\circ} = \dfrac{{295}}{ x \\ \\ }$ where x is the distance flown by the plane.
$\dfrac{{\sqrt 3 }}{2} = \dfrac{{295}}{x}$
Solving for $x$ we get,
$x = \dfrac{{295 \times 2}}{{\sqrt 3 }}$
$x = 340.63{\text{minutes}}$
Since the speed is $400$ miles per hour, the time becomes
$\Rightarrow \dfrac{{340.63}}{{400 \times 60}}$
Multiplying the terms we get,
$\Rightarrow \dfrac{{340.63}}{{24000}}$
Dividing the terms we get,
$\Rightarrow 51.0945 \\ \Rightarrow 51{\text{(approx}}{\text{.)}} \\$

$\therefore$$51$minutes from the pierre.

Note: $\sin {60^\circ} = \dfrac{{295}}{ x \\ \\ }$ where $x$ is the distance flown by the plane. Speed is $400$ miles per hour to convert for minutes multiplied by 60. While taking sine to check the diagram where angle was placed. The plane flew at an average rate of $400$ miles per mile per hour.