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AshleyVillarin
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Tyro
Asked: 2021-04-12T10:54:11+00:00 In: ,

How far apart are the ships?

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From a plane flying due east at 265 m above sea level, the angles of depression of two ships sailing due east measure 35° and 25°.

189.9m

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1 Answer

  2. acedstud

    The attached figure illustrates the question with point A as the location of the plane and points D and E are the positions of the two ships.

    \triangle{ACD} gives a right triangle, using the right triangle trigonometry, we have \tan{25^o}=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{265}{|AC|}

    \Rightarrow|AC|=\dfrac{265}{\tan{25^o}}\approx568.3 m.

    Similarly, \triangle{ABE} gives a right triangle, using the right triangle trigonometry, we have \tan{35^o}=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{265}{|AB|}

    \Rightarrow|AC|=\dfrac{265}{\tan{35^o}}\approx378.5 m.

    The distance between the two ships is |BC| which is given by |AC| – |AB| = 568.3 m – 378.5 m = 189.8 m.

    Attachment

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