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Math Help - Confusing question

  1. #1
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    Confusing question

    From a point P the ships R and V are observed bearing 325 degrees T and 047 degrees T respectively whilst from a point Q 10 km due north of P the ships R and V bear 310 degrees T and 079 degrees T respectively. Show that PR = 10 sin 50 degrees/sin 15 degrees and PV = 10 sin 79 degrees/sin 32 degrees.

    When I drew the diagram, I found out that angles 50 degrees and 79 degrees were outside the triangles. So i was just wondering how the sine rule could be applied to show the above question. All I could get was PR = 10 sin 130 degrees/sin 15 degrees and PV = 10 sin 101 degrees/ sin 32 degrees.

    Please help!
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    From a point P the ships R and V are observed bearing 325 degrees T and 047 degrees T respectively whilst from a point Q 10 km due north of P the ships R and V bear 310 degrees T and 079 degrees T respectively. Show that PR = 10 sin 50 degrees/sin 15 degrees and PV = 10 sin 79 degrees/sin 32 degrees.

    When I drew the diagram, I found out that angles 50 degrees and 79 degrees were outside the triangles. So i was just wondering how the sine rule could be applied to show the above question. All I could get was PR = 10 sin 130 degrees/sin 15 degrees and PV = 10 sin 101 degrees/ sin 32 degrees.

    Please help!
    The Sine-function has the following property:

    If 0^\circ < \alpha < 90^\circ then \sin(\alpha) = \sin(180^\circ - \alpha) and therefore

    \sin(50^\circ) = \sin(130^\circ) and

    \sin(79^\circ) = \sin(101^\circ)
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