1. ## 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.

2. Originally Posted by xwrathbringerx
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.

If $0^\circ < \alpha < 90^\circ$ then $\sin(\alpha) = \sin(180^\circ - \alpha)$ and therefore
$\sin(50^\circ) = \sin(130^\circ)$ and
$\sin(79^\circ) = \sin(101^\circ)$