A ship at A is to sail to C 56 kilometer north and 258 km east of A. After sailing North 25 degrees 10 minutes east for 120 miles to P the ships is headed toward C. Find the distance of P from C and the required course to reach C.
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I'll assume that you're working in the plane, rather than on the sphere. If that's not the case, skip this answer. Let's put some coordinates on your picture. Using $(east, north)$ coordinates, assume the ship starts at $A=(0,0)$ and sails to $P = (p_e,p_n)$. Along with the north axis, you can draw a right triangle with the lower angle $\alpha$ equal to 25 degrees 10 minutes and hypotenuse $193.1218$ (converting miles to kilometers). Now you can use $193.1218\sin\alpha$ and $193.1218\cos\alpha$ to find the coordinates $p_e$ and $p_n$ (do you see how?). Now from $P$ draw two lines, one to $C$ and one in the east direction. You can make another right triangle and using the (now known) lengths of the east and north sides you can find the heading angle and the distance to $C$. Hope this hint helps.
We are given the starting and ending positions: $A(0,0)$ and $C(258,56)$. Suppose the angle $\theta=25^0 10^'$. Then the intermediate position of $P$ is $(120 \cos(\theta), 120 \sin(\theta))=(108.609,51.030)$ and so now one can determine the distance $PC$ and the angle (heading) from $P$ to $C$