Answer to Question #237804 in Geometry for Andu

Question #237804

An airplane makes a trip from city L to city P, which is located 213 km away from city L and on the course 148⁰. He then flies 287 km on a 75 kurs course to city F. Finally, the plane returns to city L. a. Draw the LPF triangle exactly. b. Find the distance and course from city F to city L.

Expert's answer

triangle LPF

Calculates LF, ∠F, and ∠L based on given LP, PF, and ∠P

"LF = \u221a(LP^2 + PF^2 - 2LP.PF\u00b7cos(P) )"

"\u21d2LF = \u221a(213^2 + 287^2 - 2\u00b7213\u00b7287\u00b7cos(75) )\\\\\n\n\n\n\u21d2LF = 309.99075"

Round of to nearest tenth

"\u21d2LF = 310 km"

"\u2220L=arc\\space cos (\\frac{LF^\n\n2\n\n+LP\n\n^2\n\n\u2212PF\n\n^2}{\n\n2LF\u00b7LP}\n\n\n)\\\\\n\n\u21d2 \u2220L = arc\\space cos (\\frac{310\n\n^2\n\n+213\n\n^2\n\n\u2212287\n\n^2\n}\n{2\u00d7310\u00d7213}\n\n\n)\\\\\n\n\n\n\u21d2\u2220L=arc\\space cos(\\frac{59100}{\n\n132060}\n\n\n)\\\\\n\n\n\u21d2\u2220L= arc\\space cos(0.4475)\n\\\\\n\n\n=\n\n\u2220L=63.415\u00b0"

Now

"\u2220F = 180\u00b0 \u2212 \u2220L\u2212\u2220P\\\\\n\n\u21d2\u2220F = 180\u00b0 \u2212 63.415\u221275\\\\\n\n\u21d2\u2220F=41.585\u00b0"

Hence Distance between L and F is 310 km

and angle from L to F

"\u21d2360 -148-63.415-90 = 58.585\u00b0"

Now angle from F to L "= 90 - 58.585\u00b0 = 31.415\u00b0"

Hence the distance and course from city F to city L is 310 km and 31.415°