Answer to Question #237804 in Geometry for Andu

Question

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.

Answer 1

a)

triangle LPF

b)

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

$LF = √(LP^2 + PF^2 - 2LP.PF·cos(P) )$

$⇒LF = √(213^2 + 287^2 - 2·213·287·cos(75) )\\ ⇒LF = 309.99075$

Round of to nearest tenth

$⇒LF = 310 km$

 $∠L=arc\space cos (\frac{LF^ 2 +LP ^2 −PF ^2}{ 2LF·LP} )\\ ⇒ ∠L = arc\space cos (\frac{310 ^2 +213 ^2 −287 ^2 } {2×310×213} )\\ ⇒∠L=arc\space cos(\frac{59100}{ 132060} )\\ ⇒∠L= arc\space cos(0.4475) \\ = ∠L=63.415°$

 

Now 

$∠F = 180° − ∠L−∠P\\ ⇒∠F = 180° − 63.415−75\\ ⇒∠F=41.585°$

Hence Distance between L and F is 310 km

and angle from L to F 

$⇒360 -148-63.415-90 = 58.585°$

Now angle from F to L $= 90 - 58.585° = 31.415°$

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