Answer to Question #168696 in Geometry for Gideon

Question

Two men P and Q set off from a base camp R prospecting for oil. P moves 20km on a bearing of 250° and A moves 15km on a bearing of 60. Calculate the Distance of Q from P

2: bearing of Q from

Answer 1

Let distance traveled man P is a, man Q is b and distance between P and Q is c. Angle between P-man way and Q-man way is $250^0-60^0=190^0$ . So we have a triangle with two sides ("a" and "b") and an angle between them (190⁰), and a side "c" which we have to find. According law of cosines: c²=a²+b²- 2*a*b*cos(190⁰). So

$c=\sqrt{a^{2}+b^{2}-2*a*b*cos(190^0)}$

$c=\sqrt{20^{2}+15^{2}-2*20*15*cos(190^0)}$

$с\approx24.19 (km)$ - distance of Q from P.

What I need to find in task #2 is not known.