# i need help on a physics questionnnnn. helllp<3?

2 airplanes leave an airport at the same time. The velocity of the 1st airplane is 730 m/h at a heading of 25.9 (deegrees). The velocity of the 2nd is 630 m/h at a heading of 113 (deegrees) HOW FAR APART ARE THEY AFTER 2.2. h?

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• Anonymous

Your instructor wants you to use the law of cosines.

c2 = a2 + b2 – 2ab cos C.

Aloha

You need to use the law of cosines.

c^2 = a^2 + b^2 - 2ab*Cos(angle C)

c^2 = (730*2.2)^2 + (630*2.2)^2 - 2(730*2.2)(630*2.2)Cos(113-25.9)

c^2 = (1606)^2 + (1386)^2 - 2(1606)(1386)(.0506)

c = 2067.6 miles.

All you have is a geometry problem here.

You know the lengths of two sides of the triangle (730x2.2 and 630x2.2) and the angle between them (113 - 25.9).

You should be able to go from there...

(BTW - Joel is wrong - Pythagorean theorem is for right triangles)

let Theta = 113 - 25.9 = 87.1

let d1 = 730 * 2.2

let d2 = 630 * 2.2

let d12 = distance between planes

d12 = sqrt(d1^2 + d2^2 - 2*d1*d2*cos(theta)) = 2084 miles

it's been 12 years since high school but i'll give it a go. i think you need to use the cosine rule: aa=bb+cc-2bc.cosA

distance(sq)= 2579236+1587600-4047120cos 87.1

don't have a scientific calculator handy so you'll have to de the math yourself.

peace

p.s the first answer only works for right angles triangles