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?
2006-08-29 15:15:43 · 6 answers · asked by BaleriaBaleyva<33 2 in Science & Mathematics ➔ Physics
Your instructor wants you to use the law of cosines.
c2 = a2 + b2 – 2ab cos C.
Aloha
2006-08-29 15:29:34 · answer #1 · answered by Anonymous · 2⤊ 0⤋
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.
2006-08-29 22:32:58 · answer #2 · answered by something 3 · 0⤊ 0⤋
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)
2006-08-29 22:24:10 · answer #3 · answered by Art_333 2 · 0⤊ 0⤋
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
2006-08-29 22:32:35 · answer #4 · answered by none2perdy 4 · 0⤊ 0⤋
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
2006-08-29 22:27:10 · answer #5 · answered by pleiades-im-coming-home 2 · 0⤊ 0⤋
find the square root of 4500232 that is the answer using the pathagorian therom. A squared+ B squared = C squared. so 630x2.2=1920996, and 730x2.2=2579236. add thode together to get 4500232 then use calculator to find the square root of that number
2006-08-29 22:23:23 · answer #6 · answered by Joel G 2 · 0⤊ 1⤋