A plane leaves airport A and travels 580 miles to airport B on a bearing of N 34° E. The plane later leaves airport B and travels to airport C 400 miles away on a bearing of S 74° E.
Find the distance from airport A to airport C to the nearest tenth of a mile.
what is the bearing if you flew directly from airport A to airport C to the nearest degree?
I need a statement of the problem as well as well as scale drawing of situation with detailed explanation of solution which includes whether Law of Sines or Law of Cosines was used and why. The general form of any formulas used must also be provided.
A plane leaves airport A, bearing `34^@ ` East of North, and travels 580 miles to airport B. The plane then leaves airport B, at a bearing of `74^@ ` East of South, for a distance of 400 miles to airport C. What is the distance from A to C, and what bearing should you take to get from A to C?
See below for a diagram of the given information.
Now `m/_ABC=108^@ ` . (Use the fact that the two NS lines are parallel, so with `bar(AB) ` a transversal we have congruent alternate interior angles. So the angle at B is the sum of 34 and 74 degrees.)
So we have `Delta ABC ` with `AB=580,m/_B=108^@,BC=400 ` .
(1) To find AC we use the Law of Cosines. With AB=c, BC=a, and AC=b we use the form` ` `b^2=a^2+c^2-2ac"Cos" B ` . We use the Law of Cosines because the givens include two sides and the included angle.
Then `b^2=580^2+400^2-2(580)(400)"Cos" 108^@ `
So AC is approximately 800 miles. (As a check, this makes sense as it is opposite the largest angle of the triangle, so will be the longest side of the triangle.)
(2) We can use the Law of Sines to find the measure of angle A.
`a/(sin A)=b/(sin B)=c/(sin C) ` Here we know a=400, b we found to be 800, and angle B is 108 degrees.
`400/(sin A)=800/(sin 108) ` ==> `sin A=(400 sin 108)/800 `
==> ` `` ``sin A~~.4755 ` ==> `A~~ "sin"^(-1).4755 ~~28.39 `
Thus the measure of angle A is approximately 28 degrees.
The measure of the angle between `bar(AC) ` and North can be found by 34+28=62 degrees from North.
Thus the bearing from A to C is approximately ` ``"N" 62^@ "E" `
A diagram of the given information: