Please explain this question is you can...
Ok, here goes:
An airplane leaves its airport on the east coast of the US and flies in a direction of 085 degrees.
Because of bad weather, it returns to another airport 230 km to the north of it's home base.
For the return trip, it flies in a direction of 283 degrees.
What is the total distance that the airplane flew?
Thanks for you help!!!!!!!!
2007-08-15 20:53:01 · 5 answers · asked by Rebirth _2007 1 in Science & Mathematics ➔ Mathematics
Draw a diagram of the journeys and you will get a triangle.Unfortunately, his return journey will take him away from his base and in the wrong direction (W) to which he was originally going.Check the question again.
2007-08-15 21:15:38 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The given info defines a triangle ABC with C at the end of a vertical line from A (second airport, C, is 230 km north of first, A)
The origin is point A. The plane leaves A for B at an angle of 85 deg. At the point B,where the plane turns towards C the heading changes from 85 to 283deg. This makes angle ABC 77 deg (360 - 283). Since the enclosed angles in a triangle sum to 180 deg angle BCA must be 18 deg (180-77-85).
From the Sine rule BC/SinA = AC/SinB = AB/SinC,
So (230/Sin77)=BC/Sin85, or (230/Sin77)*Sin85= BC = 235.15km. Also (230/Sin77)=AB/Sin18 or (230/Sin77)*Sin18= AB = 72.94 km.
So the total is 230+235.15 +72.94 = 538.09
Draw the triangle first then check my figures.
Hope this helps
Just re-read the question and may be the answer is 308.09, which is the distance the plane flew from one airport to the next. Originally I included travelling back to its origin (extra 230km)
2007-08-15 21:23:03 · answer #2 · answered by hersheba 4 · 0⤊ 0⤋
α = 85°
β = 180° - (360° - 283° + 85°) = 18°
γ = 180° - 85° - 18° = 77°
c = outbound leg
a = return leg
c/sinγ = b/sinβ
c/sin77° = 230/sin18°
c = 230sin77°/sin18°
a/sinα = b/sinβ
a/sin85° = 230/sin18°
a = 230sin85°/sin18°
c + a = 230(sin77°/sin18° + sin85°/sin18°)
c + a = 230(sin77° + sin85°)/sin18°
c + a ≈ 230(0.97437 + 0.99619)/
c + a ≈ 230(1.97056)/0.30902
c + a ≈ 230(6.37688)
c + a ≈ 1,466.7 km
2007-08-15 21:38:14 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
the main important perspective in a triangle is often opposite of the longest part, this is b for this reason. regulation of cosines states that: a^2 = b^2 + c^2 - 2bc * cos A. this is such as : b^2 = a^2 + c^2 - 2ac * cos B. So, put in each and every of the given variables: 10^2 = 6^2 + 7^2 - 2(6)(7) * cos B. simplify - 10^2 = 6^2 + 7^2 - 2(6)(7) * cos B. a hundred = 36 + 40 9 - eighty 4(cos B) a hundred = eighty 5 - eighty 4(cos B) eighty 4(cos B) = -15 cos B = -15/eighty 4 = -5/28 Use inverse cosine for degree of perspective B - cos-a million (-5/28) = approximately a hundred.28656.
2016-12-13 09:03:00 · answer #4 · answered by melvina 3 · 0⤊ 0⤋
The problem is not clear.
2007-08-15 21:01:54 · answer #5 · answered by contemplating 5 · 0⤊ 0⤋