Question

A plane leaves airport A and travels 560 miles to airport B on a bearing of Upper N 35 degrees Upper E. The plane later leaves airport B and travels to airport C 410 miles away on a bearing of Upper S 76 degrees Upper E. Find the distance from airport A to airport C to the nearest tenth of a mile

Answer 1

d₃ = 803.9 Miles

Explanation:

d₁ = 560 Miles

R₁ = N35⁰E

d₂ = 410 Miles

R₂ = S76⁰E

d₃ = ?

α = 35⁰+76⁰ = 111⁰

We can apply the Law of Cosines as follows

d₃ = √(d₁² + d₂² - 2*d₁*d₂* Cos α)

⇒ d₃ = √(560² + 410² - 2*560*410* Cos 111⁰) = 803.9 Miles

Answer 2

d₃ = 803.9 Miles

Explanation:

d₁ = 560 Miles

R₁ = N35⁰E

d₂ = 410 Miles

R₂ = S76⁰E

d₃ = ?

α = 35⁰+76⁰ = 111⁰

We can apply the Law of Cosines as follows

d₃ = √(d₁² + d₂² - 2*d₁*d₂* Cos α)

⇒ d₃ = √(560² + 410² - 2*560*410* Cos 111⁰) = 803.9 Miles