Grade 10 Trig [+rep]

[Maki]

Two airplanes leave an airport at the same time.

One airplane travels at 355 km/h. The other airplane travels at 450 km/h.
About 2h later, they are 800 km apart.

Determine the angle between their paths.

My solution:

Using the cos law:

800^2 = 710^2 + 900 ^2 - 2 (710)(900) (cosAirport)


However, when I work this out to solve for cosAirport and ultimately angle x, I get an error in my calculator.

Diagram + solution will get you +5 rep, the answer is "about 58 degrees"

[Chili]

(you need an account to see links)

[Maki]

800^2=900^2 + 710^2 - 2(900)(710)cos(A)

640000 = 36100cosA

cosA = 640000/36100

However, I'm not sure how to find the angle from here....
Inverse cosA = error

[Raj]

800^2 = 710^2 + 900 ^2 - 2 (710)(900) (cosAirport)

640000 = 1314100 - 1278000(cosA)

subtract 134100 on both sides

-674100 = -1278000(cosA)

divide by -1278000 to get cosA all by itself

(-674100) / (-1278000) = cosA

boom you got cosA

cosA = 0.52746

inverse cosine to solve for angle of the airport

cos-1(0.52746) = A
A = 58.16 degrees

[Maki]

800^2 = 710^2 + 900 ^2 - 2 (710)(900) (cosAirport)

640000 = 1314100 - 1278000(cosA)

subtract 134100 on both sides

Why do I have to suddenly do this as opposed to solving 1314100 - 1278000, and then solving for cosA?

[Raj]

Why do I have to suddenly do this as opposed to solving 1314100 - 1278000, and then solving for cosA?

Because ya can't do that bruh

640000 = 1314100 - 1278000(cosA)

the -1278000 is being multiplied by cosA, and you can't do 1314100 - 1278000 because the -1278000 is attached to the cosA

[Maki]

Because ya can't do that bruh

640000 = 1314100 - 1278000(cosA)

the -1278000 is being multiplied by cosA, and you can't do 1314100 - 1278000 because the -1278000 is attached to the cosA

Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh smh reckless
Thanks bru