180º < θ < 360º and cosθ is positive means that θ lies in quadrant IV
On your calculator enter:
2ndF
cos
.778
=
and you should get and acute angle.
that would be the angle in standard position, so
θ = 360- angle = ..... to the nearest tenth.
check by taking cos(of your angle)
let me know what you get.
Solve for θ in the equation cos θ = 0.778 when 180º < θ < 360º. Round your answer to the nearest tenth of a degree.
6 answers
I get 38.9 and then subtracting it from 360 gets 321.1 which isn't the correct answer, apparently.
That is the answer I got,
and when I find cos 321.1 , I get .778243, which matches the given
considering that we rounded to one decimal.
Our answer is correct.
and when I find cos 321.1 , I get .778243, which matches the given
considering that we rounded to one decimal.
Our answer is correct.
Well, I've tried every possible variation for it, but it won't say it's correct. I'll ask my teacher. Thank you so much!
just a thought ... are you using the degree symbol in your answer?
321.1º
alt-167
radians is the default I think ... degrees must be specified
321.1º
alt-167
radians is the default I think ... degrees must be specified
The domain was definitely given in degrees:
180º < θ < 360º
180º < θ < 360º