Solve for θ in the equation cos θ = 0.778 when 180º < θ < 360º. Round your answer to the nearest tenth of a degree.
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.
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.
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
The domain was definitely given in degrees:
180º < θ < 360º
To solve for θ in the equation cos θ = 0.778 when 180º < θ < 360º, we can use the inverse cosine function (also known as the arccosine function).
1. First, take the inverse cosine of both sides of the equation: θ = arccos(0.778).
So, we have θ = arccos(0.778).
2. Next, we need to use a calculator to evaluate the arccosine of 0.778. Make sure your calculator is in degree mode.
If you input arccos(0.778) in your calculator, you should get the value of θ in radians.
3. Convert the value from radians to degrees by multiplying by 180/π.
Let's assume that the result of arccos(0.778) is approximately 0.710 (in radians).
θ (in degrees) = 0.710 * (180/π) ≈ 40.7°.
4. However, we need to consider the restriction 180º < θ < 360º. Since we obtained 40.7°, which is less than 180º, we need to add 360º to it.
θ (in degrees) = 40.7° + 360° = 400.7°.
5. Finally, round the answer to the nearest tenth of a degree, as requested.
Therefore, the solution for θ is approximately 400.7° when 180º < θ < 360º.