D.Atlas
As atlas is basically a book of maps
Hope this helps!
I can’t see the question (picture is blurry)
Answer:
![\bold{cos\dfrac{A}{2} = -\dfrac{1}{\sqrt3}}](https://tex.z-dn.net/?f=%5Cbold%7Bcos%5Cdfrac%7BA%7D%7B2%7D%20%3D%20-%5Cdfrac%7B1%7D%7B%5Csqrt3%7D%7D)
Step-by-step explanation:
Given that:
![cosA=-\dfrac{1}3](https://tex.z-dn.net/?f=cosA%3D-%5Cdfrac%7B1%7D3)
and
![tanA > 0](https://tex.z-dn.net/?f=tanA%20%3E%200)
To find:
![cos\dfrac{A}{2} = ?](https://tex.z-dn.net/?f=cos%5Cdfrac%7BA%7D%7B2%7D%20%3D%20%3F)
Solution:
First of all,we have cos value as negative and tan value as positive.
It is possible in the 3rd quadrant only.
will lie in the 2nd quadrant so
will be negative again.
Because Cosine is positive in 1st and 4th quadrant.
Formula:
![cos2\theta =2cos^2(\theta) - 1](https://tex.z-dn.net/?f=cos2%5Ctheta%20%3D2cos%5E2%28%5Ctheta%29%20-%201)
Here ![\theta = \frac{A}{2}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7BA%7D%7B2%7D)
![cosA =2cos^2(\dfrac{A}{2}) - 1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =cosA+1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =-\dfrac{1}3+1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =\dfrac{2}3\\\Rightarrow cos(\dfrac{A}{2}) = \pm \dfrac{1}{\sqrt3}](https://tex.z-dn.net/?f=cosA%20%3D2cos%5E2%28%5Cdfrac%7BA%7D%7B2%7D%29%20-%201%5C%5C%5CRightarrow%202cos%5E2%28%5Cdfrac%7BA%7D%7B2%7D%29%20%3DcosA%2B1%5C%5C%5CRightarrow%202cos%5E2%28%5Cdfrac%7BA%7D%7B2%7D%29%20%3D-%5Cdfrac%7B1%7D3%2B1%5C%5C%5CRightarrow%202cos%5E2%28%5Cdfrac%7BA%7D%7B2%7D%29%20%3D%5Cdfrac%7B2%7D3%5C%5C%5CRightarrow%20cos%28%5Cdfrac%7BA%7D%7B2%7D%29%20%3D%20%5Cpm%20%5Cdfrac%7B1%7D%7B%5Csqrt3%7D)
But as we have discussed,
will be negative.
So, answer is:
![\bold{cos\dfrac{A}{2} = -\dfrac{1}{\sqrt3}}](https://tex.z-dn.net/?f=%5Cbold%7Bcos%5Cdfrac%7BA%7D%7B2%7D%20%3D%20-%5Cdfrac%7B1%7D%7B%5Csqrt3%7D%7D)
Answer:
Yes, it makes it true
Step-by-step explanation:
Plug in 3 and 10 as x and y into the inequality:
y > 2x + 2
10 > 2(3) + 2
10 > 6 + 2
10 > 8
Since 10 is bigger than 8, this inequality is true.
So, (3, 10) makes the inequality true.