Answer:
![\cos(\theta)=\pm (24)/(25)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/s6b0i3uze7k9ruqxfj6bvgagz3oa4sawn5.png)
Explanation:
I don't know where
is so there is going to be two possibilities for cosine value, one being positive while the other is negative.
A Pythagorean Identity is
.
We are given
.
So we are going to input
for the
:
![\cos^2(\theta)+((7)/(25))^2=1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o4ou70dpagji7dcu579aisckpnrz0gf8db.png)
![\cos^2(\theta)+(49)/(625)=1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t2ygs4geole34wb53ad6tb6oxi3jefnd3b.png)
Subtract 49/625 on both sides:
![\cos^2(\theta)=1-(49)/(625)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ef7ncy7r4yz13tb1lmydpboqs3nug0dall.png)
Find a common denominator:
![\cos^2(\theta)=(625-49)/(625)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/c17a3g853l5hdjer0r7u10bxjvx3x1g42z.png)
![\cos^2(\theta)=(576)/(625)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/r8xz5smnigmcyv4lzzlkzcunftwwwtqiqx.png)
Square root both sides:
![\cos(\theta)=\pm \sqrt{(576)/(625)}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wholcue3k0ezm4bzm4nqmvhnlpsjrnc9t4.png)
![\cos(\theta)=\pm (√(576))/(√(625))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5a8g8mrjuxvq4ou3itbxtldnynbokfwsa1.png)
![\cos(\theta)=\pm (24)/(25)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/s6b0i3uze7k9ruqxfj6bvgagz3oa4sawn5.png)