Given that
[tex]\tan \theta=-\frac{7}{24}[/tex]
Where
[tex]\tan \theta=\frac{y}{x}[/tex]
In the second quadrant x < 0, y > 0
[tex]x=-24,y=7[/tex]
Where
[tex]r^2=x^2+y^2^{}_{}[/tex]
Substitute for x and y to find the value of r
[tex]\begin{gathered} r^2=(-24)^2+7^2=576+49=625^{} \\ r^2=625 \\ \text{square root of both sides} \\ \sqrt[]{r^2}=\sqrt[]{625} \\ r=25\text{ } \\ \text{With r}>0 \end{gathered}[/tex]
Since, it lies in the second quadrant,
[tex]\cos \theta=\frac{x}{r}[/tex]
Substitute the values of x and r
[tex]\cos \theta=-\frac{24}{25}[/tex]
Hence, the answer is
[tex]\cos \theta=-\frac{24}{25}[/tex]
