theoremGiven:-
A triangle.
To find:-
[tex]\cos \frac{\alpha}{2}[/tex]
So now we use the formula,
[tex]\cos \frac{\alpha}{2}=\pm\sqrt[]{\frac{1+\cos\alpha}{2}}[/tex]
Also,
[tex]\cos \alpha=\frac{Adjacent\text{ side}}{Hypotnews}[/tex]
So we use Pythagoras theorms,
[tex]\begin{gathered} x=\sqrt[]{24^2+7^2} \\ x=\sqrt[]{576+49} \\ x=\sqrt[\square]{625} \\ x=25 \end{gathered}[/tex]
So now we substitute,
[tex]\cos \alpha=\frac{24}{25}[/tex]
So now we get,
[tex]\pm\sqrt[]{\frac{1+\cos \alpha}{2}}=\pm\sqrt[]{\frac{1+\frac{24}{25}}{2}}=\pm\sqrt[]{\frac{\frac{49}{25}}{2}}=\pm\sqrt[]{\frac{7}{10}}[/tex]
So the solution is,
[tex]\pm\sqrt[]{\frac{7}{10}}[/tex]
