You can solve for C using the law of cosines.
P, Q, and R are the angles of the triangle and p, q, and r are the lengths of the sides of the triangle opposite of the angles. That means p = 10ft, q = 12ft, and r=5ft. The law of cosines states that:
[tex]p^2 = q^2+r^2-2qr(cosP) \\
q^2 = p^2+r^2-2pr(cosQ) \\
r^2 = p^2+q^2-2pq(cosR)[/tex]
Since we're looking for the measure of angle P, we would use the first equation, [tex]p^2 = q^2+r^2-2qr(cosP)[/tex]. Plug the values we have for the sides of the triangle into the equation and solve for P:
[tex]p^2 = q^2+r^2-2qr(cosP)\\
10^{2} = 12^{2} + 5^{2} - 2(12)(5)(cosP)\\
100 = 144 + 25 - 120cosP\\
120cosP = 69\\
cosP = 0.575\\
P = cos^{-1}(0.575)\\
P \approx 55\°[/tex]
The measure of angle P is about 55°.
