In PQR, Q = 76, PQ = 18, and QR = 27
Can you help me with 5!

Answer:
The answer is
m<p =66°21'14"
m<R =37°38'46"
PR =28,6
Step-by-step explanation:
[tex] {pr}^{2} = {18}^{2} + {27}^{2} - 2 \times 18 \times 27 \cos(76) = 817.85 \\ pr = 28.6 \\ \cos(p) = \frac{ {18}^{2} + {28.6}^{2} - { {27}^{2} } }{2 \times 18 \times 28.6 } = 0.4 \\ m < p = 66.42 \\ m < r = 180 - 76 - 66.42 = 37.58[/tex]