Respuesta :
Answer: 57°
Given : In triangle PQR, the side PQ is 17, QR is 15, and PR is 14.
We use law of cosine formula
[tex] QR^2 = PR^2 + PQ^2 - 2* PR * PQ * cos (P) [/tex]
[tex] 15^2 = 14^2 + 17^2 - 2*14 * 17 * cos(P) [/tex]
225 = 196 + 289 - 476 cos(P)
225 = 485 - 476 cos (P)
Subtract 485 on both sides
-260 = -476 cos(P)
[tex]\frac{260}{476} [/tex] = cos(P)
P = [tex] cos^{-1} (\frac{260}{476}) [/tex]
P= 56.89202
So measure of angle P is 57 degrees
