Triangle P Q R is shown. The length of P Q is 17, the length of Q R is 15, and the length of P R is 14. Law of cosines: a2 = b2 + c2 – 2bccos(A) What is the measure of AngleP to the nearest whole degree? 35° 52° 57° 72°

P = 57°

Step-by-step explanation:

Given the following :

PQ = 17

QR = 15

PR = 14

Using the cosine formula since the length of the three sides are given:

a2 = b2 + c2 – 2bccos(A)

To find P:

QR^2 = PQ^2 + PR^2 – 2(PQ)(PR)cos(P)

15^2 = 17^2 + 14^2 – 2(17)(14)cos(P)

225 = 289 + 196 - 476 cosP

476*CosP = 485 - 225

476*CosP = 260

CosP = 260/476

CosP = 0.5462184

P = Cos^-1(0.5462184)

P = 56.892029

P = 57°

it should be a and d. lmk if i'm wrong xd