In triangle PQR, m∠P = 43°, PQ = 7.5, and PR = 8.4 What is m∠R to the nearest degree

Answer 1

Using the Law of Cosines, we have:

QR^2 = PQ^2 + PR^2 - 2PQ*PR*cos(P)

where P is the angle at vertex P. Substituting the given values, we get:

QR^2 = 7.5^2 + 8.4^2 - 2(7.5)(8.4)cos(43°) ≈ 27.901

Taking the square root of both sides, we get:

QR ≈ 5.282

Now, using the Law of Cosines again for angle R, we have:

cos(R) = (PQ^2 + QR^2 - PR^2) / (2PQ*QR)

Substituting the given values, we get:

cos(R) = (7.5^2 + 5.282^2 - 8.4^2) / (2(7.5)(5.282)) ≈ 0.550

Taking the inverse cosine of both sides, we get:

R ≈ 57°

Therefore, m∠R is approximately 57 degrees.