In a triangle PQT, QR = \(\sqrt{3}cm\), PR = 3cm, PQ = \(2\sqrt{3}\)cm and PQR = 30o. Find angles P and R
A. P = 60o and R = 90o B. P = 30o and R = 120o C. P = 90o and R = 60o D. P = 60o and R 60o E. P = 45o and R = 105o
Correct Answer: A
Explanation
By using cosine formula, p2 = Q2 + R2 - 2QR cos p Cos P = \(\frac{Q^2 + R^2 - p^2}{2 QR}\) = \(\frac{(3)^2 + 2(\sqrt{3})^2 - 3^2}{2\sqrt{3}}\) = \(\frac{3 + 12 - 9}{12}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\) = 0.5 Cos P = 0.5 p = cos-1 0.5 = 60o = < P = 60o If < P = 60o and < Q = 30 < R = 180o - 90o angle P = 60o and angle R is 90o
