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In a triangle PQT, QR = \(\sqrt{3}cm\), PR = 3cm, PQ = \(...

In a triangle PQT, QR = \(\sqrt{3}cm\), PR = 3cm, PQ = \(2\sqrt{3}\)cm and PQR = 30°. 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 is A

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 = 60°

= < P = 60°

If < P = 60° and < Q = 30

< R = 180° - 90°

angle P = 60° and angle R is 90°