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

Mathematics

JAMB 1983

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 = 60^o and R = 90^o
B. P = 30^o and R = 120^o
C. P = 90^o and R = 60^o
D. P = 60^o and R 60^o
E. P = 45^o and R = 105^o

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Correct Answer: Option 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