In a triangle PQT, $QR = \sqrt{3}\text{ cm}$, PR = 3cm, $PQ = 2\sqrt{3}\text{ cm}$ and PQR = 30o. Find angles P and R

Assessment: JAMB UTME - Mathematics - 2006 Subject: General Mathematics

Question 1 Report

In a triangle PQT, $QR = \sqrt{3}\text{ cm}$, PR = 3cm, $PQ = 2\sqrt{3}\text{ cm}$ and PQR = 30^o. Find angles P and R

P = 60o and R = 90o

P = 30o and R = 120o

P = 90o and R = 60o

P = 60o and R 60o

P = 45o and R = 105o

Answer Details

By using cosine formula, p2 = Q2 + R2 - 2QR cos p
Cos P =

\frac{Q^{2} + R^{2} - p^{2}}{2 Q R}

\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

^{\circ}

= < P = 60

^{\circ}

If < P = 60

^{\circ}

and < Q = 30

< R = 180 $^{\circ}$ - 90 $^{\circ}$
angle P = 60 $^{\circ}$ and angle R is 90 $^{\circ}$

In a triangle PQT, \(QR = \sqrt{3}\text{ cm}\), PR = 3cm, \(PQ = 2\sqrt{3}\text{ cm}\) and PQR = 30o. Find angles P and R