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

P = 60^o and R = 90^o

P = 30^o and R = 120^o

P = 90^o and R = 60^o

P = 60^o and R 60^o

P = 45^o and R = 105^o

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°

See all Mathematics questions & answers

See more JAMB questions & answers

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

Similar Questions

Aptitude Tests

In a triangle PQT, QR = √3cm\sqrt{3}cm, PR = 3cm, PQ = 2√32\sqrt{3}cm and PQR = 30°. Find angles P and R

Similar Questions

Aptitude Tests

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