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
Verified Correct Answer: A.
Step-by-Step Mathematical Solution
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
Source: Myschool Past Questions Archive (Record #32478)