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