Q is 32 km away from P on a bearing 042o and R is 25km from P on a bearing of 132o. Calculate the bearing of R from Q.
A
122o
B
184o
C
190o
D
226o
Verified Correct Answer: B.
Step-by-Step Mathematical Solution
△PQR is a right-angle triangle
/QR/\(^2\) = 32\(^2\) + 25\(^2\) = 1024 + 625 = 1649
/QR/ = \(\sqrt{1649}\) = 40.6Km
Let < PQR = \theta
\(\frac{25}{sin \theta}\) = \(\frac{40.6}{sin 90}\)
Sin \theta = \(\frac{25 Sin 90}{40.6}\) = 38º
Bearing of R from Q = 270 - (48 + 38) = 184º
Source: Myschool Past Questions Archive (Record #36280)