P, q and r are points on the same horizontal plane. The bearing of q from p is 150 and the bearing of r from q is 060 if line p and q is equal to 5 and line q and r is 3 find the bearing of r from p, correct to the nearest degree. (Waec)
Some one should help me
According to your description I made triangle PQR ,
so that PQ = 5, QR = 3 and with basic geometry found angle Q = 90°
using the tangent ratio,
tan QPR = 3/5
so angle P of the triangle = 30.96°
let me know what your final answer is
To find the bearing of point r from point p, we need to add the bearings of q from p and r from q.
Given that the bearing of q from p is 150 degrees and the bearing of r from q is 060 degrees, we can start by calculating the final bearing of q from p.
Since the bearing of r from q is 060 degrees, it means that the angle between line qr and the north direction is 60 degrees counterclockwise. Similarly, the bearing of q from p is 150 degrees, which means the angle between line pq and the north direction is 150 degrees counterclockwise.
To find the angle between line pr and the north direction, we need to add the angles of 60 degrees and 150 degrees.
150° + 60° = 210°
So, the angle between line pr and the north direction is 210 degrees counterclockwise. The bearing of r from p is the complement of this angle, which is 180 degrees - 210 degrees = -30 degrees.
Since the bearing is always measured clockwise from the north direction, we convert -30 degrees to a positive angle:
360 degrees - 30 degrees = 330 degrees
Therefore, the bearing of r from p is 330 degrees (rounded to the nearest degree).
Hence, the bearing of r from p is 330 degrees.