Answer: I don't have the capability to create or share images. However, I can provide the step-by-step solution to the problem as text:
To solve this problem, we can use the cosine and sine rules in trigonometry. Let's first draw a diagram to represent the situation:
R
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732 km |
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Q----------------P
a) To find the distance from Q to R, we need to first find the length of the side QR of the triangle PQR. We can do this by using the cosine rule:
cos(QPR) = (QP^2 + QR^2 - PR^2) / (2 * QP * QR)
Since angle QPR is a right angle, we have:
cos(QPR) = 0
Substituting the given values, we get:
0 = (156^2 + QR^2 - 732^2) / (2 * 156 * QR)
Simplifying and solving for QR, we get:
QR = sqrt(732^2 - 156^2) / 2.4 ≈ 269.3 km
Therefore, the distance from Q to R is approximately 269.3 km.
b) To find the total distance flown, we need to add up the distances PQ and QR. We have:
PQ = 156 km (given)
QR ≈ 269.3 km (from part a)
Total distance = PQ + QR ≈ 425.3 km
Therefore, the total distance flown is approximately 425.3 km.
c) To determine the bearing of point R from point Q, we need to find the angle QRS. We can do this by using the sine rule:
sin(QRS) / QR = sin(QRP) / PR
Since angle QRP is a right angle, we have:
sin(QRP) = PQ / PR
Substituting the given values, we get:
sin(QRS) / QR = 156 / 732
Simplifying and solving for sin(QRS), we get:
sin(QRS) = (156 / 732) * QR ≈ 0.203
Using the inverse sine function, we find that:
QRS ≈ 11.75°
Since the plane turned to the right at an angle in excess of 90°, the bearing of point R from point Q is:
Bearing of R from Q = 270° + QRS ≈ 281.75°
Therefore, the bearing of point R from point Q is approximately 281.75°.
Explanation: