Certainly! Let's solve this problem step-by-step.
### Given Data:
1. First Leg of the Journey:
- Bearing: [tex]\( 035^\circ \)[/tex]
- Time: [tex]\( 1\frac{1}{2} \)[/tex] hours or 1.5 hours
- Speed: 600 km/h
2. Second Leg of the Journey:
- Bearing: [tex]\( 130^\circ \)[/tex]
- Time: [tex]\( 1\frac{1}{2} \)[/tex] hours or 1.5 hours
- Speed: 400 km/h
### Solution:
#### Step 1: Calculate the distance for each leg of the journey
1. Distance for the first leg:
[tex]\[
\text{Distance first leg} = \text{Speed} \times \text{Time} = 600 \, \text{km/h} \times 1.5 \, \text{h} = 900 \, \text{km}
\][/tex]
2. Distance for the second leg:
[tex]\[
\text{Distance second leg} = \text{Speed} \times \text{Time} = 400 \, \text{km/h} \times 1.5 \, \text{h} = 600 \, \text{km}
\][/tex]
#### Step 2: Convert the bearing angles to radians
[tex]\[
\text{Bearing first leg} = 35^\circ = \frac{35 \pi}{180} \, \text{radians}
\][/tex]
[tex]\[
\text{Bearing second leg} = 130^\circ = \frac{130 \pi}{180} \, \text{radians}
\][/tex]
#### Step 3: Calculate the coordinates
1. Coordinates of airport B relative to A:
[tex]\[
x_{B} = \text{Distance first leg} \times \sin(35^\circ) = 900 \times \sin(35^\circ) = 516.22 \, \text{km}
\][/tex]
[tex]\[
y_{B} = \text{Distance first leg} \times \cos(35^\circ) = 900 \times \cos(35^\circ) = 737.24 \, \text{km}
\][/tex]
2. Coordinates of airport C relative to A:
[tex]\[
x_{C} = x_{B} + (\text{Distance second leg} \times \sin(130^\circ)) = 516.22 + 600 \times \sin(130^\circ) = 975.85 \, \text{km}
\][/tex]
[tex]\[
y_{C} = y_{B} + (\text{Distance second leg} \times \cos(130^\circ)) = 737.24 + 600 \times \cos(130^\circ) = 351.56 \, \text{km}
\][/tex]
#### Step 4: Calculate the distance from A to C
[tex]\[
\text{Distance AC} = \sqrt{x_C^2 + y_C^2} = \sqrt{(975.85)^2 + (351.56)^2} = 1037.24 \, \text{km}
\][/tex]
#### Step 5: Calculate the bearing of C from A
[tex]\[
\text{Bearing CA} = \tan^{-2}\left(\frac{x_C}{y_C}\right) = \tan^{-1}\left(\frac{975.85}{351.56}\right) = 70.19^\circ
\][/tex]
Summarized results:
(a) Distance from A to C: 1037.24 km
(b) Bearing of C from A: 70.19°
