An aircraft flies round a triangular course. The fist leg is 200km on a bearing of 115degrees .the second leg is 150km on a bearing of 230degrees. How long is the third leg of the course and what bearing must The aircraft fly
make your sketch
On mine, using the cosine law, I get
x^2 = 200^2 + 150^2 - 2(200)(15)cos65°
..x = 192.72.. km
then by the sine law,
sinø/150 = sin65/192.72..
ø = 44.86..°
I will let you translate that angle into your bearing notation
other way.... using vectors and using standard trig notation
bearing of 115degrees ---> 25°
bearing of 230degrees ---> 220°
vector r = (200cos-25, 200sin-25) + (150cos220, 150sin220)
= (66.3548..., -180.9417...)
magnitude = √(66.3548...)^2 + (-180.9417...)^2 ) = 192.72 , just as before
angle...
tanx = -180.9417.../66.3548...
x = -68.86° translate that into your bearing notation
Thanks I really appreciate
All angles are measured CW from +y-axis.
D = 200[115o]+150[230o]
D = (200*sin115+150*sin230)+(200*cos115+150*cos230)i
D = 66.4-181i = 193km[20o].
Bearing(direction) = 180+20 = 200o.
To find the length of the third leg of the triangular course and the bearing the aircraft must fly, we need to break down the given information and use trigonometry.
1. First, let's draw the diagram of the triangular course with the given information:
B (230°)
.
/ \
/ \
/ \
A (115°)..........C..........A'
200km 150km
2. From the information given, we have:
- Leg AC is 200 km long and on a bearing of 115°.
- Leg CA' is 150 km long and on a bearing of 230°.
3. We need to find the length of leg C'A' and the bearing the aircraft must fly.
4. To find the length of C'A', we can use the law of cosines, which relates the lengths of the sides of a triangle to the cosine of one of its angles:
C'C² = CA'² + A'C² - 2 * CA' * A'C * cos(∠CA'C)
In this case, we know CA' = 150 km, A'C = 200 km, and ∠CA'C = 180° - 230° = -50°.
Since cosine is commutative, cos(-50°) = cos(50°), we can calculate C'C.
C'C² = (150)² + (200)² - 2 * 150 * 200 * cos(50°)
C'C² = 22500 + 40000 - 60000 * cos(50°)
C'C² = 62500 - 60000 * cos(50°)
Taking the square root of both sides gives us:
C'C = √(62500 - 60000 * cos(50°))
Using a calculator, we find that C'C ≈ 98.36 km (rounded to two decimal places).
5. To find the bearing the aircraft must fly, we can use trigonometry. Since we have the sides AC (200 km) and C'A' (98.36 km), we can use the tangent function to find the bearing. The tangent of an angle is the ratio of the opposite side to the adjacent side.
tan(∠AC'A') = opposite/adjacent = C'A'/AC
tan(∠AC'A') = 98.36/200
∠AC'A' = arctan(98.36/200)
Using a calculator, we find that ∠AC'A' ≈ 26.3° (rounded to one decimal place).
6. The bearing is the angle between the course line and the reference direction (usually true north). It is measured clockwise from the reference direction.
Since the first leg (AC) is on a bearing of 115°, we need to add the angle ∠AC'A' (26.3°) to determine the final bearing.
Final bearing = 115° + 26.3° ≈ 141.3° (rounded to one decimal place).
Therefore, the length of the third leg of the course is approximately 98.36 km, and the aircraft must fly on a bearing of approximately 141.3°.