A jet leaves a runway whose bearing is N26E from the control tower. After flying 4 miles, the jet turns 99 degrees and flies on a bearing of s55e for 6 miles. At that time what is the bearing from the jet from the control tower. I have no idea how to solve this please help!
How far did the jet go North?
4 cos 26 - 6 cos 55
= .1537 north of origin
How far did the jet go East?
4 sin 26 + 6 sin 55
= 6.67 east of origin
tan angle north of east = .1537/6.67
angle North of East = 1.32 degrees
angle East of North = 90 - 1.32 = 88.7 degrees.
That is from the control tower to the Jet
You have a typo and I can not tell which is from and which is to.
from the jet TO the control tower would be
South 88.7 West
To solve this problem, we can use trigonometry and vector addition. Here's how you can approach it step by step:
1. Draw a diagram:
- Start by drawing a line representing the runway, indicating the N26E bearing.
- Mark the starting point as the control tower and label it as A.
- Mark the point where the jet turns and label it as B.
- Draw a line segment from A to B to represent the first leg of the jet's path (4 miles).
- From point B, draw a line representing the second leg, indicating the S55E bearing.
- Mark the end point of the second leg and label it as C.
2. Measure the angles:
- Measure the angle between the runway and the first leg of the jet's path (N26E).
- Measure the angle between the first and second legs of the jet's path (99 degrees).
- Measure the angle between the second leg of the jet's path and the desired bearing from the control tower (which we need to find).
3. Determine the bearing from the control tower:
- Start by finding the direction of the jet's path in relation to the control tower.
- To do this, add the bearing angle of the first leg (N26E) to the turn angle (99 degrees).
- In this case, N26E is equivalent to a 64-degree angle (since the bearing is measured clockwise from the north direction).
- So, 64 degrees (from the first leg) + 99 degrees (from the turn) = 163 degrees.
4. Convert the bearing to compass notation:
- Since the bearing from the control tower is greater than 90 degrees, we subtract the angle from 360 degrees to get the final bearing.
- 360 degrees - 163 degrees = 197 degrees.
Therefore, the bearing from the jet to the control tower, at the end of the second leg, is S17W (South 17 degrees West), or simply 197 degrees in compass notation.
To solve this problem, we can use the concept of vector addition and trigonometry. Let's break down the given information step by step:
1. The jet leaves a runway with a bearing of N26E from the control tower. This means the angle between the runway and the north direction is 26 degrees to the east.
2. After flying 4 miles, the jet turns 99 degrees to the south-east (bearing S55E) and flies for a further 6 miles.
Now, let's find the resulting bearing from the control tower after these movements:
Step 1:
Since the jet flies 4 miles with a bearing of N26E, it means it moves 4 miles in the direction indicated by this bearing. We can represent this movement as a vector:
```
Vector 1: 4 miles in the N26E direction
```
Step 2:
After the turn, the jet flies on a bearing of S55E for 6 miles. This means it moves 6 miles in the direction indicated by this bearing. Again, we can represent this movement as a vector:
```
Vector 2: 6 miles in the S55E direction
```
Now, we need to find the total displacement vector to determine the resulting bearing. We can add Vector 1 and Vector 2 by considering their components in the north and east directions.
The north component of each vector can be calculated as follows (use cosine function):
```
North component of Vector 1 = 4 * cos(116) ≈ -1.7335
North component of Vector 2 = 6 * cos(45) ≈ 4.2426
```
The east component of each vector can be calculated as follows (use sine function):
```
East component of Vector 1 = 4 * sin(116) ≈ 3.5595
East component of Vector 2 = 6 * sin(45) ≈ 4.2426
```
Now, let's add the north and east components separately to get the total displacement vector:
North component of total displacement = (-1.7335) + 4.2426 ≈ 2.5091
East component of total displacement = 3.5595 + 4.2426 ≈ 7.8021
Using the north and east components, we can calculate the magnitude of the total displacement vector:
```
Magnitude of total displacement = sqrt((2.5091)^2 + (7.8021)^2) ≈ 8.286 miles
```
Finally, we can find the resulting bearing from the control tower by calculating the angle between the total displacement vector and the north direction using the arctan function:
```
Bearing from the control tower = arctan((7.8021) / (2.5091)) ≈ 72.2 degrees
```
Therefore, at that time, the bearing from the jet to the control tower is approximately N72.2E.