From A to B, a private plane flies 2.6 hours at 110 mph whose bearing is N70E
 
 It turns at point B and continues another 2.8 hours at the same speed, but on a bearing of S20E
 
  to point C.  Round each answer to three decimal places.  

A) At the end of this time, how far is the plane from its starting point?   
 Miles

B) On what bearing (from due north) is the plane from its original location?  
 o

We can solve this problem by breaking it into two parts: the distance and direction (bearing) of each leg of the flight, and then combining them using trigonometry. 

### Part A: Distance Calculation

1. **First leg (A to B):**
   - The plane flies 2.6 hours at 110 mph on a bearing of N70E.
   - Distance \( d_1 \) traveled in the first leg is:
     \[
     d_1 = 110 \, \text{mph} \times 2.6 \, \text{hours} = 286 \, \text{miles}
     \]
   - Since the bearing is N70E, the direction forms an angle of 70° from the north (or 20° from the east). The components of this distance in the x (eastward) and y (northward) directions are:
     \[
     x_1 = d_1 \times \cos(70^\circ) = 286 \times \cos(70^\circ)
     \]
     \[
     y_1 = d_1 \times \sin(70^\circ) = 286 \times \sin(70^\circ)
     \]

2. **Second leg (B to C):**
   - The plane flies 2.8 hours at 110 mph on a bearing of S20E.
   - Distance \( d_2 \) traveled in the second leg is:
     \[
     d_2 = 110 \, \text{mph} \times 2.8 \, \text{hours} = 308 \, \text{miles}
     \]
   - The bearing S20E is 20° east of south, meaning it forms a 20° angle with the south (or 70° from the east). The components of this distance in the x (eastward) and y (southward) directions are:
     \[
     x_2 = d_2 \times \cos(20^\circ) = 308 \times \cos(20^\circ)
     \]
     \[
     y_2 = -d_2 \times \sin(20^\circ) = -308 \times \sin(20^\circ) \, \text{(since it's southward)}
     \]

3. **Total displacement:**
   The total displacement of the plane from the starting point is found by summing the x and y components:
   \[
   x_{\text{total}} = x_1 + x_2
   \]
   \[
   y_{\text{total}} = y_1 + y_2
   \]
   The straight-line distance (using the Pythagorean theorem) is:
   \[
   D = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}
   \]

### Part B: Bearing Calculation

The bearing of the plane from the original location can be found by calculating the angle of the resultant displacement vector from the north:
\[
\theta = \arctan \left( \frac{x_{\text{total}}}{y_{\text{total}}} \right)
\]
If the final x-component is positive and y-component is negative, the bearing will be measured clockwise from north, which will give us the bearing SθE.

---

Let me calculate these values for you.### Part A: Distance Calculation
The total distance the plane is from its starting point after both legs of the flight is approximately **420.309 miles**.

### Part B: Bearing Calculation
The bearing of the plane from its original location (measured clockwise from due north) is approximately **67.121°**.

Let me know if you'd like more details or have any questions!

---

Here are five related questions to expand your understanding:
1. How would the total distance change if the plane flew at a higher speed?
2. What happens to the bearing if the second leg's direction was S70E instead of S20E?
3. How would the total distance differ if the plane flew a shorter second leg (e.g., 2 hours instead of 2.8)?
4. Can we determine how long the entire journey took in minutes?
5. What would be the x and y components if the plane flew westward instead of eastward in the first leg?

**Tip:** When calculating bearings, always ensure you're measuring the angle from due north or south, adjusting for east/west directions accordingly.

From A to B, a private plane flies 2.6 hours at 110 mph whose bearing is N70E. It turns at point B and continues another 2.8 hours at the same speed, but on a bearing of S20E to point C. Round each answer to three decimal places. A) At the end of this time, how far is the plane from its starting point? B) On what bearing (from due north) is the plane from its original location?

Learn how to calculate the total distance and bearing of a plane after two flight legs, using trigonometric principles and vector decomposition. This problem involves key concepts like bearings, distance, and the Pythagorean theorem. Suitable for Grades 10-12.

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