2. Emefa walked $$ 2 t \mathrm{~km} $$ from a village, P to $$ v_{i_{ ext {sit }}} $$ a friend in another village, Q on a bearing of $$ 065^{\circ} $$. After spending some time with her friend, she continued to a nearby town, R , $$ 3 t \mathrm{~km} $$ away on a bearing of $$ 155^{\circ} $$. If the distance between P and R is $$ 6 \sqrt{13} \mathrm{~km} $$ : (a) illustrate the information in a diagram; (b) calculate, correct to the nearest whole number, the: (i) value of $$ t $$; (ii) bearing of R from P .

Question

2. Emefa walked $$ 2 t \mathrm{~km} $$ from a village, P to $$ v_{i_{	ext {sit }}} $$ a friend in another village, Q on a bearing of $$ 065^{\circ} $$. After spending some time with her friend, she continued to a nearby town, R , $$ 3 t \mathrm{~km} $$ away on a bearing of $$ 155^{\circ} $$. If the distance between P and R is $$ 6 \sqrt{13} \mathrm{~km} $$ : (a) illustrate the information in a diagram; (b) calculate, correct to the nearest whole number, the: (i) value of $$ t $$; (ii) bearing of R from P .

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### **Given:**
- **From Village P to Village Q:**
  - Distance: $$2t$$ km
  - Bearing: $$065^\circ$$
  
- **From Village Q to Town R:**
  - Distance: $$3t$$ km
  - Bearing: $$155^\circ$$
  
- **Distance between Village P and Town R (PR):** $$6\sqrt{13}$$ km

---

### **Part (b)(i): Calculating the Value of $$t$$**

**1. **Establish a Coordinate System:**
   - Let’s place Village P at the origin $$(0,0)$$.
   - Bearings are measured clockwise from the north (positive y-axis).

**2. **Determine Coordinates of Village Q:**
   - **Bearing of $$065^\circ$$:**
     - $$x_Q = 2t \cdot \sin(65^\circ)$$
     - $$y_Q = 2t \cdot \cos(65^\circ)$$
     
   - **Approximations:**
     - $$\sin(65^\circ) \approx 0.9063$$
     - $$\cos(65^\circ) \approx 0.4226$$
     
   - **Coordinates:**
     - $$x_Q \approx 2t \cdot 0.9063 = 1.8126t$$
     - $$y_Q \approx 2t \cdot 0.4226 = 0.8452t$$

**3. **Determine Coordinates of Town R:**
   - **Bearing of $$155^\circ$$:**
     - $$x_R = x_Q + 3t \cdot \sin(155^\circ)$$
     - $$y_R = y_Q + 3t \cdot \cos(155^\circ)$$
     
   - **Approximations:**
     - $$\sin(155^\circ) = \sin(180^\circ - 25^\circ) = \sin(25^\circ) \approx 0.4226$$
     - $$\cos(155^\circ) = -\cos(25^\circ) \approx -0.9063$$
     
   - **Coordinates:**
     - $$x_R \approx 1.8126t + 3t \cdot 0.4226 = 3.0805t$$
     - $$y_R \approx 0.8452t + 3t \cdot (-0.9063) = -1.8737t$$

**4. **Calculate Distance PR:**
   - $$PR = \sqrt{(x_R)^2 + (y_R)^2} = 6\sqrt{13}$$
   - Squaring both sides:
     $$
     (3.0805t)^2 + (-1.8737t)^2 = (6\sqrt{13})^2
     $$
     $$
     9.489t^2 + 3.508t^2 = 468
     $$
     $$
     13.0t^2 = 468
     $$
     $$
     t^2 = \frac{468}{13} = 36
     $$
     $$
     t = \sqrt{36} = 6
     $$

**Answer:**  
**(b)(i) The value of $$ t $$ is **6 km**.

---

### **Part (b)(ii): Calculating the Bearing of R from P**

**1. **Determine Coordinates with $$ t = 6 $$:**
   - $$x_R = 3.0805 \times 6 = 18.483$$ km
   - $$y_R = -1.8737 \times 6 = -11.242$$ km

**2. **Calculate the Bearing:**
   - The bearing is the angle measured clockwise from the north.

**3. **Compute the Angle:**
   - $$\tan(\theta) = \frac{x_R}{|y_R|} = \frac{18.483}{11.242} \approx 1.642$$
   - $$\theta = \arctan(1.642) \approx 58^\circ$$

**4. **Determine the Bearing:**
   - Since $$y_R$$ is negative and $$x_R$$ is positive, Town R is in the southeast quadrant.
   - Bearing $$ = 180^\circ - 58^\circ = 122^\circ $$

**Answer:**  
**(b)(ii) The bearing of R from P is approximately **$$122^\circ$$**.
**(b)(i) The value of $$ t $$ is 6 km.** **(b)(ii) The bearing of R from P is approximately $$122^\circ$$.**

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Solución

Given:

Part (b)(i): Calculating the Value of

Part (b)(ii): Calculating the Bearing of R from P

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