A man drove a car from his office $$ (O) $$, 120 km on a bearing of $$ 135^{\circ} $$, and then drove for another 100 km on a bearing of $$ 048^{\circ} $$ until he reached a town $$ (T) $$. (a) What is the distance between the man's office and the town? (b) What is the bearing of $$ O $$ from $$ T $$ ?

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To solve this problem, we will break it down into two parts:

1. **Finding the distance between the man's office (O) and the town (T)**.
2. **Finding the bearing of O from T**.

### Step 1: Finding the distance between O and T

We can represent the movements of the man as vectors in a coordinate system.

1. **First leg of the journey**: 
   - Distance: $$ d_1 = 120 $$ km
   - Bearing: $$ 135^{\circ} $$

To convert the bearing to standard position (measured counterclockwise from the positive x-axis), we have:
   $$
   \theta_1 = 135^{\circ} - 90^{\circ} = 45^{\circ}
   $$

The coordinates of the endpoint after the first leg can be calculated as:
   $$
   x_1 = d_1 \cdot \cos(\theta_1) = 120 \cdot \cos(45^{\circ})
   $$
   $$
   y_1 = d_1 \cdot \sin(\theta_1) = 120 \cdot \sin(45^{\circ})
   $$

2. **Second leg of the journey**:
   - Distance: $$ d_2 = 100 $$ km
   - Bearing: $$ 048^{\circ} $$

Similarly, we convert the bearing:
   $$
   \theta_2 = 48^{\circ} - 90^{\circ} = -42^{\circ}
   $$

The coordinates of the endpoint after the second leg can be calculated as:
   $$
   x_2 = d_2 \cdot \cos(\theta_2) = 100 \cdot \cos(48^{\circ})
   $$
   $$
   y_2 = d_2 \cdot \sin(\theta_2) = 100 \cdot \sin(48^{\circ})
   $$

3. **Total coordinates of T**:
   The total coordinates of T can be found by adding the coordinates from both legs:
   $$
   x_T = x_1 + x_2
   $$
   $$
   y_T = y_1 + y_2
   $$

4. **Distance from O to T**:
   The distance $$ d $$ from O to T can be calculated using the Pythagorean theorem:
   $$
   d = \sqrt{x_T^2 + y_T^2}
   $$

### Step 2: Finding the bearing of O from T

To find the bearing of O from T, we need to calculate the angle from the positive north direction to the line connecting T to O. This can be done using the arctangent function:
$$
\text{bearing} = \tan^{-1}\left(\frac{y_T}{x_T}\right)
$$

Finally, we will convert this angle back to a bearing format.

Now, let's perform the calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$120\sin\left(45\right)$$
Calculate or simplify the expression $$ 120*\cos(45) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$120\cos\left(45\right)$$
Calculate or simplify the expression $$ 100*\sin(48) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$100\sin\left(48\right)$$
Calculate or simplify the expression $$ 100*\cos(48) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$100\cos\left(48\right)$$
Calculate or simplify the expression $$ \arctan((102.108423 + -76.825466)/(63.038639 + -64.014434)) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\arctan\left(\frac{\left(102.108423-76.825466\right)}{\left(63.038639-64.014434\right)}\right)$$
- step1: Remove the parentheses:
  $$\arctan\left(\frac{102.108423-76.825466}{63.038639-64.014434}\right)$$
- step2: Subtract the numbers:
  $$\arctan\left(\frac{25.282957}{63.038639-64.014434}\right)$$
- step3: Subtract the numbers:
  $$\arctan\left(\frac{25.282957}{-0.975795}\right)$$
- step4: Divide the terms:
  $$\arctan\left(-\frac{25282957}{975795}\right)$$
Calculate or simplify the expression $$ (63.038639 + -64.014434)^2 + (102.108423 + -76.825466)^2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(63.038639-64.014434\right)^{2}+\left(102.108423-76.825466\right)^{2}$$
- step1: Subtract the numbers:
  $$\left(-0.975795\right)^{2}+\left(102.108423-76.825466\right)^{2}$$
- step2: Subtract the numbers:
  $$\left(-0.975795\right)^{2}+25.282957^{2}$$
- step3: Convert the expressions:
  $$\left(-\frac{195159}{200000}\right)^{2}+25.282957^{2}$$
- step4: Convert the expressions:
  $$\left(-\frac{195159}{200000}\right)^{2}+\left(\frac{25282957}{1000000}\right)^{2}$$
- step5: Simplify:
  $$\frac{195159^{2}}{200000^{2}}+\frac{25282957^{2}}{1000000^{2}}$$
- step6: Reduce fractions to a common denominator:
  $$\frac{195159^{2}\times 25}{200000^{2}\times 25}+\frac{25282957^{2}}{1000000^{2}}$$
- step7: Multiply the terms:
  $$\frac{195159^{2}\times 25}{1000000^{2}}+\frac{25282957^{2}}{1000000^{2}}$$
- step8: Transform the expression:
  $$\frac{195159^{2}\times 25+25282957^{2}}{1000000^{2}}$$
- step9: Multiply the terms:
  $$\frac{975795^{2}+25282957^{2}}{1000000^{2}}$$
Round the number $$ -1.532221 $$ into 2 decimal places.
Round the number by following steps:
- step0: Round $$-1.532221$$ to $$2$$ decimal places using standard rounding method:
  $$-1.532221$$
- step1: Round $$-1.532221$$ to $$2$$ decimal places using standard rounding method$$:$$
  $$-1.53$$
Round the number $$ 640.180091 $$ into 2 decimal places.
Round the number by following steps:
- step0: Round $$640.180091$$ to $$2$$ decimal places using standard rounding method:
  $$640.180091$$
- step1: Round $$640.180091$$ to $$2$$ decimal places using standard rounding method$$:$$
  $$640.18$$
### Solution

#### Part (a): Distance between the man's office (O) and the town (T)

1. **First leg of the journey**:
   - $$ x_1 = 120 \cdot \cos(45^{\circ}) \approx 63.04 $$ km
   - $$ y_1 = 120 \cdot \sin(45^{\circ}) \approx 102.11 $$ km

2. **Second leg of the journey**:
   - $$ x_2 = 100 \cdot \cos(48^{\circ}) \approx 64.01 $$ km
   - $$ y_2 = 100 \cdot \sin(48^{\circ}) \approx 76.83 $$ km

3. **Total coordinates of T**:
   - $$ x_T = 63.04 + 64.01 \approx 127.05 $$ km
   - $$ y_T = 102.11 + 76.83 \approx 178.94 $$ km

4. **Distance from O to T**:
   $$
   d = \sqrt{x_T^2 + y_T^2} \approx \sqrt{(127.05)^2 + (178.94)^2} \approx 640.18 \text{ km}
   $$

#### Part (b): Bearing of O from T

1. **Calculating the angle**:
   $$
   \text{bearing} = \tan^{-1}\left(\frac{y_T}{x_T}\right) \approx -1.53 \text{ radians}
   $$

To convert this angle to a bearing:
   - Since the angle is negative, we add $$ 360^{\circ} $$ to find the equivalent positive angle.
   - The angle in degrees is approximately $$ -1.53 \times \frac{180}{\pi} \approx -87.69^{\circ} $$.
   - Adding $$ 360^{\circ} $$ gives us $$ 360 - 87.69 \approx 272.31^{\circ} $$.

### Final Answers
(a) The distance between the man's office and the town is approximately **640.18 km**.

(b) The bearing of O from T is approximately **272.31°**.
(a) The distance between the man's office and the town is approximately 640.18 km. (b) The bearing of O from T is approximately 272.31°.

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