7. 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 (b) Whn's office and the town? (b) What is the bearing of $$ O $$ from $$ T $$ ?

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To solve this problem, we can use the concept of vectors and trigonometry.

Let's denote the office as point $$ O $$, the town as point $$ T $$, and the car's starting position as point $$ S $$.

Given:
- Distance from $$ O $$ to $$ S $$ is 120 km on a bearing of $$ 135^{\circ} $$.
- Distance from $$ S $$ to $$ T $$ is 100 km on a bearing of $$ 048^{\circ} $$.

We need to find:
(a) The distance between the office $$ O $$ and the town $$ T $$.
(b) The bearing of $$ O $$ from $$ T $$.

Let's break it down step by step:

1. Find the vector from $$ O $$ to $$ S $$:
   - The vector from $$ O $$ to $$ S $$ can be represented as $$ \vec{OS} = 120 	ext{ km} \angle 135^{\circ} $$.

2. Find the vector from $$ S $$ to $$ T $$:
   - The vector from $$ S $$ to $$ T $$ can be represented as $$ \vec{ST} = 100 	ext{ km} \angle 048^{\circ} $$.

3. Find the vector from $$ O $$ to $$ T $$:
   - The vector from $$ O $$ to $$ T $$ is the sum of the vectors from $$ O $$ to $$ S $$ and from $$ S $$ to $$ T $$:
     $$ \vec{OT} = \vec{OS} + \vec{ST} $$.

4. Calculate the magnitude of $$ \vec{OT} $$ to find the distance between $$ O $$ and $$ T $$.

5. Calculate the angle between $$ \vec{OT} $$ and the positive x-axis to find the bearing of $$ O $$ from $$ T $$.

Let's calculate the vectors and the distance between $$ O $$ and $$ T $$ step by step.
Calculate the value by following steps:
- step0: Calculate:
  $$120\cos\left(135ight)+100\cos\left(48ight)$$
- step1: Calculate the trigonometric value:
  $$-183.544974$$
Calculate or simplify the expression $$ 120 * \sin(135) + 100 * \sin(48) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$120\sin\left(135ight)+100\sin\left(48ight)$$
- step1: Calculate the trigonometric value:
  $$-66.221224$$
Calculate or simplify the expression $$ \sqrt{(-183.544974)^2 + (-66.221224)^2} $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{\left(-183.544974ight)^{2}+\left(-66.221224ight)^{2}}$$
- step1: Convert the expressions:
  $$\sqrt{\left(-\frac{91772487}{500000}ight)^{2}+\left(-66.221224ight)^{2}}$$
- step2: Convert the expressions:
  $$\sqrt{\left(-\frac{91772487}{500000}ight)^{2}+\left(-\frac{8277653}{125000}ight)^{2}}$$
- step3: Add the numbers:
  $$\sqrt{\frac{91772487^{2}+16	imes 8277653^{2}}{500000^{2}}}$$
- step4: Use the properties of radicals:
  $$\frac{\sqrt{91772487^{2}+16	imes 8277653^{2}}}{\sqrt{500000^{2}}}$$
- step5: Simplify the expression:
  $$\frac{\sqrt{91772487^{2}+16	imes 8277653^{2}}}{500000}$$
The distance between the office $$ O $$ and the town $$ T $$ is approximately 195.13 km.

Now, let's calculate the bearing of $$ O $$ from $$ T $$ by finding the angle between $$ \vec{OT} $$ and the positive x-axis.
Calculate the value by following steps:
- step0: Calculate:
  $$\arctan\left(\frac{-183.544974}{\left(-66.221224ight)}ight)$$
- step1: Remove the parentheses:
  $$\arctan\left(\frac{-183.544974}{-66.221224}ight)$$
- step2: Divide the terms:
  $$\arctan\left(\frac{91772487}{33110612}ight)$$
The bearing of $$ O $$ from $$ T $$ is approximately 1.2245 radians or 69.5 degrees.
The distance between the office $$ O $$ and the town $$ T $$ is approximately 195.13 km. The bearing of $$ O $$ from $$ T $$ is approximately 69.5 degrees.

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