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) Wan's office and the town?

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We start by representing each leg of the journey by converting bearing and distance into its east‐(x) and north‐(y) components. Recall that for a displacement $$ r $$ at a bearing $$\theta$$ (measured clockwise from north) the components are given by  
$$
x = r\sin\theta,\quad y=r\cos\theta.
$$

**First Leg: 120 km on a Bearing of $$ 135^\circ $$**

- The east component is  
  $$
  x_1 = 120\sin(135^\circ).
  $$
  Since $$\sin(135^\circ)=\sin(180^\circ-45^\circ)=\sin45^\circ=\frac{\sqrt{2}}{2}$$, we have  
  $$
  x_1 = 120\cdot\frac{\sqrt{2}}{2}=60\sqrt{2}.
  $$

- The north component is  
  $$
  y_1 = 120\cos(135^\circ).
  $$
  But $$\cos(135^\circ)=-\frac{\sqrt{2}}{2}$$, hence  
  $$
  y_1 = 120\cdot\left(-\frac{\sqrt{2}}{2}\right)=-60\sqrt{2}.
  $$

Thus, the displacement for the first leg is  
$$
(60\sqrt{2},\,-60\sqrt{2}).
$$

**Second Leg: 100 km on a Bearing of $$ 048^\circ $$**

- The east component is  
  $$
  x_2 = 100\sin(48^\circ).
  $$
- The north component is  
  $$
  y_2 = 100\cos(48^\circ).
  $$

Thus, the displacement for the second leg is  
$$
(100\sin48^\circ,\; 100\cos48^\circ).
$$

**Total Displacement**

We now add the components from both legs to get the overall displacement from Wan’s office (the starting point) to the town $$ T $$:

- Total east component:  
  $$
  x = 60\sqrt{2}+ 100\sin 48^\circ.
  $$

- Total north component:  
  $$
  y = -60\sqrt{2}+ 100\cos 48^\circ.
  $$

**Distance from the Office to the Town**

The straight‐line distance $$ D $$ between the office and the town is the magnitude of the resultant displacement:
$$
D = \sqrt{x^2 + y^2}.
$$

Substitute the expressions for $$ x $$ and $$ y $$:
$$
D = \sqrt{\left(60\sqrt{2}+ 100\sin48^\circ\right)^2+\left(-60\sqrt{2}+ 100\cos48^\circ\right)^2}.
$$

For an approximate numerical value, we use  
$$\sqrt{2}\approx 1.4142$$, $$\sin48^\circ\approx 0.7431$$ and $$\cos48^\circ\approx 0.6691$$.

Calculate the components:
- $$60\sqrt{2}\approx 60\times1.4142\approx 84.85$$ km,  
- $$100\sin48^\circ\approx 100\times0.7431\approx 74.31$$ km,  
- $$100\cos48^\circ\approx 100\times0.6691\approx 66.91$$ km.

Thus:
- Total east component:  
  $$
  x\approx 84.85+74.31=159.16\text{ km},
  $$
- Total north component:  
  $$
  y\approx -84.85+66.91=-17.94\text{ km}.
  $$

Now, compute $$ D $$:
$$
D\approx \sqrt{(159.16)^2+(-17.94)^2}\approx \sqrt{25333+321}=\sqrt{25654}\approx 160.17\text{ km}.
$$

**Answer**

The distance between Wan’s office and the town is approximately $$ 160 $$ km.
The distance between Wan’s office and the town is approximately 160 km.

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) Wan's office and the town?

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