Hello, Dee!

1.A hiker walks due north from $\displaystyle A$ and after 8km reaches $\displaystyle B.$

She then walks a further 8km on a bearing of 120° to $\displaystyle C.$

Find: (a) the distance from $\displaystyle A$ to $\displaystyle C$, and (b) the bearing of $\displaystyle C$ from $\displaystyle A.$ Code:

:
: 120°
B o
| * 8
| 60° *
| *
8 | o C
| *
| *
| *
A o

Triangle $\displaystyle ABC$ is equilateral.

We have all the information we need.

2. A helicopter flies on a bearing of 200° from A to B, where AB = 70km.

It then flies on a bearing of 150° from B to C, where C is due south of A.

Find the distance of C from A. Code:

P
:
:
o A
* |
Q 70 * 20° |
: * |
: 20° * |
: * |
B o 130° | x
: * |
: 30° * |
: * |
R * 30° |
* |
o C

Major angle $\displaystyle PAB = 200^o \quad\Rightarrow\quad \angle BAC = 20^o = \angle QBA$

Angle $\displaystyle QBC = 150^o \quad\Rightarrow\quad \angle RBC = 30^o = \angle BCA$

Hence: .$\displaystyle \angle ABC = 130^o$

Let $\displaystyle x = AC.$

Law of Sines: .$\displaystyle \frac{x}{\sin130^o} \:=\:\frac{70}{\sin30^o}$