Town Y is 9 km due north of town Z. Town X is 8km from Y, 5km from Z and somewhere to the west of the line YZ.

A) Draw triangle XYZ and find angle YZX

B) During an earthquake , town X moves due south until it is due west of Z. Find how far it has moved.

2. Originally Posted by <3<3
Town Y is 9 km due north of town Z. Town X is 8km from Y, 5km from Z and somewhere to the west of the line YZ.

A) Draw triangle XYZ and find angle YZX

B) During an earthquake , town X moves due south until it is due west of Z. Find how far it has moved.
A) Use law of cosines:

$z^2=x^2+y^2-2xy \cos Z$

$8^2=9^2+5^2-2(9)(5) \cos Z$

etc....

3. Hello, <3<3!

I assume you made sketches . . .

Town $Y$ is 9 km due north of town $Z.$
Town $X$ is 8km from $Y$, 5km from $Z$ and somewhere to the west of the line $YZ.$

A) Draw $\Delta XYZ$ and find $\angle YZX$
Code:
* Y
* |
8  *   |
*     |
*       | 9
*         |
X *           |
*        |
5 *   θ |
*  |
* Z

Let $\theta = \angle YZX$

Law of Cosines: . $\cos\theta \:=\:\frac{5^2 + 9^2 - 8^2}{2(5)(9)} \:=\:0.46666...$

Therefore: . $\theta \:\approx\:62.2^o$

B) During an earthquake, town $X$ moves due south until it is due west of $Z.$
Find how far it has moved.
Code:
* Y
* |
8  *   |
*     |
*       | 9
*         |
X *    5      |
:  *   62.2°|
:     *     |
:  27.8° *  |
W * - - - - - * Z

Town $X$ has moved directly south to $W.$

Since $\angle YZX = 62.2^o$, then: . $\angle XZW = 27.8^o$

In right triangle $XWZ\!:\;\;\sin27.8^o \:=\:\frac{XW}{5} \quad\Rightarrow\quad XW \:=\:5\sin27.8^o \:=\:2.3319...$

Therefore, town $X$ moved about 2.3 km.

4. Originally Posted by <3<3
Town Y is 9 km due north of town Z. Town X is 8km from Y, 5km from Z and somewhere to the west of the line YZ.

A) Draw triangle XYZ and find angle YZX

B) During an earthquake , town X moves due south until it is due west of Z. Find how far it has moved.
$\left( a \right){\text{ In }}\Delta {\text{XYZ, use the cosine law to find }}\angle {\text{YZX (}} = \angle {\text{Z):}} \hfill \\$

$\cos Z = \frac{{x^2 + y^2 - z^2 }}
{{2xy}} \hfill \\$

$\cos Z = \frac{{\left( 9 \right)^2 + \left( 5 \right)^2 - \left( 8 \right)^2 }}
{{2 \times 9 \times 5}} \hfill \\$

$Z = 62.2^\circ \hfill \\$

$\left( b \right){\text{ Town X has moved south, let us say, to point P, which is west of Z}}{\text{.}} \hfill \\$

${\text{In }}\Delta {\text{XPZ, }}\angle XZP = 90^\circ - 62.2^\circ = 27.8^\circ \hfill \\$

${\text{So, the distance moved by town X is }} = XP \hfill \\$

${\text{In }}\Delta {\text{XPZ, }}\frac{{XP}}
{{XZ}} = \sin 27.8^\circ \hfill \\$

$\frac{{XP}}
{5} = 0.46638664 \hfill \\$

$XP = 2.3{\text{ km}}{\text{.}} \hfill \\$

${\text{Distance moved by town X is 2}}{\text{.3 km}}{\text{.}} \hfill \\
$

5. Am I seeing double?!