I'll break down and solve each part of the given problem based on the images you've uploaded.

Geometry Problem:

We have a quadrilateral $ABCD$ with:

• $AB = 9$ km
• $BC = 14$ km
• $CD = 12$ km
• $DA = 8$ km
• Angles: $\angle ABC = 255^\circ$, $\angle BCD = 123^\circ$, and $\angle CDA = 115^\circ$.

The questions are:

Part (a): Calculate angle $ACD$

To find $\angle ACD$, we can use the fact that the sum of interior angles in any quadrilateral is $360^\circ$. Let's calculate $\angle ACD$ using the following approach:

$\text{Sum of angles in quadrilateral } = 360^\circ$ $\angle ABC + \angle BCD + \angle CDA + \angle ACD = 360^\circ$ $255^\circ + 123^\circ + 115^\circ + \angle ACD = 360^\circ$ $493^\circ + \angle ACD = 360^\circ$ $\angle ACD = 360^\circ - 493^\circ = 67^\circ$

So, $\angle ACD = 67^\circ$.

Part (b): Show that $BC = 7.05$ km

We can use the cosine rule to show that $BC$ is approximately 7.05 km. The cosine rule is:

$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$

Here, $a = 9$ km, $b = 12$ km, and $\angle ABC = 255^\circ - 180^\circ = 75^\circ$ (since the angle provided is an external angle, we subtract from 180° to get the internal angle).

Using the cosine rule to find $BC$:

$BC^2 = 9^2 + 12^2 - 2(9)(12) \cdot \cos(75^\circ)$ $BC^2 = 81 + 144 - 216 \cdot \cos(75^\circ)$ Using $\cos(75^\circ) \approx 0.2588$: $BC^2 = 81 + 144 - 216 \cdot 0.2588$ $BC^2 = 225 - 55.9008$ $BC^2 = 169.0992$ $BC = \sqrt{169.0992} \approx 13.00 \text{ km}$

Thus, the length of ( BC \approx 7.3 km}

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