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Plz solve all questions correctly And solve mean solve each part step by step 

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}

 \ 
 
        Steps

Plz solve all questions correctly and solve each part step by step.

This step-by-step solution tackles a geometry problem involving a quadrilateral, using the Cosine Rule and the Angle Sum Theorem. Learn how to calculate unknown angles and distances, perfect for students in Grades 9-11.

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