Sure, I'd be happy to go through the steps for both parts of the problem.
### Part A: Find sin(2X) and compare with sin(LA)
To find [tex]\(\sin 2X\)[/tex], we first need to find [tex]\(\sin(X)\)[/tex], where [tex]\(X\)[/tex] is the angle in [tex]\( \triangle XYZ\)[/tex] opposite side [tex]\( YZ \)[/tex].
1. Calculate [tex]\(\sin(X)\)[/tex]:
Given:
[tex]\[
\sin(X) = \frac{YZ}{XZ}
\][/tex]
Substituting the given lengths [tex]\(YZ = 5\)[/tex] and [tex]\(XZ = 5.59\)[/tex],
[tex]\[
\sin(X) = \frac{5}{5.59} \approx 0.894
\][/tex]
2. Calculate [tex]\(\cos(X)\)[/tex]:
Using the identity [tex]\(\cos^2(X) = 1 - \sin^2(X)\)[/tex],
[tex]\[
\cos(X) = \sqrt{1 - \sin^2(X)} = \sqrt{1 - (0.894)^2} = \sqrt{1 - 0.799} \approx \sqrt{0.201} \approx 0.447
\][/tex]
3. Calculate [tex]\(\sin(2X)\)[/tex]:
Using the double-angle formula [tex]\(\sin(2X) = 2 \sin(X) \cos(X)\)[/tex],
[tex]\[
\sin(2X) = 2 \cdot 0.894 \cdot 0.447 \approx 0.8
\][/tex]
Next, we compare [tex]\(\sin(2X)\)[/tex] to [tex]\(\sin(LA)\)[/tex]. Since triangle ACB is a dilation of triangle XYZ by a scale factor of 2, the angles in the triangle ACB remain the same as those in triangle XYZ. Therefore,
[tex]\[
\sin(LA) = \sin(X) \approx 0.894
\][/tex]
### Part B: Find [tex]\(CB\)[/tex] and [tex]\(BA\)[/tex]
Since triangle ACB is a dilated version of triangle XYZ by a scale factor of 2, the sides of triangle ACB are twice the corresponding sides of triangle XYZ.
1. Calculate [tex]\(CB\)[/tex]:
Given that [tex]\(CB\)[/tex] corresponds to side [tex]\(YZ\)[/tex],
[tex]\[
CB = 2 \times YZ = 2 \times 5 = 10
\][/tex]
2. Calculate [tex]\(BA\)[/tex]:
Given that [tex]\(BA\)[/tex] corresponds to side [tex]\(XZ\)[/tex],
[tex]\[
BA = 2 \times XZ = 2 \times 5.59 \approx 11.18
\][/tex]
### Summary of Results
- [tex]\(\sin(X) \approx 0.894\)[/tex]
- [tex]\(\cos(X) \approx 0.447\)[/tex]
- [tex]\(\sin(2X) \approx 0.8\)[/tex]
- [tex]\(\sin(LA) \approx 0.894\)[/tex]
- [tex]\(CB = 10\)[/tex]
- [tex]\(BA = 11.18\)[/tex]
Therefore, we have successfully calculated [tex]\(\sin(2X)\)[/tex], compared it to [tex]\(\sin(LA)\)[/tex] and found the lengths of [tex]\(CB\)[/tex] and [tex]\(BA\)[/tex] in the dilated triangle ACB.
