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Use the Law of Cosines to solve the triangle.  Show all parts for the solution to the triangle.
Solve all the missing parts – write all values together including known/shown values as solution.
ABC-abc

Caution on using Law of Sines any place here.

To solve triangle \( ABC \) using the Law of Cosines, we are given:

- \( AB = c = ? \)
- \( AC = 8 \)
- \( BC = 15 \)
- \( \angle C = 40^\circ \)

### Step 1: Use the Law of Cosines to Find Side \( c \)
The Law of Cosines states:
\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]
where:
- \( a = AC = 8 \)
- \( b = BC = 15 \)
- \( \angle C = 40^\circ \)

Plugging in these values:
\[
c^2 = 8^2 + 15^2 - 2 \cdot 8 \cdot 15 \cdot \cos(40^\circ)
\]

1. Calculate \( 8^2 = 64 \).
2. Calculate \( 15^2 = 225 \).
3. Calculate \( 2 \cdot 8 \cdot 15 = 240 \).

Now, substitute these values:
\[
c^2 = 64 + 225 - 240 \cdot \cos(40^\circ)
\]

Using \( \cos(40^\circ) \approx 0.7660 \):
\[
c^2 = 64 + 225 - 240 \cdot 0.7660
\]
\[
c^2 = 289 - 183.84
\]
\[
c^2 = 105.16
\]
\[
c = \sqrt{105.16} \approx 10.26
\]

So, \( c \approx 10.26 \).

### Step 2: Use the Law of Cosines to Find \( \angle A \)
Now that we know all three sides (\( a = 8 \), \( b = 15 \), \( c \approx 10.26 \)), we can use the Law of Cosines to find \( \angle A \):
\[
\cos(A) = \frac{b^2 + c^2 - a^2}{2bc}
\]

Plugging in the values:
\[
\cos(A) = \frac{15^2 + 10.26^2 - 8^2}{2 \cdot 15 \cdot 10.26}
\]
1. Calculate \( 15^2 = 225 \).
2. Calculate \( 10.26^2 \approx 105.16 \).
3. Calculate \( 8^2 = 64 \).
4. Calculate \( 2 \cdot 15 \cdot 10.26 \approx 307.8 \).

Now substitute:
\[
\cos(A) = \frac{225 + 105.16 - 64}{307.8}
\]
\[
\cos(A) = \frac{266.16}{307.8} \approx 0.865
\]

Using the inverse cosine function:
\[
A \approx \cos^{-1}(0.865) \approx 30^\circ
\]

### Step 3: Find \( \angle B \) Using the Angle Sum in a Triangle
Since we now have two angles (\( \angle A \approx 30^\circ \) and \( \angle C = 40^\circ \)), we can find \( \angle B \):
\[
B = 180^\circ - A - C
\]
\[
B = 180^\circ - 30^\circ - 40^\circ
\]
\[
B = 110^\circ
\]

### Final Solution
The solution for triangle \( ABC \) is:

- \( AB = c \approx 10.26 \)
- \( AC = 8 \)
- \( BC = 15 \)
- \( \angle A \approx 30^\circ \)
- \( \angle B \approx 110^\circ \)
- \( \angle C = 40^\circ \)

Would you like more details or have any questions?

---

Here are five related questions for further understanding:

1. How does the Law of Cosines differ from the Pythagorean Theorem?
2. Why can't we use the Law of Sines directly to find \( \angle A \) or \( \angle B \) here?
3. How would we solve for missing parts if we had only two sides and one non-included angle?
4. What role does rounding play in finding precise values in trigonometric calculations?
5. Can we verify \( \angle B \) by using the Law of Sines after finding \( c \)?

**Tip:** When solving a triangle, confirm all angles add up to 180 degrees as a quick check for accuracy.

Use the Law of Cosines to solve the triangle. Show all parts for the solution to the triangle. Solve all the missing parts – write all values together including known/shown values as solution.

Learn how to solve triangle ABC with sides AC = 8, BC = 15, and angle C = 40°. Using the Law of Cosines, we find the missing side and angles, and provide a detailed solution, suitable for Grades 10-12.

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