A triangle ABC is such that AB = 10 cm,BC = 12 cm and angle ABC = 100 , Calculate the length of AC. (5 marks)

To find the length of side \( AC \) in triangle \( ABC \), we can apply the Law of Cosines. The Law of Cosines states:

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

Where:
- \( c \) is the side opposite angle \( C \) (which is \( AC \) in this case),
- \( a \) and \( b \) are the other two sides,
- \( C \) is the included angle.

Given:
- \( AB = 10 \, \text{cm} \) (side \( b \)),
- \( BC = 12 \, \text{cm} \) (side \( a \)),
- \( \angle ABC = 100^\circ \).

Let's plug these values into the formula:

\[
AC^2 = 10^2 + 12^2 - 2 \times 10 \times 12 \times \cos(100^\circ)
\]

Now, calculate step by step:

1. \( 10^2 = 100 \)
2. \( 12^2 = 144 \)
3. \( 2 \times 10 \times 12 = 240 \)
4. \( \cos(100^\circ) \) is approximately \(-0.1736\).

So,

\[
AC^2 = 100 + 144 - 240 \times (-0.1736)
\]

\[
AC^2 = 244 + 41.664
\]

\[
AC^2 = 285.664
\]

Finally, taking the square root:

\[
AC \approx \sqrt{285.664} \approx 16.9 \, \text{cm}
\]

So, the length of side \( AC \) is approximately **16.9 cm**.

Would you like further details or have any questions?

Here are 5 related questions to further explore this topic:
1. How would the length of \( AC \) change if angle \( \angle ABC \) was \( 60^\circ \) instead of \( 100^\circ \)?
2. What is the formula for the area of triangle \( ABC \) given these sides and angle?
3. Can the Law of Sines be used to solve this problem? If so, how?
4. How would you find the other angles of the triangle given this data?
5. What is the range of possible values for angle \( \angle ABC \) to make triangle \( ABC \) valid?

**Tip:** In trigonometry, the Law of Cosines is particularly useful when dealing with non-right triangles, especially when two sides and the included angle are known.

Learn how to calculate the length of side AC in triangle ABC using the Law of Cosines. Given AB = 10 cm, BC = 12 cm, and angle ABC = 100°, follow a step-by-step solution to find AC. Suitable for Grades 10-12.

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