Answer:
5.4 units
Explanation:
To find side c in triangle ABC, we can use the Law of Cosines, which states:
\[ c^2 = a^2 + b^2 - 2ab \cdot \cos(A) \]
Given that \( A = 23° \), \( a = 10 \), and \( b = 13 \), let's calculate c:
\[ c^2 = 10^2 + 13^2 - 2 \cdot 10 \cdot 13 \cdot \cos(23°) \]
Now, calculate the value inside the cosine function:
\[ \cos(23°) \approx 0.9205 \]
\[ c^2 = 100 + 169 - 2 \cdot 10 \cdot 13 \cdot 0.9205 \]
\[ c^2 = 100 + 169 - 240.13 \]
\[ c^2 = 28.87 \]
Finally, take the square root of both sides to find c:
\[ c \approx \sqrt{28.87} \approx 5.37 \]
So, side c is approximately 5.4 units (rounded to the nearest tenth).