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In triangle ABC, m/A = 23°, a = 10, and b= 13. Find c to the nearest tenth.

User Nyks
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1 Answer

2 votes

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).

User RidgeA
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