Question

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in \\(\triangle abc\\), \\(a = 34\\) inches, \\(b = 40\\) inches and \\(c = 53\\) inches. find the measure of \\(\angle c\\) to the nearest 10th of a degree.
answer attempt 1 out of 3
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Explanation:

Step1: Identify the Law to Use

We use the Law of Cosines, which is \( c^{2}=a^{2}+b^{2}-2ab\cos(C) \). We need to solve for \( \angle C \), so we rearrange the formula to \( \cos(C)=\frac{a^{2}+b^{2}-c^{2}}{2ab} \).

Step2: Substitute the Values

Given \( a = 34 \), \( b = 40 \), and \( c = 53 \). Substitute these into the formula:
\( \cos(C)=\frac{34^{2}+40^{2}-53^{2}}{2\times34\times40} \)
First, calculate the squares: \( 34^{2}=1156 \), \( 40^{2}=1600 \), \( 53^{2}=2809 \).
Then, calculate the numerator: \( 1156 + 1600 - 2809 = 2756 - 2809=-53 \).
The denominator: \( 2\times34\times40 = 2720 \).
So, \( \cos(C)=\frac{-53}{2720}\approx - 0.019485 \).

Step3: Find the Angle

To find \( C \), we take the inverse cosine: \( C=\cos^{-1}(-0.019485) \). Using a calculator, we get \( C\approx91.1^{\circ} \) (to the nearest tenth of a degree).

Answer:

\( 91.1^{\circ} \)