Use the Law of Cosines to solve the triangle. Round your answers to two decimal places.

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asked Mar 2, 2023 41.9k views

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Despertar · Answer 1 · 2023-03-07T02:34:00+0000

ANSWER:

A = 34.09°

B = 39.84°

C = 106.07°

Explanation:

The Law of Cosines in its general form has the following form:

$a^2=b^2+c^2-2bc\cos A$

We apply the law of cosines for each angle and then solve for each of them as follows:

$\begin{gathered} a^(2)=b^(2)+c^(2)-2bc\cos(A) \\ \\ \cos(A)=(b^2+c^2-a^2)/(2bc) \\ \\ A=\cos^(-1)\:\left((b^2+c^2-a^2)/(2bc)\right) \\ \\ \text{ We replacing} \\ \\ A=\cos^(-1)\left((16^2+24^2-14^2)/(2\left(16\right)\left(24\right))\right)\: \\ \\ A=\cos^(-1)\left((53)/(64)\right)\:=34.09\degree \\ \\ \\ b^2=a^2+c^2-2ac\cos(B) \\ \\ \cos(B)=(a^2+c^2-b^2)/(2ac) \\ \\ B=\cos^(-1)\left((a^2+c^2-b^2)/(2ac)\right) \\ \\ \text{ We replacing} \\ \\ B=\cos^(-1)\left((14^2+24^2-16^2)/(2\left(14\right)\left(24\right))\right)\: \\ \\ B=\cos^(-1)\left((43)/(56)\right) \\ \\ B=39.84\degree \\ \\ \\ c^2=a^2+b^2-2ab\cos(C) \\ \\ \cos(C)=(a^2+b^2-c^2)/(2ab) \\ \\ C=\cos^(-1)\left((a^2+b^2-c^2)/(2ab)\right) \\ \\ \text{ We replacing:} \\ \\ C=\cos^(-1)\left((14^2+16^2-24^2)/(2\left(14\right)\left(16\right))\right)\: \\ \\ C=\cos^(-1)\left(-(31)/(112)\right) \\ \\ C=106.07\degree \end{gathered}$

Use the Law of Cosines to solve the triangle. Round your answers to two decimal places.

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