Question

select the correct answer.
to prepare for a triathlon, amanda starts from position a and rides her bike along a straight road for 12 miles to reach position b. at b, she turns left and rides along another straight road for 15 miles to reach position c. at c, she turns left again and rides 20 miles along a straight road to return to a. in △abc, what are m∠a, m∠b, and m∠c, respectively?
a. 48.35°, 94.94°, 36.71°
b. 35.41°, 67.87°, 76.72°
c. 51.05°, 70.66°, 58.29°
d. 15.97°, 81.89°, 82.14°

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines is $c^{2}=a^{2}+b^{2}-2ab\cos C$, where $a, b, c$ are the side - lengths of a triangle and $C$ is the angle opposite to side $c$. Let $AB = c = 12$, $BC=a = 15$, and $AC = b = 20$.
First, find $\cos A$:
\[

$$\begin{align*} a^{2}&=b^{2}+c^{2}-2bc\cos A\\ \cos A&=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ &=\frac{20^{2}+12^{2}-15^{2}}{2\times20\times12}\\ &=\frac{400 + 144-225}{480}\\ &=\frac{319}{480}\approx0.6646 \end{align*}$$

\]
$A=\cos^{-1}(0.6646)\approx48.35^{\circ}$

Step2: Find $\cos B$

\[

$$\begin{align*} b^{2}&=a^{2}+c^{2}-2ac\cos B\\ \cos B&=\frac{a^{2}+c^{2}-b^{2}}{2ac}\\ &=\frac{15^{2}+12^{2}-20^{2}}{2\times15\times12}\\ &=\frac{225 + 144 - 400}{360}\\ &=\frac{-31}{360}\approx - 0.0861 \end{align*}$$

\]
$B=\cos^{-1}(-0.0861)\approx94.94^{\circ}$

Step3: Find $\cos C$

\[

$$\begin{align*} c^{2}&=a^{2}+b^{2}-2ab\cos C\\ \cos C&=\frac{a^{2}+b^{2}-c^{2}}{2ab}\\ &=\frac{15^{2}+20^{2}-12^{2}}{2\times15\times20}\\ &=\frac{225+400 - 144}{600}\\ &=\frac{481}{600}\approx0.8017 \end{align*}$$

\]
$C=\cos^{-1}(0.8017)\approx36.71^{\circ}$

Answer:

A. $48.35^{\circ},94.94^{\circ},36.71^{\circ}$