Answer:
The angles of ABC from smallest to largest: 52.83°, 57.53°, 69.64°
Explanation:
Given:
AC = 18in, AB = 20 in, BC = 17 in
To find:
the angles of ABC from smallest to largest
To determine the angles, we will apply cosine rules:
[tex]a^2\text{ = b}^2\text{ + c}^2\text{ - 2bcCosA}[/tex][tex]\begin{gathered} c\text{ = side opposite angle C}=\text{ AB} \\ c\text{ = 20} \\ b\text{ = side opposite angle B = AC} \\ b\text{ = 18} \\ a\text{ = side oppsite angle A = BC} \\ a\text{ = 17} \end{gathered}[/tex][tex]\begin{gathered} 17^2\text{ = 18}^2\text{ + 20}^2-\text{ 2\lparen18\rparen\lparen20\rparen cos A} \\ \\ 289\text{ = 724 -720cosA} \\ \\ -435\text{ = -720cosA} \\ \\ cosA\text{ = }\frac{-435}{-720}\text{ = 0.6042} \\ \\ A\text{ = cos}^{-1}(0.6042) \\ \\ A=\text{ 52.83\degree} \end{gathered}[/tex][tex]\begin{gathered} To\text{ get B, we will use the formula:} \\ b^2\text{ = a}^2\text{ + c}^2\text{ - 2acCosB} \\ \\ 18^2\text{ = 17}^2\text{ + 20}^2\text{ - 2\lparen17\rparen\lparen20\rparen cosB} \\ \\ 324\text{ = 689 -680cosB} \\ \\ -365\text{ = -680cosB} \\ \\ cosB\text{ = }\frac{-365}{-680}=\text{ 0.5368} \\ \\ B\text{ = cos}^{-1}(0.5368) \\ \\ B=\text{ 57.53\degree} \end{gathered}[/tex][tex]\begin{gathered} To\text{ get C, wewill apply sum of angles in a triangle:} \\ A+\text{ B}+\text{ C}=\text{ 180\degree} \\ \\ 52.83°\text{ + 57.53\degree +}C=\text{ 180} \\ \\ 110.36\text{ + C = 180} \\ \\ C\text{ = 180 - 110.36} \\ \\ C=\text{ 69.64\degree} \end{gathered}[/tex]
The angles of ABC from smallest to largest: 52.83°, 57.53°, 69.64°
