Question

(G.5.b, 1pt) Given: ARST, RS = 14 in, ST = 10 in, TR = 16 in Order the interior angles of ARST in order from smallest to largest. ZS ZR ZT Smallest Middle Largest

Answer 1

we have this formula to solve triangle angles from the sides

we need to identify who each is by comparing the pictures

so A=R , B=S, C=T , a=10, b=16 , c=14

the, replace

$\begin{gathered} \cos A=\frac{b^2+c^2-a^2}{2bc} \\ \cos R=\frac{16^2+14^2-10^2}{2(16)(14)} \\ \cos R=\frac{352}{448} \\ \\ R=\cos ^{-1}(\frac{11}{14}) \\ R=38.21^{\circ} \end{gathered}$

now S or B

$\begin{gathered} \cos B=\frac{a^2+c^2-b^2}{2ac} \\ \cos S=\frac{10^2+14^2-16^2}{2(10)(14)} \\ \cos S=\frac{40}{280} \\ S=\cos ^{-1}(\frac{1}{7}) \\ S=81.78^{\circ} \end{gathered}$

and finally C or T

$\begin{gathered} C=180-A-B \\ T=180-R-S \\ T=180-38.21-81.78 \\ T=60.01^{\circ} \end{gathered}$

next, the order ir

$\begin{gathered} R=38.21^{\circ} \\ T=60.01^{\circ} \\ S=81.78^{\circ} \end{gathered}$