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

now S or B

`\begin{gathered} \cos B=(a^2+c^2-b^2)/(2ac) \\ \cos S=(10^2+14^2-16^2)/(2(10)(14)) \\ \cos S=(40)/(280) \\ S=\cos ^(-1)((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}`