[tex]\begin{gathered} \\ 63.61\text{ \degree ,63.61 \degree and 52.78 \degree} \end{gathered}[/tex]
Explanation
to solve this we can use the law of cosines,it says
[tex]\begin{gathered} a^2=b^2+c^2-2bc\text{ *cos\lparen A\rparen} \\ b^2=a^2+c^2-2ac*\text{cos}\operatorname{\lparen}B\operatorname{\rparen} \\ c^2=a^2+b^2-2ab*\text{cos}\operatorname{\lparen}C\operatorname{\rparen} \end{gathered}[/tex]
Step 1
hence, let
[tex]\begin{gathered} a=\text{ 18} \\ b=18 \\ c=16 \end{gathered}[/tex]
now, let's find the angles
a) angle A)
[tex]\begin{gathered} a^{2}=b^{2}+c^{2}-2bc\text{cos\operatorname{\lparen}A\operatorname{\rparen}} \\ 18^2=18^2+16^2-2\left(18\right)\left(16\right)cos\lparen A) \\ 0=16^2-576*Cos\left(A\right) \\ -256=-576*Cos\left(A\right) \\ divide\text{ both sides by -576} \\ \frac{-256}{-576}=\frac{-576Cos(A)}{-576} \\ 0.444=cos\left(A\right? \\ \cos^{-1}\mleft(0.444\mright)=A \\ A=63.61\text{ \degree} \end{gathered}[/tex]
so
the first angle is 63.61 °
b) angle B)
[tex]\begin{gathered} b^2=a^2+c^2-2ac\text{cos}\operatorname{\lparen}B\operatorname{\rparen} \\ 18^2=18^2+16^2-2\left(18\right)\left(16\right)cos\lparen B) \\ 0=16^2-576*Cos\left(B\right) \\ -256=-576*Cos\left(B\right) \\ divide\text{ both sides by -576} \\ \frac{-256}{-576}=\frac{-576Cos(B)}{-576} \\ 0.444=cos\left(B\right? \\ \cos^{-1}\mleft(0.444\mright)=B \\ B=63.61\text{ \degree} \end{gathered}[/tex]
so,the second angle is 63.61
c)angle C)
[tex]\begin{gathered} c^2=a^2+b^2-2ab\text{cos}\operatorname{\lparen}C\operatorname{\rparen} \\ 16^2=18^2+18^2-2\left(18\right)\left(18\right)cos\lparen C) \\ 256=648-648cos\lparen C) \\ 256-648=-648*Cos\left(C\right) \\ divide\text{ both sides by -648} \\ \frac{-392}{-648}=\frac{-648Cos(C)}{-648} \\ 0.604=cos\left(C\right? \\ \cos^{-1}\mleft(0.604\mright)=C \\ C=52.78\text{ \degree} \end{gathered}[/tex]
so, the entire answer is
[tex]\begin{gathered} \\ 63.61\text{ \degree ,63.61 \degree and 52.78 \degree} \end{gathered}[/tex]
I hope this helps you