Consider the triangle with sides a=1.6, b=2.3 and c=1.2 feet. You can use cosine theorem twice to find two unknown angles:
1.
[tex] b^2=a^2+c^2-2ac\cos \angle B,\\ (2.3)^2=(1.6)^2+(1.2)^2-2\cdot 1.6\cdot 1.2\cos \angle B,\\ 5.29=2.56+1.44-3.84\cos \angle B,\\ 3.84\cos \angle B=4-5.29=-1.29,\\\cos \angle B=-\dfrac{1.29}{3.84} =-0.336[/tex]
Then you can conclude that ∠B is obtuse (because [tex] \cos \angle B<0 [/tex]) and m∠B=110°.
2.
[tex] c^2=a^2+b^2-2ab\cos \angle C,\\ (1.2)^2=(1.6)^2+(2.3)^2-2\cdot 1.6\cdot 2.3\cos \angle C,\\ 1.44=2.56+5.29-7.36\cos \angle C,\\ 7.36\cos \angle C=7.85-1.44=6.41,\\\cos \angle C=\dfrac{6.41}{7.36} =0.871[/tex].
Then you can conclude that ∠C is acute (because [tex] \cos \angle C>0 [/tex]) and m∠C=29°.
3. m∠A+m∠B+m∠C=180°, then m∠A=180°-110°-29°=41°.
Answer: m∠A=41° (∠A is opposite to the side a=1.6), m∠B=110° (∠B is opposite to the side b=2.3) and m∠C=29° (∠C is opposite to the side c=1.2)
