Answer
$a=3.0$ $;$ $b=4.0$ $;$ $c\approx3.25$
$\angle A\approx47.51^{\circ}$ $;$ $\angle B\approx79.49^{\circ}$ $;$ $\angle C=53^{\circ}$
Work Step by Step
$a=3.0$ $,$ $b=4.0$ $,$ $\angle C=53^{\circ}$
Find side $c$ by using the formula $c^{2}=a^{2}+b^{2}-2ab\cos C$, obtained from the Law of Cosines. Substitute the known values into the formula and solve for $c$:
$c^{2}=(3.0)^{2}+(4.0)^{2}-2(3.0)(4.0)\cos53^{\circ}$
$c=\sqrt{(3.0)^{2}+(4.0)^{2}-2(3.0)(4.0)\cos53^{\circ}}$
$c=\sqrt{9+16-24\cos53^{\circ}}\approx3.25$
Find angle $A$ by using the formula $a^{2}=b^{2}+c^{2}-2bc\cos A$, obtained from the Law of Cosines. Substitute the known values into the formula and solve for $\angle A$:
$(3.0)^{2}=(4.0)^{2}+(3.25)^{2}-2(4.0)(3.25)\cos A$
$2(4.0)(3.25)\cos A=16+10.5625-9$
$26\cos A=17.5625$
$\cos A=\dfrac{17.5625}{26}$
$A=\cos^{-1}\Big(\dfrac{17.5625}{26}\Big)\approx47.51^{\circ}$
Two angles are now known. Since $\angle A+\angle B+\angle C=180^{\circ}$, substitute the known angles into the formula and solve for $\angle B$:
$\angle B=180^{\circ}-53^{\circ}-47.51^{\circ}\approx79.49^{\circ}$
