Question

Question: solve ABC below using either the law of sines or law of cosines. Round to the nearest tenth.

Answer 1

First we calculate for a with cosines rule

`\begin{gathered} a^2=b^2+c^2-2bcCosA \\ b=\text{ 29yd, c= 23yd, A= 99}\degree \\ a^2=(29)^2+(23)^2-2(29)(23)Cos(99) \\ a^2=841^{}+529^{}-1334*(-0.1564) \\ a^2=841^{}+529^{}-1334*(-0.1564) \\ a^2=841^{}+529^{}+208.68 \\ a^2=1578.686 \\ a=\text{ }\sqrt[]{1578.686} \\ a=\text{ 39.73} \end{gathered}`

Then we calculate angle B with sine rule

`\begin{gathered} (a)/(\sin A)=\text{ }(b)/(\sin B) \\ A=\text{ 99}\degree,\text{ a= 39.73 yd, b= 29 yd} \\ (39.73)/(\sin 99)=\text{ }(29)/(\sin B) \\ \text{Cross multiplying} \\ \sin \text{ B= }(29\sin 99)/(39.73) \\ \sin \text{ B= }(29*0.987)/(39.73) \\ \sin \text{ B= 0.720} \\ B=\text{ }\sin ^(-1)(0.720) \\ B=\text{ 46.05}\degree \end{gathered}`

Then we find the last angle C

`\begin{gathered} A\text{ + B + C= 180}\degree \\ 99\text{ + 46.05}\degree\text{ + C= 180}\degree \\ C=\text{ 180}\degree-\text{ 99}\degree-\text{ 46.05}\degree \\ C=\text{ 34.95}\degree \end{gathered}`

Final answers:

Side

a= 39.7 yd (nearest tenth)

Angles

B= 46.1 degrees (nearest tenth)

C= 35.0 degrees (nearest tenth)