Question

Find the remaining angle(s) and side(s) of each triangle, if it (they) exists. If no triangle exists, say “No triangle.”$a=1,b=\frac{1}{2},c=\frac{4}{3}$

Short answer

The remaining sides and angles are$$\begin{gathered} A=64 \\ B=26.7 \\ C=89.3 \end{gathered}$$

Step-by-step solution

Step 1.  Given information    

The given data is$a=1,b=\frac{1}{2},c=\frac{4}{3}$

Step 2. Find the angle A

Use the Law of cosines, $\cos A=\frac{b^2+c^2-a^2}{2bc}$

Here, $a=1,b=\frac{1}{2},c=\frac{4}{3}$

$$\begin{gathered} cosA=\frac{b^2+c^2-a^2}{2bc} \\ \cos A=\frac{{\left({\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{4}{3}}\right)}^2-1^2}{2*{\displaystyle \frac{1}{2}}*{\displaystyle \frac{4}{3}}} \\ \cos A=\frac{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{16}{9}}-1}{{\displaystyle \frac{4}{3}}} \\ \cos A=\frac{{\displaystyle \frac{9+48-36}{36}}}{{\displaystyle \frac{4}{3}}} \\ \cos A=\frac{21}{36}\times \frac{3}{4} \\ \cos A=0.43 \\ A=64 \end{gathered}$$

Step 3. Find the angle B

Use the Law of Sines, $\frac{\sin A}{a}=\frac{SinB}{b}=\frac{SinC}{c}$

Here, $$\begin{gathered} \frac{\sin A}{a}=\frac{SinB}{b}=\frac{SinC}{c} \\ \frac{\sin 64}{1}=\frac{SinB}{0.5} \\ SinB=0.5*\sin 64 \\ SinB=0.44 \\ B=26.7 \end{gathered}$$$A=64,a=1,b=0.5$

Step 4. Find the angle C

Since the angle A, B, and c are the angel of a triangle,

$$\begin{gathered} A+B+C=180 \\ 64+26.7+C=180 \\ C=89.3 \end{gathered}$$