Solution
Let the triangle be
First We will find angle B
Note 1: The sine rule
Using the sine rule
[tex]\begin{gathered} a=12 \\ A=37^{\circ} \\ b=16.1 \\ B=\text{?} \end{gathered}[/tex][tex]\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b} \\ \frac{\sin37}{12}=\frac{\sin B}{16.1} \\ \text{cross multiply} \\ 12\times\sin B=16.1\times\sin 37 \\ 12\sin B=16.1\times\sin 37 \\ \sin B=\frac{16.1\times\sin37}{12} \\ \sin B=0.8074351561 \\ B=\sin ^{-1}(0.8074351561)_{} \end{gathered}[/tex]
Now, notice that
[tex]\begin{gathered} \sin ^{-1}(0.8074351561)_{}=53.84609085 \\ \text{and} \\ \sin ^{-1}(0.8074351561)_{}=180-53.84609085 \\ \sin ^{-1}(0.8074351561)_{}=126.1539092 \end{gathered}[/tex]
Now we return back to
[tex]\begin{gathered} B=\sin ^{-1}(0.8074351561)_{} \\ B=53.84609085 \\ B=54\text{ (to the nearest degr}ees) \\ \text{and } \\ B=\sin ^{-1}(0.8074351561)_{} \\ B=126.1539092 \\ B=126\text{ ( to the nearest degr}ees) \\ Thus, \\ \\ B=54,126\text{ (to the nearest degr}ees) \end{gathered}[/tex]
Since we have two values of B, then this measurement produce two (2) triangles
The two triangles will be
We have already computed angle C and C' for the two triangles
The workings is as shown below
Note 2: Sum of angles in a triangle is 180 degrees
[tex]\begin{gathered} For\triangle\text{ABC} \\ 37+54+C=180 \\ 91+C=180 \\ C=180-91 \\ C=89^{\circ} \\ \text{Similarly} \\ For\triangle\text{A'B'C'} \\ 37+126+C^{\prime}=180 \\ 163+C^{\prime}=180 \\ C^{\prime}=17 \end{gathered}[/tex]
To find the sides c and c'
First Triangle
[tex]\begin{gathered} For\triangle\text{ABC} \\ u\sin g\text{ the sine rule} \\ \frac{a}{\sin A}=\frac{c}{\sin C} \\ \frac{12}{\sin37}=\frac{c}{\sin89} \\ \text{cross multiply} \\ c\times\sin 37=12\times\sin 89 \\ c=\frac{12\times\sin89}{\sin37} \\ c=19.93664478 \\ c=19.9\text{ ( to the nearest tenth)} \end{gathered}[/tex]
Second Triangle
[tex]\begin{gathered} For\triangle\text{A'B'C'} \\ u\sin g\text{ the sine rule} \\ \frac{a}{\sin A}=\frac{c^{\prime}}{\sin C^{\prime}} \\ \frac{12}{\sin37}=\frac{c^{\prime}}{\sin 17} \\ \text{cross multiply} \\ c^{\prime}\times\sin 37=12\times\sin 17 \\ c^{\prime}=\frac{12\times\sin 17}{\sin 37} \\ c^{\prime}=5.829798728 \\ c^{\prime}=5.8\text{ ( to the nearest tenth)} \end{gathered}[/tex]
