Question

Please click the image, this is one of my homework questions and I would like guidance, thank you!

Answer 1

Let the triangle be

First We will find angle B

Note 1: The sine rule

Using the sine rule

`\begin{gathered} a=12 \\ A=37^(\circ) \\ b=16.1 \\ B=\text{?} \end{gathered}`
`\begin{gathered} (\sin A)/(a)=(\sin B)/(b) \\ (\sin37)/(12)=(\sin B)/(16.1) \\ \text{cross multiply} \\ 12*\sin B=16.1*\sin 37 \\ 12\sin B=16.1*\sin 37 \\ \sin B=(16.1*\sin37)/(12) \\ \sin B=0.8074351561 \\ B=\sin ^(-1)(0.8074351561)_{} \end{gathered}`

Now, notice that

`\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}`

Now we return back to

`\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}`

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

`\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}`

To find the sides c and c'

First Triangle

`\begin{gathered} For\triangle\text{ABC} \\ u\sin g\text{ the sine rule} \\ (a)/(\sin A)=(c)/(\sin C) \\ (12)/(\sin37)=(c)/(\sin89) \\ \text{cross multiply} \\ c*\sin 37=12*\sin 89 \\ c=(12*\sin89)/(\sin37) \\ c=19.93664478 \\ c=19.9\text{ ( to the nearest tenth)} \end{gathered}`

Second Triangle

`\begin{gathered} For\triangle\text{A'B'C'} \\ u\sin g\text{ the sine rule} \\ (a)/(\sin A)=(c^(\prime))/(\sin C^(\prime)) \\ (12)/(\sin37)=(c^(\prime))/(\sin 17) \\ \text{cross multiply} \\ c^(\prime)*\sin 37=12*\sin 17 \\ c^(\prime)=(12*\sin 17)/(\sin 37) \\ c^(\prime)=5.829798728 \\ c^(\prime)=5.8\text{ ( to the nearest tenth)} \end{gathered}`