# Given b=2,\ a=3, and B=40 (degrees), determine whether this information

Given $$\displaystyle{b}={2},\ {a}={3}$$, and $$\displaystyle{B}={40}$$ (degrees), determine whether this information results in one triangle, two triangles, or no triangle at all. Solve any resulting triangles.

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Step 1
Given thet, the values are $$\displaystyle{b}={2},\ {a}={3}$$ and $$\displaystyle{B}={40}^{{\circ}}$$
Recall that, the sine formula is $$\displaystyle{\frac{{{a}}}{{{\sin{{A}}}}}}={\frac{{{b}}}{{{\sin{{B}}}}}}={\frac{{{c}}}{{{\sin{{c}}}}}}$$
$$\displaystyle{\frac{{{\sin{{A}}}}}{{{a}}}}={\frac{{{\sin{{B}}}}}{{{b}}}}$$
$$\displaystyle{\sin{{A}}}={a}{\left({\frac{{{\sin{{B}}}}}{{{b}}}}\right)}$$
$$\displaystyle{A}={{\sin}^{{-{1}}}{\left({\frac{{{a}\times{\sin{{B}}}}}{{{b}}}}\right)}}$$
Step 2
Substitute $$\displaystyle{a}={3},\ {b}={2}$$ and $$\displaystyle{B}={40}^{{\circ}}$$ in $$\displaystyle{A}={{\sin}^{{-{1}}}{\left({\frac{{{a}\times{\sin{{B}}}}}{{{b}}}}\right)}}$$
$$\displaystyle{A}={{\sin}^{{-{1}}}{\left({\frac{{{3}\times{\sin{{\left({40}^{{\circ}}\right)}}}}}{{{2}}}}\right)}}$$
$$\displaystyle={{\sin}^{{-{1}}}{\left({0.9642}\right)}}$$
$$\displaystyle={1.3023}\times{\frac{{{180}^{{\circ}}}}{{\pi}}}$$
$$\displaystyle={74.6162}^{{\circ}}$$
$$\displaystyle{A}={74.6}^{{\circ}}$$
or
$$\displaystyle{A}={180}^{{\circ}}-{74.6}^{{\circ}}$$
$$\displaystyle{A}={105.4}^{{\circ}}$$
$$\displaystyle{\left(\because{105.4}^{{\circ}}+{40}^{{\circ}}{<}{180}^{{\circ}}\right)}$$</span>
Step 3
Let $$\displaystyle{A}={74.6}^{{\circ}}$$ then the angle of C is,
$$\displaystyle{C}={180}^{{\circ}}-{74.6}^{{\circ}}-{40}^{{\circ}}$$
Use the sine formula to find the value of c.
$$\displaystyle{\frac{{{\sin{{B}}}}}{{{b}}}}={\frac{{{\sin{{C}}}}}{{{c}}}}$$
$$\displaystyle{c}={\frac{{{b}{\sin{{C}}}}}{{{\sin{{A}}}}}}$$
$$\displaystyle{c}={\frac{{{2}{\sin{{65.4}}}^{{\circ}}}}{{{\sin{{40}}}^{{\circ}}}}}$$
$$\displaystyle{c}={2.83}$$
Thus, the possible values of the triangle is
$$\displaystyle{a}={3},\ {b}={2},\ {c}={2.83}$$ and $$\displaystyle{A}={74.6}^{{\circ}},\ {B}={40}^{{\circ}},\ {C}={65.4}^{{\circ}}$$
Step 4
Let $$\displaystyle{A}={105.4}^{{\circ}}$$ then the angle of C is,
$$\displaystyle{C}={180}^{{\circ}}-{105.4}^{{\circ}}-{40}^{{\circ}}$$
$$\displaystyle={34.6}^{{\circ}}$$
Use the sine formula to find the value of c.
$$\displaystyle{\frac{{{\sin{{B}}}}}{{{b}}}}={\frac{{{\sin{{C}}}}}{{{c}}}}$$
$$\displaystyle{c}={\frac{{{b}{\sin{{C}}}}}{{{\sin{{A}}}}}}$$
$$\displaystyle{c}={\frac{{{2}{\sin{{34.6}}}^{{\circ}}}}{{{\sin{{40}}}^{{\circ}}}}}$$
$$\displaystyle{c}={1},{77}$$
Thus, the possible values of the triangle is
$$\displaystyle{a}={3},\ {b}={2},\ {c}={1.77}$$ and $$\displaystyle{A}={105.4}^{{\circ}},\ {B}={40}^{{\circ}},\ {C}={34.6}^{{\circ}}$$
Therefore, the given values are possible for teo triangles.