Step 1

This problem can be solved using Law of sines of triangle. The Law of Sines is the relationship between the sides and angles of non-right triangles.

Simply, it states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides and angles in a given triangle.

for the triangle \(\displaystyle\triangle{A}{B}{C}\) shown in figure with side lengths a, b, c the law of sines is given as,

\(\displaystyle{\frac{{{a}}}{{{\sin{{A}}}}}}={\frac{{{b}}}{{{\sin{{B}}}}}}={\frac{{{c}}}{{{\sin{{C}}}}}}\)

Step 2

In the question, side \(\displaystyle{b}={4},\ {c}={5},\ {B}={30}^{{\circ}}\)

applying law of sines,

\(\displaystyle{\frac{{{4}}}{{{\sin{{30}}}^{{\circ}}}}}={\frac{{{5}}}{{{\sin{{C}}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{4}}}{{{0.5}}}}={\frac{{{5}}}{{{\sin{{C}}}}}}\)

\(\displaystyle\Rightarrow{8}={\frac{{{5}}}{{{\sin{{C}}}}}}\)

\(\displaystyle\Rightarrow{\sin{{C}}}={\frac{{{5}}}{{{8}}}}={0.625}\)

\(\displaystyle\Rightarrow{C}={{\sin}^{{-{1}}}{\left({0.625}\right)}}={38.68}^{{\circ}}\)

The sum of angles of triangle \(\displaystyle={180}^{{\circ}}\)

therefore

\(\displaystyle\angle{A}={180}-{38.68}-{30}\)

\(\displaystyle\angle{A}={11.32}^{{\circ}}\)

now using Law of sines we can find length of third side.

\(\displaystyle{\frac{{{a}}}{{{\sin{{A}}}}}}={\frac{{{b}}}{{{\sin{{B}}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{a}}}{{{\sin{{111.32}}}}}}={\frac{{{4}}}{{{\sin{{30}}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{a}}}{{{0.931}}}}={\frac{{{4}}}{{{0.5}}}}\)

\(\displaystyle\Rightarrow{a}={\frac{{{4}}}{{{0.5}}}}\times{0.931}={7.45}\)

Step 3

Now let us see if we can have another triangle. We have \(\displaystyle{B}={30}^{{\circ}}\) given

Now subtract already calculated C from \(\displaystyle{180}^{{\circ}}\)

\(\displaystyle\Rightarrow{C}_{{{2}}}={180}^{{\circ}}-{30}^{{\circ}}-{141.32}^{{\circ}}\)

therefore the third angle will be,

\(\displaystyle{A}_{{{2}}}={180}^{{\circ}}-{30}^{{\circ}}-{141.32}^{{\circ}}\)

\(\displaystyle{A}_{{{2}}}={8.68}^{{\circ}}\)

\(\displaystyle{\frac{{{a}_{{{2}}}}}{{{\sin{{A}}}}}}={\frac{{{b}}}{{{\sin{{B}}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{a}_{{{2}}}}}{{{\sin{{8.68}}}}}}={\frac{{{4}}}{{{\sin{{30}}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{a}_{{{2}}}}}{{{0.15}}}}={\frac{{{4}}}{{{0.5}}}}\)

\(\displaystyle\Rightarrow{a}_{{{2}}}={\frac{{{4}}}{{{0.5}}}}\times{0.15}={1.207}\)

Step 4

Therefore there are two triangles possible for the given measurements \(\displaystyle{b}={4},\ {c}={5},\ {B}={30}^{{\circ}}\)

Triangle 1

\(\displaystyle{C}_{{{1}}}={38.68}^{{\circ}}\)

\(\displaystyle{A}_{{{1}}}={111.32}^{{\circ}}\)

\(\displaystyle{a}_{{{1}}}={7.45}\)

Triangle 2

\(\displaystyle{C}_{{{2}}}={141.32}^{{\circ}}\)

\(\displaystyle{A}_{{{2}}}={8.68}^{{\circ}}\)

\(\displaystyle{a}_{{{2}}}={1.207}\)

This problem can be solved using Law of sines of triangle. The Law of Sines is the relationship between the sides and angles of non-right triangles.

Simply, it states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides and angles in a given triangle.

for the triangle \(\displaystyle\triangle{A}{B}{C}\) shown in figure with side lengths a, b, c the law of sines is given as,

\(\displaystyle{\frac{{{a}}}{{{\sin{{A}}}}}}={\frac{{{b}}}{{{\sin{{B}}}}}}={\frac{{{c}}}{{{\sin{{C}}}}}}\)

Step 2

In the question, side \(\displaystyle{b}={4},\ {c}={5},\ {B}={30}^{{\circ}}\)

applying law of sines,

\(\displaystyle{\frac{{{4}}}{{{\sin{{30}}}^{{\circ}}}}}={\frac{{{5}}}{{{\sin{{C}}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{4}}}{{{0.5}}}}={\frac{{{5}}}{{{\sin{{C}}}}}}\)

\(\displaystyle\Rightarrow{8}={\frac{{{5}}}{{{\sin{{C}}}}}}\)

\(\displaystyle\Rightarrow{\sin{{C}}}={\frac{{{5}}}{{{8}}}}={0.625}\)

\(\displaystyle\Rightarrow{C}={{\sin}^{{-{1}}}{\left({0.625}\right)}}={38.68}^{{\circ}}\)

The sum of angles of triangle \(\displaystyle={180}^{{\circ}}\)

therefore

\(\displaystyle\angle{A}={180}-{38.68}-{30}\)

\(\displaystyle\angle{A}={11.32}^{{\circ}}\)

now using Law of sines we can find length of third side.

\(\displaystyle{\frac{{{a}}}{{{\sin{{A}}}}}}={\frac{{{b}}}{{{\sin{{B}}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{a}}}{{{\sin{{111.32}}}}}}={\frac{{{4}}}{{{\sin{{30}}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{a}}}{{{0.931}}}}={\frac{{{4}}}{{{0.5}}}}\)

\(\displaystyle\Rightarrow{a}={\frac{{{4}}}{{{0.5}}}}\times{0.931}={7.45}\)

Step 3

Now let us see if we can have another triangle. We have \(\displaystyle{B}={30}^{{\circ}}\) given

Now subtract already calculated C from \(\displaystyle{180}^{{\circ}}\)

\(\displaystyle\Rightarrow{C}_{{{2}}}={180}^{{\circ}}-{30}^{{\circ}}-{141.32}^{{\circ}}\)

therefore the third angle will be,

\(\displaystyle{A}_{{{2}}}={180}^{{\circ}}-{30}^{{\circ}}-{141.32}^{{\circ}}\)

\(\displaystyle{A}_{{{2}}}={8.68}^{{\circ}}\)

\(\displaystyle{\frac{{{a}_{{{2}}}}}{{{\sin{{A}}}}}}={\frac{{{b}}}{{{\sin{{B}}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{a}_{{{2}}}}}{{{\sin{{8.68}}}}}}={\frac{{{4}}}{{{\sin{{30}}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{a}_{{{2}}}}}{{{0.15}}}}={\frac{{{4}}}{{{0.5}}}}\)

\(\displaystyle\Rightarrow{a}_{{{2}}}={\frac{{{4}}}{{{0.5}}}}\times{0.15}={1.207}\)

Step 4

Therefore there are two triangles possible for the given measurements \(\displaystyle{b}={4},\ {c}={5},\ {B}={30}^{{\circ}}\)

Triangle 1

\(\displaystyle{C}_{{{1}}}={38.68}^{{\circ}}\)

\(\displaystyle{A}_{{{1}}}={111.32}^{{\circ}}\)

\(\displaystyle{a}_{{{1}}}={7.45}\)

Triangle 2

\(\displaystyle{C}_{{{2}}}={141.32}^{{\circ}}\)

\(\displaystyle{A}_{{{2}}}={8.68}^{{\circ}}\)

\(\displaystyle{a}_{{{2}}}={1.207}\)