Step 1

For a triangle with sides a, b, c and angles A, B, C the law of consines is defined as:

\(\displaystyle{c}^{{{2}}}={a}^{{{2}}}+{b}^{{{2}}}-{2}{a}{b}\times{\cos{{\left({C}\right)}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({C}\right)}}}={\frac{{{a}^{{{2}}}+{b}^{{{2}}}-{c}^{{{2}}}}}{{{2}{a}{b}}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({C}\right)}}}={\frac{{{4}^{{{2}}}+{6}^{{{2}}}-{9}^{{{2}}}}}{{{2}\times{4}\times{6}}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({C}\right)}}}=-{0.60416}\)

\(\displaystyle\Rightarrow{C}={{\cos}^{{-{1}}}{\left(-{0.60416}\right)}}\) (Because \(\displaystyle{\cos{{\left({C}\right)}}}\) is negative, C is an obtuse angle.)

\(\displaystyle\Rightarrow\angle{C}={127.17}^{{\circ}}\approx{127}^{{\circ}}\)

Step 2

Now we will use law ofsines to find angle A.

\(\displaystyle{\frac{{{\sin{{\left({A}\right)}}}}}{{{a}}}}={\frac{{{\sin{{\left({C}\right)}}}}}{{{c}}}}\)

\(\displaystyle\Rightarrow{\sin{{\left({A}\right)}}}={\frac{{{4}{\sin{{\left({127}^{{\circ}}\right)}}}}}{{{9}}}}\)

\(\displaystyle\Rightarrow{\sin{{\left({A}\right)}}}={0.35494}\)

\(\displaystyle\Rightarrow{A}={\arcsin{{\left({0.35494}\right)}}}\)

\(\displaystyle\Rightarrow\angle{A}={20.7898}^{{\circ}}\approx{21}^{{\circ}}\)

Now \(\displaystyle\angle{B}-{180}^{{\circ}}-\angle{A}-\angle{C}\)

\(\displaystyle={180}^{{\circ}}-{127}^{{\circ}}-{21}^{{\circ}}\)

\(\displaystyle\angle{B}={32}^{{\circ}}\)

Step 3

2nd Triangle: \(\displaystyle{a}={4},\ {b}={7},\ {c}={6}\)

First we will find angle B using law of cosines.

\(\displaystyle{b}^{{{2}}}={a}^{{{2}}}+{c}^{{{2}}}-{2}{a}{c}\times{\cos{{\left({B}\right)}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({B}\right)}}}={\frac{{{a}^{{{2}}}+{c}^{{{2}}}-{b}^{{{2}}}}}{{{2}{a}{c}}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({B}\right)}}}={\frac{{{4}^{{{2}}}+{6}^{{{2}}}-{7}^{{{2}}}}}{{{2}\times{4}\times{6}}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({B}\right)}}}={0.0625}\)

\(\displaystyle\Rightarrow{B}={{\cos}^{{-{1}}}{\left({0.0625}\right)}}\)

\(\displaystyle\Rightarrow\angle{B}={86.4164}^{{\circ}}\approx{86}^{{\circ}}\)

Step 4

Now we will find A by using law of sines.

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

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

\(\displaystyle{\sin{{\left({A}\right)}}}={\frac{{{4}{\sin{{\left({86}^{{\circ}}\right)}}}}}{{{7}}}}\)

\(\displaystyle{\sin{{\left({A}\right)}}}={0.57003}\)

\(\displaystyle{A}={\arcsin{{\left({0.57003}\right)}}}\)

\(\displaystyle\angle{A}={34.7528}^{{\circ}}\approx{35}^{{\circ}}\)

Now \(\displaystyle\angle{C}={180}^{{\circ}}-\angle{B}-\angle{A}\)

\(\displaystyle={180}^{{\circ}}-{86}^{{\circ}}-{35}^{{\circ}}={59}^{{\circ}}\)

\(\displaystyle\angle{C}={59}^{{\circ}}\)

For a triangle with sides a, b, c and angles A, B, C the law of consines is defined as:

\(\displaystyle{c}^{{{2}}}={a}^{{{2}}}+{b}^{{{2}}}-{2}{a}{b}\times{\cos{{\left({C}\right)}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({C}\right)}}}={\frac{{{a}^{{{2}}}+{b}^{{{2}}}-{c}^{{{2}}}}}{{{2}{a}{b}}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({C}\right)}}}={\frac{{{4}^{{{2}}}+{6}^{{{2}}}-{9}^{{{2}}}}}{{{2}\times{4}\times{6}}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({C}\right)}}}=-{0.60416}\)

\(\displaystyle\Rightarrow{C}={{\cos}^{{-{1}}}{\left(-{0.60416}\right)}}\) (Because \(\displaystyle{\cos{{\left({C}\right)}}}\) is negative, C is an obtuse angle.)

\(\displaystyle\Rightarrow\angle{C}={127.17}^{{\circ}}\approx{127}^{{\circ}}\)

Step 2

Now we will use law ofsines to find angle A.

\(\displaystyle{\frac{{{\sin{{\left({A}\right)}}}}}{{{a}}}}={\frac{{{\sin{{\left({C}\right)}}}}}{{{c}}}}\)

\(\displaystyle\Rightarrow{\sin{{\left({A}\right)}}}={\frac{{{4}{\sin{{\left({127}^{{\circ}}\right)}}}}}{{{9}}}}\)

\(\displaystyle\Rightarrow{\sin{{\left({A}\right)}}}={0.35494}\)

\(\displaystyle\Rightarrow{A}={\arcsin{{\left({0.35494}\right)}}}\)

\(\displaystyle\Rightarrow\angle{A}={20.7898}^{{\circ}}\approx{21}^{{\circ}}\)

Now \(\displaystyle\angle{B}-{180}^{{\circ}}-\angle{A}-\angle{C}\)

\(\displaystyle={180}^{{\circ}}-{127}^{{\circ}}-{21}^{{\circ}}\)

\(\displaystyle\angle{B}={32}^{{\circ}}\)

Step 3

2nd Triangle: \(\displaystyle{a}={4},\ {b}={7},\ {c}={6}\)

First we will find angle B using law of cosines.

\(\displaystyle{b}^{{{2}}}={a}^{{{2}}}+{c}^{{{2}}}-{2}{a}{c}\times{\cos{{\left({B}\right)}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({B}\right)}}}={\frac{{{a}^{{{2}}}+{c}^{{{2}}}-{b}^{{{2}}}}}{{{2}{a}{c}}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({B}\right)}}}={\frac{{{4}^{{{2}}}+{6}^{{{2}}}-{7}^{{{2}}}}}{{{2}\times{4}\times{6}}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({B}\right)}}}={0.0625}\)

\(\displaystyle\Rightarrow{B}={{\cos}^{{-{1}}}{\left({0.0625}\right)}}\)

\(\displaystyle\Rightarrow\angle{B}={86.4164}^{{\circ}}\approx{86}^{{\circ}}\)

Step 4

Now we will find A by using law of sines.

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

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

\(\displaystyle{\sin{{\left({A}\right)}}}={\frac{{{4}{\sin{{\left({86}^{{\circ}}\right)}}}}}{{{7}}}}\)

\(\displaystyle{\sin{{\left({A}\right)}}}={0.57003}\)

\(\displaystyle{A}={\arcsin{{\left({0.57003}\right)}}}\)

\(\displaystyle\angle{A}={34.7528}^{{\circ}}\approx{35}^{{\circ}}\)

Now \(\displaystyle\angle{C}={180}^{{\circ}}-\angle{B}-\angle{A}\)

\(\displaystyle={180}^{{\circ}}-{86}^{{\circ}}-{35}^{{\circ}}={59}^{{\circ}}\)

\(\displaystyle\angle{C}={59}^{{\circ}}\)