In a right triangle you use te Pythagorean theoren:
[tex]c^2=a^{_{}2}+b^2[/tex]
As you have the value of sides b and c, solve the equation fot a:
[tex]\begin{gathered} a^2=c^2-b^2 \\ a=\sqrt[]{c^2-b^2} \end{gathered}[/tex]
b= 4
c= 7
[tex]a=\sqrt[]{7^2-4^2}=\sqrt[]{49-16}=\sqrt[]{33}[/tex]
Trigonometric functions for angle B.
Being a the adjacent and b the opposite, c is the hypotenuse. To rationalizing the denominator multiply the fraction by a fraction with the radical in numerator and denominator:
[tex]\begin{gathered} \sin B=\frac{b}{c}=\frac{4}{7} \\ \end{gathered}[/tex][tex]\cos B=\frac{a}{c}=\frac{\sqrt[]{33}}{7}[/tex][tex]\tan B=\frac{b}{a}=\frac{4}{\sqrt[]{33}}\cdot\frac{\sqrt[]{33}}{\sqrt[]{33}}=\frac{4\cdot\sqrt[]{33}}{33}[/tex][tex]\csc B=\frac{c}{b}=\frac{7}{4}[/tex][tex]\sec B=\frac{c}{a}=\frac{7}{\sqrt[]{33}}\cdot\frac{\sqrt[]{33}}{\sqrt[]{33}}=\frac{7\cdot\sqrt[]{33}}{33}[/tex][tex]\cot B=\frac{a}{b}=\frac{\sqrt[]{33}}{4}[/tex]
