Answer
$b=\sqrt {13}$
$\sin B=\frac{\sqrt {13}}{7}$
$\cos B=\frac{6}{7}$
$\tan B=\frac{\sqrt {13}}{6}$
$\csc B=\frac{7\sqrt {13}}{13}$
$\sec B=\frac{7}{6}$
$\cot B=\frac{6\sqrt {13}}{13}$
Work Step by Step
$b=\sqrt {c^{2}-a^{2}}=\sqrt {7^{2}-6^{2}}=\sqrt {13}$
$\sin B=\frac{\text{side opposite}}{\text{hypotenuse}}=\frac{b}{c}=\frac{\sqrt {13}}{7}$
$\cos B=\frac{\text{side adjacent}}{\text{hypotenuse}}=\frac{a}{c}=\frac{6}{7}$
$\tan B=\frac{\text{side opposite}}{\text{side adjacent}}=\frac{b}{a}=\frac{\sqrt {13}}{6}$
The cosecant, secant, and cotangent ratios are reciprocals of the sine, cosine, and tangent values, respectively, so we have
$\csc B=\frac{7}{\sqrt {13}}=\frac{7\sqrt {13}}{13}$
$\sec B=\frac{7}{6}$
$\cot B=\frac{6}{\sqrt {13}}=\frac{6\sqrt {13}}{13}$
