Answer
b = $\sqrt 3$
$\sin$$B$ = $\frac{\sqrt 3}{2}$
$\cos$$B$ = $\frac{1}{2}$
$\tan$$B$ = $\frac{\sqrt 3}{1}$
$\csc$$B$ = $\frac{2\sqrt 3}{3}$
$\sec$$B$ = 2
$\cot$$B$ =$\frac{1}{\sqrt 3}$ = $\frac{\sqrt 3}{3}$
Work Step by Step
a = 1
c = 2
Pythagorean Theorem: c$^{2}$ = a$^{2}$ + b$^{2}$
2$^{2}$ = 1$^{2}$ + b$^{2}$
4 = 1 + b$^{2}$
b$^{2}$ = 3
b = $\sqrt 3$
$\sin$$B$ = $\frac{opposite}{hypotenuse}$ = $\frac{b}{c}$ = $\frac{\sqrt 3}{2}$
$\cos$$B$ = $\frac{adjacent}{hypotenuse}$ = $\frac{a}{c}$ = $\frac{1}{2}$
$\tan$$B$ = $\frac{opposite}{adjacent}$ = $\frac{b}{a}$ = $\frac{\sqrt 3}{1}$
$\csc$$B$ = $\frac{hypotenuse}{opposite}$ = $\frac{c}{b}$ = $\frac{2}{\sqrt 3}$ = $\frac{2\sqrt 3}{3}$
$\sec$$B$ = $\frac{hypotenuse}{adjacent}$ = $\frac{c}{a}$ = $\frac{2}{1}$ = 2
$\cot$$B$ = $\frac{adjacent}{opposite}$ = $\frac{a}{b}$ = $\frac{1}{\sqrt 3}$ = $\frac{\sqrt 3}{3}$
