## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$c \approx 8.25$ $A \approx 14.0°$ $B \approx 76.0°$
Since we already have the measures of two of the sides of the triangle, we can use the Pythagorean theorem to figure out $c$, the length of the hypotenuse: $a^2 + b^2 = c^2$ Let's plug in the values we are given: $2^2 + 8^2 = c^2$ Evaluate the exponents: $4 + 64 = c^2$ Add on the left side of the equation: $68 = c^2$ Take the square root of both sides of the equation: $c = \sqrt {68} \approx 8.25$ To find the value of $A$, we can use one of the trig identities. Let us use the tangent identity: $\tan A = \frac{\text{opposite}}{\text{adjacent}}$ = $\frac{a}{b}$ We know that $a$ is $2$ and $b$ is $8$. Let's plug in these values into our formula so that we can find $A$: tan $A$ = $\frac{2}{8}$ Take the inverse tangent of both sides of the equation: $A$ = tan$^{-1} \frac{2}{8}$ Evaluate to solve for $A$, remembering to round off to one decimal place: $A \approx 14.0°$ To find $B$, we can use the tangent identity again: tan $B = \frac{opposite}{adjacent}$ = $\frac{b}{a}$ We plug in the values we know into this formula: tan $B$ = $\frac{8}{2}$ Take the inverse tangent of both sides of the equation: $B$ = tan$^{-1} \frac{8}{2}$ Evaluate to solve for $B$, rounding the answer off to one decimal place: $B \approx 76.0°$