Answer
$c \approx 8.25$
$A \approx 14.0°$
$B \approx 76.0°$
Work Step by Step
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:
$B \approx 76.0°$
