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

$a \approx 4.47$ $A \approx 48.2°$ $B \approx 41.8°$
Since we already have the measures of two of the sides of the triangle, we can use the Pythagorean theorem to figure out $a$, one of the sides of the right triangle: $a^2 + b^2 = c^2$ Let's plug in the values we are given: $a^2 + 4^2 = 6^2$ Evaluate the exponents: $a^2 + 16 = 36$ Subtract $16$ from each side of the equation: $a^2 = 20$ Take the square root of both sides of the equation: $a = \sqrt {20}=2\sqrt5$ Evaluate the square root, rounding off to two decimal places: $a \approx 4.47$ To find the value of $A$, we can use one of the trig identities. Let us use the cosine identity: $\cos A = \frac{adjacent}{hypotenuse}$ = $\frac{b}{c}$ We know that $b$ is $4$ and $c$ is $6$. Let's plug in these values into our formula so that we can find $A$: $\cos A$ = $\frac{4}{6}$ Take the inverse cosine of both sides of the equation: $A$ = cos$^{-1} \frac{4}{6}$ Evaluate to solve for $A$, remembering to round off to one decimal place: $A \approx 48.2°$ To find $B$, we can use the sine identity: $\sin B = \frac{\text{opposite}}{\text{hypotenuse}}$ = $\frac{b}{c}$ We plug in the values we know into this formula: $\sin B$ = $\frac{4}{6}$ Take the inverse cosine of both sides of the equation: $B$ = $\sin^{-1} \frac{4}{6}$ Evaluate to solve for $B$, rounding the answer off to one decimal place: $B \approx 41.8°$