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

$b =\sqrt{21}\approx 4.58$ $A \approx 23.6°$ $B \approx 66.4°$
Since we already have the measures of two of the sides of the triangle, we can use the Pythagorean theorem to figure out $b$, one of the sides of the right triangle: $a^2 + b^2 = c^2$ Let's plug in the values we are given: $2^2 + b^2 = 5^2$ Evaluate the exponents: $4 + b^2 = 25$ Subtract $4$ from each side of the equation: $b^2 = 21$ Take the square root of both sides of the equation: $b = \sqrt {21}\approx 4.58$ To find the value of $A$, we can use one of the trig identities. Let us use the sine identity: $\sin A = \frac{\text{opposite}}{\text{hypotenuse}}$ = $\frac{a}{c}$ We know that $a$ is $2$ and $c$ is $5$. Let's plug in these values into our formula so that we can find $A$: $\sin A$ = $\frac{2}{5}$ Take the inverse sine of both sides of the equation: $A = \sin^{-1} \frac{2}{5}$ Evaluate to solve for $A$, remembering to round off to one decimal place: $A \approx 23.6°$ To find $B$, we can use the cosine identity: $\cos B = \frac{\text{adjacent}}{\text{hypotenuse}}$ = $\frac{a}{c}$ We plug in the values we know into this formula: $\cos B$ = $\frac{2}{5}$ Take the inverse cosine of both sides of the equation: $B = \cos^{-1} \frac{2}{5}$ Evaluate to solve for $B$, rounding the answer off to one decimal place: $B \approx 66.4°$