Answer
$a \approx 4.47$
$A \approx 48.2°$
$B \approx 41.8°$
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 $a$, 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:
$B \approx 41.8°$
