Answer
sin θ = $\frac{opposite}{hypotenuse}$ = $\frac{2\sqrt {13}}{13}$
cos θ = $\frac{adjacent}{hypotenuse}$ = $\frac{3\sqrt {13}}{13}$
tan θ = $\frac{opposite}{adjacent}$ = $\frac{2}{3}$
cot θ = $\frac{adjacent}{opposite}$ = $\frac{3}{2}$
sec θ = $\frac{hypotenuse}{adjacent}$ = $\frac{\sqrt {13}}{3}$
csc θ = $\frac{hypotenuse}{opposite}$ = $\frac{\sqrt {13}}{2}$
Work Step by Step
First, we need to find the value of the hypotenuse using the Pythagorean theorem, which states that $a^2 + b^2 = c^2$, where $a$ is the length of the side adjacent to the given angle, $b$ is the length of the side opposite the angle, and $c$ is the length of the hypotenuse.
In this triangle, $a$ is $3$ and $b$ is $2$, so let's plug these values into the formula to find the length of the hypotenuse:
$3^2 + 2^2 = c^2$
Evaluate the exponents:
$9 + 4 = c^2$
Add on the left side of the equation:
$c^2 = 13$
Take the square root of both sides of the equation:
$c = \sqrt {13}$
Now that we have the lengths of all sides of the right triangle, let's plug in the values into the formulas for the trigonometric functions:
sin θ = $\frac{opposite}{hypotenuse}$ = $\frac{2}{\sqrt {13}}$
We can't leave a radical in the denominator, so we multiply both the denominator and numerator by $\sqrt {13}$:
sin θ = $\frac{opposite}{hypotenuse}$ = $\frac{2\sqrt {13}}{13}$
cos θ = $\frac{adjacent}{hypotenuse}$ = $\frac{3}{\sqrt {13}}$
We can't leave a radical in the denominator, so we multiply both the denominator and numerator by $\sqrt {13}$:
cos θ = $\frac{adjacent}{hypotenuse}$ = $\frac{3\sqrt {13}}{13}$
tan θ = $\frac{opposite}{adjacent}$ = $\frac{2}{3}$
cot θ = $\frac{adjacent}{opposite}$ = $\frac{3}{2}$
sec θ = $\frac{hypotenuse}{adjacent}$ = $\frac{\sqrt {13}}{3}$
csc θ = $\frac{hypotenuse}{opposite}$ = $\frac{\sqrt {13}}{2}$
