Question

suppose abc is a right triangle with sides a, b, and c and right angle at c. find the unknown side length using the pythagorean theorem and then find the values of the six trigonometric functions for angle b.
a = 5, c = 13
the unknown side length b is .
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
sin b =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
cos b =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
tan b =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
csc b =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
sec b =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
cot b =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find side length b using Pythagorean theorem

By the Pythagorean theorem $c^{2}=a^{2}+b^{2}$, so $b = \sqrt{c^{2}-a^{2}}$. Substitute $a = 5$ and $c = 13$: $b=\sqrt{13^{2}-5^{2}}=\sqrt{169 - 25}=\sqrt{144}=12$.

Step2: Find $\sin B$

$\sin B=\frac{b}{c}$, substituting $b = 12$ and $c = 13$, we get $\sin B=\frac{12}{13}$.

Step3: Find $\cos B$

$\cos B=\frac{a}{c}$, substituting $a = 5$ and $c = 13$, we get $\cos B=\frac{5}{13}$.

Step4: Find $\tan B$

$\tan B=\frac{b}{a}$, substituting $a = 5$ and $b = 12$, we get $\tan B=\frac{12}{5}$.

Step5: Find $\csc B$

$\csc B=\frac{c}{b}$, substituting $b = 12$ and $c = 13$, we get $\csc B=\frac{13}{12}$.

Step6: Find $\sec B$

$\sec B=\frac{c}{a}$, substituting $a = 5$ and $c = 13$, we get $\sec B=\frac{13}{5}$.

Step7: Find $\cot B$

$\cot B=\frac{a}{b}$, substituting $a = 5$ and $b = 12$, we get $\cot B=\frac{5}{12}$.

Answer:

The unknown side length b is $12$.
$\sin B=\frac{12}{13}$
$\cos B=\frac{5}{13}$
$\tan B=\frac{12}{5}$
$\csc B=\frac{13}{12}$
$\sec B=\frac{13}{5}$
$\cot B=\frac{5}{12}$