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 = 3, c = 5 complete the table by using the names of the sides to express each trigonometric function as a ratio. sin b = \boxed{} csc b = \boxed{} cos b = \boxed{} sec b = \boxed{} tan b = \boxed{} cot b = \boxed{} the unknown side length: (simplify your answer, using any radicals. use integers or fractions for any numbers in the expression.) sin b = \frac{4}{5} (simplify your answer, using any radicals. use integers or fractions for any numbers in the expression.) cos b = \frac{3}{5} (simplify your answer, using any radicals. use integers or fractions for any numbers in the expression.) tan b = \frac{4}{3} (simplify your answer, using any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find unknown side $b$

In right triangle $ABC$ with right angle at $C$, $c$ is the hypotenuse ($c=5$), $a=AC=3$. Use Pythagorean theorem:
$$a^2 + b^2 = c^2$$
$$3^2 + b^2 = 5^2$$
$$9 + b^2 = 25$$
$$b^2 = 25 - 9 = 16$$
$$b = BC = 4$$

Step2: Define sides for $\angle B$

• Opposite to $\angle B$: $AC = a = 3$
• Adjacent to $\angle B$: $BC = b = 4$
• Hypotenuse: $AB = c = 5$

Step3: Calculate $\sin B$

Sine is opposite over hypotenuse.
$$\sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AC}{AB} = \frac{3}{5}$$

Step4: Calculate $\cos B$

Cosine is adjacent over hypotenuse.
$$\cos B = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{4}{5}$$

Step5: Calculate $\tan B$

Tangent is opposite over adjacent.
$$\tan B = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{BC} = \frac{3}{4}$$

Step6: Calculate $\csc B$

Cosecant is reciprocal of sine.
$$\csc B = \frac{1}{\sin B} = \frac{AB}{AC} = \frac{5}{3}$$

Step7: Calculate $\sec B$

Secant is reciprocal of cosine.
$$\sec B = \frac{1}{\cos B} = \frac{AB}{BC} = \frac{5}{4}$$

Step8: Calculate $\cot B$

Cotangent is reciprocal of tangent.
$$\cot B = \frac{1}{\tan B} = \frac{BC}{AC} = \frac{4}{3}$$

Answer:

Unknown side length: $4$
$\sin B = \frac{3}{5}$
$\cos B = \frac{4}{5}$
$\tan B = \frac{3}{4}$
$\csc B = \frac{5}{3}$
$\sec B = \frac{5}{4}$
$\cot B = \frac{4}{3}$

Side ratio equivalents:
$\sin B = \frac{AC}{AB}$, $\cos B = \frac{BC}{AB}$, $\tan B = \frac{AC}{BC}$
$\csc B = \frac{AB}{AC}$, $\sec B = \frac{AB}{BC}$, $\cot B = \frac{BC}{AC}$