Question

Find sin(B), cos(B) and tan(B) for the right triangle ABC shown. Please show your work below.

Answer 1

Hello!

These are the following formulas to calculate sine, cosine, and tangent.

sin = opposite / hypotenuse

cos = adjacent / hypotenuse

tan = opposite / adjacent

To find sin(B), we need to find the side opposite of angle B, and the hypotenuse.

The opposite of angle B is 12, and the hypotenuse is 13.

sin(B) = 12/13

sin^-1(12/13) = 67.3801... This can be rounded to 67.38 degrees. (Tenths place: 67.4 degrees)

To find cos(B), we need to find the side adjacent of angle B, and the hypotenuse.

The adjacent side of angle B is 5, and the hypotenuse is 13.

cos(B) = 5/13

cos^-1(5/13) = 67.3801... It can be rounded to 67.38 degrees. Rounded to the tenths place is 67.4 degrees.

To find cos(B), we need to find the side adjacent and opposite of angle B.

The adjacent side of angle B is 5, and the opposite is 12.

tan(B) = (12/5)

tan^-1(12/5) = 67.3801... This can be rounded to about 67.38 degrees. Rounded to the tenths place is 67.4 degrees.

Therefore, the value of sin(B) is 67.38 degrees, the value of cos(B) is 67.38 degrees, and finally the value of tan(B) is 67.38 degrees.

Answer 2

Answer:

[tex] sin B=\frac[12}{13}[/tex]

$cos B=frac{5}{13}$

$tan B=\frac{12}{5}$

Step-by-step explanation:

We are given  a right triangle ABC

Perpendicular side=AC=12 units

Hypotenuse=AB=13 units

Base=BC=5 units

$sin B=\frac{opposite \;side \;to \;angle B}{hypotenuse}$

[tex] sin B=\frac[12}{13}[/tex]

$cos B=\frac{base}{hypotenuse}$

$cos B=frac{5}{13}$

$tan B=\frac{perpendicular}{base}$

$tan B=\frac{12}{5}$