Question

Find the values of the six trigonometric functions of an angle in standard position if the point with coordinates (12, 5) lies on its terminal side.

Answer 1

Using A^2+B^2=C^2
12^2+5^2=169
Hypotenuse= 13

Sin(Ø)= 5/13
Cos(Ø)= 12/13
Tan(Ø)= 5/12

csc(Ø)=13/5
sec(Ø)=13/12
cot(Ø)=12/5

Answer 2

Answer:

Part 1) $sin(\beta)=\frac{5}{13}$

Part 2) $cos(\beta)=\frac{12}{13}$

Part 3) $tan(\beta)=\frac{5}{12}$

Part 4) $cot(\beta)=\frac{12}{5}$

Part 5) $sec(\beta)=\frac{13}{12}$

Part 6) $csc(\beta)=\frac{13}{5}$

Step-by-step explanation:

see the attached figure to better understand the problem

In the right triangle ABC

we have

$AC=12\ units, BC=5\ units$

Applying the Pythagoras Theorem

Find the length side AB (hypotenuse)

$AB^{2}=AC^{2}+BC^{2}$

substitute the values

$AB^{2}=12^{2}+5^{2}$

$AB^{2}=169}$

$AB=13\ units$

Part 1) Find the $sin(\beta)$

$sin(\beta)=\frac{BC}{AB}$

substitute the values

$sin(\beta)=\frac{5}{13}$

Part 2) Find the $cos(\beta)$

$cos(\beta)=\frac{AC}{AB}$

substitute the values

$cos(\beta)=\frac{12}{13}$

Part 3) Find the $tan(\beta)$

$tan(\beta)=\frac{BC}{AC}$

substitute the values

$tan(\beta)=\frac{5}{12}$

Part 4) Find the $cot(\beta)$

$cot(\beta)=\frac{AC}{BC}$

substitute the values

$cotan(\beta)=\frac{12}{5}$

Part 5) Find the $sec(\beta)$

$sec(\beta)=\frac{1}{cos(\beta)}$

substitute the values

$sec(\beta)=\frac{13}{12}$

Part 6) Find the $csc(\beta)$

$csc(\beta)=\frac{1}{sin(\beta)}$

substitute the values

$csc(\beta)=\frac{13}{5}$