1 Answer

Nathanael Demacon · Answer 1 · 2022-09-10T23:03:15+0000

Answer:

$\sin \theta = 0.153$ ,
$\cos \theta = 0.988$ ,
$\tan \theta = 0.155$ ,
$\cot \theta = 6.462$ ,
$\sec \theta = 1.012$ ,
$\csc \theta = 6.538$

Explanation:

Let be the point
(x,y) , the six trigonometric functions of the angle are represented by following formulas:

$\sin \theta = \frac{y}{\sqrt{x^(2)+y^(2)}}$ (1)

$\cos \theta = \frac{x}{\sqrt{x^(2)+y^(2)}}$ (2)

$\tan \theta = (\sin \theta)/(\cos \theta) = (y)/(x)$ (3)

$\cot \theta = (1)/(\tan \theta) = (x)/(y)$ (4)

$\sec \theta = (1)/(\cos \theta) = \frac{\sqrt{x^(2)+y^(2)}}{x}$ (5)

$\csc \theta = (1)/(\sin \theta) = \frac{\sqrt{x^(2)+y^(2)}}{y}$ (6)

If we know that
x = 84 and
y = 13 , then the values of the six trigonometric functions is:

$\sin \theta = \frac{y}{\sqrt{x^(2)+y^(2)}}$

$\sin \theta = 0.153$

$\cos \theta = \frac{x}{\sqrt{x^(2)+y^(2)}}$

$\cos \theta = 0.988$

$\tan \theta = (y)/(x)$

$\tan \theta = 0.155$

$\cot \theta = (x)/(y)$

$\cot \theta = 6.462$

$\sec \theta = \frac{\sqrt{x^(2)+y^(2)}}{x}$

$\sec \theta = 1.012$

$\csc \theta = \frac{\sqrt{x^(2)+y^(2)}}{y}$

$\csc \theta = 6.538$

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