Question

Given the following trigonometric ratio, enumerate the meaning ratio ​

Answer 1

Answer:

The trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

Step-by-step explanation:

From Trigonometry we know the following definitions for each trigonometric ratio:

Sine

$\sin \theta = \frac{y}{h}$ (1)

Cosine

$\cos \theta = \frac{x}{h}$ (2)

Tangent

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$ (3)

Cotangent

$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{y}$ (4)

Secant

$\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}$ (5)

Cosecant

$\csc \theta = \frac{1}{\sin \theta} = \frac{h}{y}$ (6)

Where:

$x$ - Adjacent leg.

$y$ - Opposite leg.

$h$ - Hypotenuse.

The length of the hypotenuse is determined by the Pythagorean Theorem:

$h = \sqrt{x^{2}+y^{2}}$

If $y = AC$ and $x = BC$, then the trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$