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Lisa [10]
3 years ago
12

Please need help on this having a hard time with both of these

Mathematics
1 answer:
Arlecino [84]3 years ago
8 0
Question 1 - the first choice is right
Question 2- the first choice also
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Given the following trigonometric ratio, enumerate the meaning ratio ​
Likurg_2 [28]

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}

6 0
3 years ago
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