The ratio of the side adjacent am acute angle and the hypotenuse. Adjacent/hypotenuse

Mathematics

1 answer:

diamong [38]3 years ago

7 0

Answer: The answer is cosine of that acute angle.

Step-by-step explanation: We are to find the ratio of the adjacent side of an acute angle to the hypotenuse.

In the attached figure, we draw a right-angled triangle ABC, where ∠ABC is a right angle, and ∠ACB is an acute angle.

Now, side adjacent to ∠ACB is BC, which is the base with respect to this particular angle, and AC is the hypotenuse.

Now, the ratio is given by

$\dfrac{\textup{adjacent side}}{\textup{hypotenuse}}=\dfrac{BC}{AC}=\dfrac{\textup{base}}{\textup{hypotenuse}}=\cos\textup{ of angle }ACB.$

Thus, the ratio is cosine of the acute angle.

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Using simpler trigonometric identities, the given identity was proven below.

<h3>How to solve the trigonometric identity?</h3>

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$sec(x) = \frac{1}{cos(x)} \\\\tan(x) = \frac{sin(x)}{cos(x)}$

Then the identity can be rewritten as:

$sec^4(x) - sen^2(x) = tan^4(x) + tan^2(x)\\\\\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\$

Now we can multiply both sides by cos⁴(x) to get:

$\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\\\\\cos^4(x)*(\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}) = cos^4(x)*( \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)})\\\\1 - cos^2(x) = sin^4(x) + cos^2(x)*sin^2(x)\\\\1 - cos^2(x) = sin^2(x)*sin^2(x) + cos^2(x)*sin^2(x)$