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In the following right triangle, find $\sin A, \cos A, \tan A$, $\sin B, \cos B, $ and $\tan B.$

I'm confused by the $x$'s

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2 Answers 2

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Pythagorean Theorem: $$BC=\sqrt{AB^2-AC^2}=\sqrt{9x^2-x^2}=\sqrt{8x^2}=2\sqrt{2}x$$

$$\sin\angle A=\frac{BC}{AB}=\frac{2\sqrt{2}x}{3x}=\frac{2\sqrt{2}}{3}$$

$$\cos\angle A=\frac{AC}{AB}=\frac{x}{3x}=\frac{1}{3}$$

You can continue similarly.

$$\tan\angle A=\frac{2\sqrt{2}x}{x}, \sin\angle B=\frac{x}{3x}, \cos\angle B=\frac{2\sqrt{2}x}{3x}, \tan\angle B=\frac{x}{2\sqrt{2}x}$$

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The $x$ refers to the length of $AC$. The $3x$ says that the length of $BC$ is three times the length of $AC$. (I'm assuming $B$ is the unlabelled angled).

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