3 years ago

I need help on this test. Will give Brainliest to whoever answers first! Pt. 7

Mathematics

1 answer:

Gre4nikov [31]3 years ago

6 0

I think the first question is A)

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Right triangle XYZ has right angle Z. If the sin(X)=1213<br> , what is the cos(X)

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Given:

Right triangle XYZ has right angle Z.

$\sin(x)=\dfrac{12}{13}$

To find:

The value of $\cos x$ .

Solution:

We know that,

$\sin^2(x)+\cos^2(x)=1$

$\cos^2(x)=1-\sin^2(x)$

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For a triangle, all trigonometric ratios are positive. So,

$\cos(x)=\sqrt{1-\sin^2x}$

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$\cos(x)=\sqrt{1-(\dfrac{12}{13})^2}$

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$\cos(x)=\sqrt{\dfrac{169-144}{169}}$

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On further simplification, we get

$\cos(x)=\dfrac{\sqrt{25}}{\sqrt{169}}$

$\cos(x)=\dfrac{5}{13}$

Therefore, the required value is $\cos(x)=\dfrac{5}{13}$ .

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