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1 year ago

Triangle JKL is transformed to create triangle J'K'L'. The angles in both triangles are shown.

Mathematics

1 answer:

Viefleur [7K]1 year ago

5 0

The correct statement is AngleL = 25° AngleL' = 25°.

We have given that,

Triangle JKL is transformed to create triangle J'K'L'.

We have to choose the correct option.

<h3>What is the information about the transformed triangle?</h3>

The following information should be considered:

In a rigid transformation, the image & pre-image are congruent.

Reflection, translation, and rotation are rigid transformations.

In a non-rigid transformation, the image and pre-image are similar.

Dilation is a non rigid transformation.

In a rigid or nonrigid transformation, the corresponding angles are the same.

If the corresponding sides are the same, then it is a rigid transformation.

If the corresponding sides are proportional, then it is a nonrigid transformation.

It can be a rigid or a nonrigid transformation based on whether the corresponding side lengths have the same measures.

Therefore option 3 is correct.

To learn more transformation of the triangle visit:

brainly.com/question/17429689

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3 years ago

Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0

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If you're using the app, try seeing this answer through your browser: brainly.com/question/2787701

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$\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\
\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\
\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}$

Square both sides:

$\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\
\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\
\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}$

$\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\
\mathsf{17\,sin^2\,\theta=1}\\\\
\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}$

Since $\mathsf{sin\,\theta}$ is positive, you can discard the negative sign. So,

$\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}$

Substitute this value back into $\mathsf{(i)}$ to find $\mathsf{cos\,\theta:}$

$\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}$

I hope this helps. =)

Tags: <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

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