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fredd [130]
3 years ago
10

How do you do this problems?

Mathematics
1 answer:
rewona [7]3 years ago
5 0

x = 6

Step-by-step explanation:

It's based on similarity of triangles

DEF ~ △XYZ

\implies \mathsf{ \frac{DE }{ XY } =  \frac{EF}{YZ} =  \frac{ FD}{ZX}}

The sides should be taken in order, like the triangles start with DE and XY so, we'll proportionate DE and XY and then the next two sides and so on.

I'm using \mathsf{ \frac{DE }{XY } =  \frac{EF}{YZ} }to get the value of x.

[You can use any two fractions out of the three given above. ]

\implies \mathsf{ \frac{10}{x - 1} =  \frac{8}{4}  }

\implies \mathsf{ \frac{10}{x - 1} =  2  }

\implies \mathsf{ 10 =  2 (x - 1) }

\implies \mathsf{ 10 =  2 x - 2 }

\implies \mathsf{ 10  + 2=  2 x  }

\implies \mathsf{ 1 2=  2 x  }

\implies \mathsf{ x = 6 }

Hence, the value of x is 6.

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Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0
eduard
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2787701

_______________


\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>

7 0
3 years ago
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