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rjkz [21]
2 years ago
12

What are the ratios of sine, cosine, and tangent for angle Y? sin(Y) = StartFraction X Z Over X Y EndFraction; cos(Y) = StartFra

ction Y Z Over X Z EndFraction; tan(Y) = StartFraction Y Z Over X Y EndFraction sin(Y) = StartFraction X Y Over X Z EndFraction; cos(Y) = StartFraction X Z Over X Y EndFraction; tan(Y) = StartFraction Y Z Over X Z EndFraction sin(Y) = StartFraction X Z Over X Y EndFraction; cos(Y) = StartFraction Y Z Over X Y EndFraction; tan(Y) = StartFraction X Z Over Y Z EndFraction sin(Y) = StartFraction Y Z Over X Y EndFraction; cos(Y) = StartFraction X Z Over X Y EndFraction; tan(Y) = StartFraction X Z Over Y Z EndFraction
Mathematics
2 answers:
Sholpan [36]2 years ago
7 0

Answer:

\sin Y= \frac{XZ}{XY}

\cos Y= \frac{YZ}{XY}

\tan Y= \frac{XZ}{YZ}

Step-by-step explanation:

Given

See attachment for triangle

Required

Find \sin, \cos and \tan of angle Y

For angle Y:

Opposite = XZ

Adjacent = YZ

Hypotenuse = XY

The \sin of an angle is calculated as:

\sin\theta = \frac{Opposite}{Hypotenuse}

So:

\sin Y= \frac{XZ}{XY}

The \cos of an angle is calculated as:

\cos\theta = \frac{Adjacent}{Hypotenuse}

So:

\cos Y= \frac{YZ}{XY}

The \tan of an angle is calculated as:

\tan\theta = \frac{Opposite}{Adjacent}

So:

\tan Y= \frac{XZ}{YZ}

12345 [234]2 years ago
4 0

Answer:

c

Step-by-step explanation:

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let's say she adds "x" lbs of water, to get "y" lbs for the 10% mixture.

how much is 25% of 10lbs?  well, (25/100) * 10, or 2.5.

the water has no sugar syrup in it, so is just pure water, so the amoun of syrup in it is 0%, how much is 0% of "x" lbs?  well, (0/100) * x, or 0.00x, which is just 0.

how much is 10% of "y" lbs?  well (10/100) * y, or 0.10y.

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we also know that the syrup amount in that is also 2.5 + 0.00x = 0.10y, thus


\bf \begin{array}{lccclll}
&\stackrel{lbs}{syrup}&\stackrel{concentration~\%}{syrup}&\stackrel{concentration}{amount}\\
&------&------&------\\
\textit{25\% syrup}&10&0.25&2.5\\
\textit{pur water}&x&0.00&0.00x\\
------&------&------&------\\
\textit{10\% mixture}&y&0.10&0.10y
\end{array}
\\\\\\
\begin{cases}
10+x=\boxed{y}\\
2.5+0.00x=0.10y\\
----------\\
2.5 = 0.10\left( \boxed{10+x} \right)
\end{cases}
\\\\\\
2.5=1 + 0.10x\implies 1.5=0.10x
\\\\\\
\cfrac{1.5}{0.10}=x\implies 15=x
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