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2 years ago

If the hypotenuse of a 30-60-90 triangle has length 15, then the shorter leg has length

Mathematics

1 answer:

umka2103 [35]2 years ago

3 0

Step-by-step explanation:

Given : If the hypotenuse of a 30-60-90 triangle has length 15 then the shorter leg has length 5.5 , 6.5 , 7.5.

To find : What is the length of the other leg and hypotenuse ?

Solution :

For a 30-60-90 triangle, sides are in the ratio $1:\sqrt{3}:2$

where, 1 is the shorter side,

$\sqrt{3}$ the other side

and 2 the hypotenuse.

Short leg = 5.5

Other leg = $5.5\times \sqrt3=5.5\sqrt3$

Hypotenuse = $5.5\times 2=11$

Short leg = 6.5

Other leg = $6.5\times \sqrt3=6.5\sqrt3$

Hypotenuse = $6.5\times 2=13$

Short leg = 7.5

Other leg = $7.5\times \sqrt3=7.5\sqrt3$

Hypotenuse = $7.5\times 2=15$

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
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with that template in mind, let's see $\bf f(x)=\sqrt{x}\qquad 
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&\quad B
\end{array}$

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