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diamong [38]
3 years ago
6

Millennium Park has an outdoor concert theater. Before a concert, the area reserved for special seating is roped off in the shap

e of a triangle as shown below. How can the converse of the Pythagorean theorem help you determine whether the roped off area is in the shape of a right triangle? (2 points)
Mathematics
1 answer:
nordsb [41]3 years ago
3 0

Answer:

The converse of Pythagorean Theorem states:

<em>If the square length of the longes side of a triangle is equal to the sum of the squares of the other two sides, then it's a right triangle.</em>

<em />

We use the converse of Pythagorean Theorem to prove that a triangle is a right triangle. It's like the inverse use of the theorem. Often, this theorem is used to find the length of a side of a triangle when we already know it's a right triangle. However, when we don't know that is a right triangle, we use the converse to prove it.

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Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas.
Sedaia [141]
\large\begin{array}{l} \textsf{Solve the equation for x:}\\\\ &#10;\mathsf{(tan\,x)^2+2\,tan\,x-4.76=0}\\\\\\ \textsf{Substitute}\\\\ &#10;\mathsf{tan\,x=t\qquad(t\in \mathbb{R})}\\\\\\ \textsf{so the equation &#10;becomes}\\\\ \mathsf{t^2+2t-4.76=0}\quad\Rightarrow\quad\begin{cases} &#10;\mathsf{a=1}\\\mathsf{b=2}\\\mathsf{c=-4.76} \end{cases} &#10;\end{array}


\large\begin{array}{l} \textsf{Using &#10;the quadratic formula:}\\\\ \mathsf{\Delta=b^2-4ac}\\\\ &#10;\mathsf{\Delta=2^2-4\cdot 1\cdot (-4.76)}\\\\ &#10;\mathsf{\Delta=4+19.04}\\\\ \mathsf{\Delta=23.04}\\\\ &#10;\mathsf{\Delta=\dfrac{2\,304}{100}}\\\\ &#10;\mathsf{\Delta=\dfrac{\diagup\!\!\!\! 4\cdot 576}{\diagup\!\!\!\! 4\cdot&#10; 25}}\\\\ \mathsf{\Delta=\dfrac{24^2}{5^2}} \end{array}

\large\begin{array}{l}&#10; \mathsf{\Delta=\left(\dfrac{24}{5}\right)^{\!2}}\\\\ &#10;\mathsf{\Delta=(4.8)^2}\\\\\\ &#10;\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\ &#10;\mathsf{t=\dfrac{-2\pm\sqrt{(4.8)^2}}{2\cdot 1}}\\\\ &#10;\mathsf{t=\dfrac{-2\pm 4.8}{2}}\\\\ \mathsf{t=\dfrac{\diagup\!\!\!\! &#10;2\cdot (-1\pm 2.4)}{\diagup\!\!\!\! 2}}\\\\\mathsf{t=-1\pm 2.4} &#10;\end{array}

\large\begin{array}{l} \begin{array}{rcl} &#10;\mathsf{t=-1-2.4}&~\textsf{ or }~&\mathsf{t=-1+2.4}\\\\ &#10;\mathsf{t=-3.4}&~\textsf{ or }~&\mathsf{t=1.4} \end{array} &#10;\end{array}


\large\begin{array}{l} \textsf{Both &#10;are valid values for t. Substitute back for }\mathsf{t=tan\,x:}\\\\ &#10;\begin{array}{rcl} \mathsf{tan\,x=-3.4}&~\textsf{ or &#10;}~&\mathsf{tan\,x=1.4} \end{array}\\\\\\ \textsf{Take the inverse &#10;tangent function:}\\\\ \begin{array}{rcl} &#10;\mathsf{x=tan^{-1}(-3.4)+k\cdot \pi}&~\textsf{ or &#10;}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi}\\\\ &#10;\mathsf{x=-tan^{-1}(3.4)+k\cdot \pi}&~\textsf{ or &#10;}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi} \end{array}\\\\\\ &#10;\textsf{where k in an integer.} \end{array}

__________


\large\begin{array}{l}&#10; \textsf{Now, restrict x values to the interval &#10;}\mathsf{[0,\,2\pi]:}\\\\ \bullet~~\textsf{For }\mathsf{k=0:}\\\\ &#10;\begin{array}{rcl} &#10;\mathsf{x=-tan^{-1}(3.4)


\large\begin{array}{l}&#10; \bullet~~\textsf{For }\mathsf{k=1:}\\\\ \begin{array}{rcl} &#10;\mathsf{x=-tan^{-1}(3.4)+\pi}&~\textsf{ or &#10;}~&\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}\\\\\\ &#10;\boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+\pi} &#10;\end{array}}\textsf{ is in the 2}^{\mathsf{nd}}\textsf{ quadrant.}\\\\ &#10;\mathsf{x\approx 1.86~rad~~(106.39^\circ)}\\\\\\ &#10;\boxed{\begin{array}{c}\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}}\textsf{&#10; is in the 3}^{\mathsf{rd}}\textsf{ quadrant.}\\\\ \mathsf{x\approx &#10;4.09~rad~~(234.46^\circ)}\\\\\\ \end{array}


\large\begin{array}{l}&#10; \bullet~~\textsf{For }\mathsf{k=2:}\\\\ \begin{array}{rcl} &#10;\mathsf{x=-tan^{-1}(3.4)+2\pi}&~\textsf{ or &#10;}~&\mathsf{x=tan^{-1}(1.4)+2\pi>2\pi~~\textsf{(discard)}} &#10;\end{array}\\\\\\ \boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+2\pi} &#10;\end{array}}\textsf{ is in the 4}^{\mathsf{th}}\textsf{ quadrant.}\\\\ &#10;\mathsf{x\approx 5.00~rad~~(286.39^\circ)} \end{array}


\large\begin{array}{l}&#10; \textsf{Solution set:}\\\\ &#10;\mathsf{S=\left\{tan^{-1}(1.4);\,-tan^{-1}(3.4)+\pi;\,tan^{-1}(1.4)+\pi;\,-tan^{-1}(3.4)+2\pi\right\}}&#10; \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2071152</span>


\large\textsf{I hope it helps.}


Tags: <em>trigonometric trig quadratic equation tangent tan solve inverse symmetry parity odd function</em>

6 0
3 years ago
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