Since this is an obtuse triangle, Point O is not equidistant from A, B, and C. Point O is not on the perpendicular bisectors, so the third statement is true. Point O is equidistant from AB, BC, and CA because these lines are pressed against the circpe in a mannered way.

7 0

3 years ago

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IMPORTANT <br><br>[how do you]Find sin theta if tan theta = 3/4

zvonat [6]

$\bf tan(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{adjacent}{4}}\impliedby \textit{let's find the \underline{hypotenuse}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{3^2+4^2}\implies c=5
\\\\\\
sin(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{hypotenuse}{5}}$

bear in mind that, though the square has two valid roots, one negative and one positive, the hypotenuse is just a radius distance, and therefore is never negative.

5 0

4 years ago