Find Theta

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Find Theta

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Suppose you have a triangle (not a right triangle) wher one angle is (theta). the length of the opposite to theta is 5. the other two sides have length 4 and 6.

so i have all the sides. picturing it like a normal triangle.. 5 is on the bottom, for theta is at the tip of the traingle with 4 and 6 on either side...now how can i find the angle theta? Theta would be the angle correct?

Guest Jun 18, 2014

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$a=4\quad b=6\quad c=5 \quad \mbox{ angle } \theta\mbox{ ? }$

$\begin{array}{l} c^2=a^2+b^2-2ab*\cos{(\theta) } \\\\ \cos{(\theta)} = \dfrac{a^2+b^2-c^2}{2ab}\\\\ \cos{(\theta)} = \dfrac{4^2+6^2-5^2}{2*4*6}\\\\ \cos{(\theta)} = \dfrac{16+36-25}{48}\\\\ \cos{(\theta)} = \dfrac{27}{48}\\\\ \cos{(\theta)} = 0.5625\\\\ \textcolor[rgb]{1,0,0}{\theta=55.7711336722\ensuremath{^\circ}} \end{array}$

heureka Jun 18, 2014

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heureka · Answer 1 · 2014-06-18T06:26:55

$a=4\quad b=6\quad c=5 \quad \mbox{ angle } \theta\mbox{ ? }$

$\begin{array}{l} c^2=a^2+b^2-2ab*\cos{(\theta) } \\\\ \cos{(\theta)} = \dfrac{a^2+b^2-c^2}{2ab}\\\\ \cos{(\theta)} = \dfrac{4^2+6^2-5^2}{2*4*6}\\\\ \cos{(\theta)} = \dfrac{16+36-25}{48}\\\\ \cos{(\theta)} = \dfrac{27}{48}\\\\ \cos{(\theta)} = 0.5625\\\\ \textcolor[rgb]{1,0,0}{\theta=55.7711336722\ensuremath{^\circ}} \end{array}$

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