How do you find the exact value of the third side given $△ R S T$ $triangle RST$ , RS=9, $S T = 9 \sqrt{3}$ $ST=9sqrt3$ , $m ∠ S = \frac{5 π}{6}$ $mangleS=(5pi)/6$ ?

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How do you find the exact value of the third side given $△ R S T$ $triangle RST$ , RS=9, $S T = 9 \sqrt{3}$ $ST=9sqrt3$ , $m ∠ S = \frac{5 π}{6}$ $mangleS=(5pi)/6$ ?

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Answer 1 · 2017-01-15T13:03:36

$= 9 \sqrt{7}$ $=9sqrt(7)$

Explanation:

You would apply the cosine theorem to calculate the third side:

$¯ ¯¯¯¯ ¯ R T = \sqrt{{¯ ¯¯¯ ¯ R S}^{2} + {¯ ¯¯¯ ¯ S T}^{2} - 2 \cdot ¯ ¯¯¯ ¯ R S \cdot ¯ ¯¯¯ ¯ S T \cdot cos ˆ S}$ $bar(RT)=sqrt(bar(RS)^2+bar(ST)^2-2*bar(RS)*bar(ST)*cos hatS)$

$= \sqrt{9^{2} + {(9 \sqrt{3})}^{2} - 2 \cdot 9 \cdot 9 \sqrt{3} \cdot cos (\frac{5}{6} π)}$ $=sqrt(9^2+(9sqrt(3))^2-2*9*9sqrt3*cos(5/6pi))$

$=sqrt(81+243-cancel162^81sqrt(3)*(-sqrt(3)/cancel2)$

$=sqrt(324+243)=sqrt(567)=9sqrt(7)$

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How do you find the exact value of the third side given $△ R S T$ $triangle RST$ , RS=9, $S T = 9 \sqrt{3}$ $ST=9sqrt3$ , $m ∠ S = \frac{5 π}{6}$ $mangleS=(5pi)/6$ ?

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How do you find the exact value of the third side given triangle RST△RSTtriangle RST, RS=9, ST=9sqrt3ST=9√3ST=9sqrt3, mangleS=(5pi)/6m∠S=5π6mangleS=(5pi)/6?

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How do you find the exact value of the third side given $△ R S T$ $triangle RST$ , RS=9, $S T = 9 \sqrt{3}$ $ST=9sqrt3$ , $m ∠ S = \frac{5 π}{6}$ $mangleS=(5pi)/6$ ?