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# Trig

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Suppose we have $$PQ = 6$$$$QR = 7$$, and $$PR = 9$$, as in the picture below:

Find the length of the median from $$R$$ to $$\overline{PQ}$$.

Jul 18, 2021

#1
+63
0

I won't directly solve this problem, but hint: try using sin on triangle RPM. You don't need theta, just pull up a chart of all sin values and compare the results. After that, use theta to calculate side d.

Jul 18, 2021
edited by PBJcatalinasandwich  Jul 18, 2021
#2
+1425
+2

PR = a = 9          QR = b = 7           PQ = c = 6          RM = d = ?

d = 1/2[sqrt(2*a2 + 2*b2 - c2)]

d = √224 / 2 ≈ 7.483

Jul 19, 2021
#3
+26122
+1

Suppose we have $$PQ=6$$, $$QR=7$$, and $$PR=9$$, as in the picture below:

Find the length of the median from $$R$$ to $$\overline{PQ}$$.

cos Rule:

$$\begin{array}{|rcll|} \hline 9^2&=&(2*3)^2+7^2-2*2*3*7*\cos(Q) \\ 9^2&=&6^2+7^2-2*2*3*7*\cos(Q) \\ 2*2*3*7*\cos(Q) &=& 6^2+7^2-9^2 \\ 2*2*3*7*\cos(Q) &=& 4 \quad | \quad :2 \\ 2*3*7*\cos(Q) &=& 2 \qquad (1) \\ \hline \end{array}$$

cos Rule:

$$\begin{array}{|rcll|} \hline d^2&=&3^2+7^2-2*3*7*\cos(Q) \\ 2*3*7*\cos(Q) &=& 3^2+7^2-d^2 \\ 2*3*7*\cos(Q) &=& 58-d^2 \qquad (2) \\ \hline \end{array}\\ \begin{array}{|lrcll|} \hline (1)=(2):& 2 &=& 58-d^2 \\ & d^2&=& 58-2 \\ & d^2&=& 56 \\ & d^2&=& 4*14 \\ & \mathbf{d}&=&\mathbf{ 2\sqrt{14} } \\ \hline \end{array}$$

Jul 19, 2021
edited by heureka  Jul 19, 2021
edited by heureka  Jul 19, 2021