Trig

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.

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

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:

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