In quadrilateral $QRTS$, we have $QR = 11$, $QS = 9$, and $ST=2$. Sides $\overline{RQ}$ and $\overline{ST}$ are extended past $Q$ and $S$, respectively, to meet at point $P$. If $PS = 8$ and $PQ = 5$, then what is $RT$?
+129630 We can use the Law of Cosines to find angle QPS....and we have
QS^2 = QP^2 + SP^2 - 2(QS)(QP)cosQPS
9^2 = 5^2 + 8^2 - 2(5)(8)cosQPS
81 = 25 + 64 - 80cosQPS
[ 81 - 25 - 64 ] = -80cosQPS
cosQPS = [ -8] / -80
cosQPS = 1/10 and using the cosine inverse, m< QPS =
arccos [1/10] = QPS = about 84.26°
And we can use this Law again to find RT
Note : PR = PQ + QR = 5 + 11 = 16
And PT = PS + ST = 8 + 2 = 10
And angle QPS = angle RPT ....so we have
RT^2 = PR^2 + PT^2 - 2(16)(10)cosRPT
RT^2 = 16^2 + 10^2 - 2(16)(10)cos(84.26)
RT^2 = 16^2 + 10^2 - (320)*(1/10)
RT^2 = 256 + 100 - 32 = 324
RT = 18
