+537 Right triangle ABC is inscribed in equilateral triangle PQR, as shown. Given PB = 2, PC = 5, and QC = 3, then find QA.
Diagram: https://latex.artofproblemsolving.com/miscpdf/utullifn.pdf?t=1577916255464.
I am just adding the pic. (Melody).
+118687 I may not have done this the best way but this was my method:
1) Find BC
2) find angle BCP
3) Find angle ACQ
4) Find angfle CAQ
5) Now you have 3 angles and one side of triangle ACQ so you can find AQ
This is a teaching answer. Please do not override it.
MathCuber, if you have questions/problems then please ask.
+118687 You can find BC using cosine rule. You have 2 sides and the included angle so you can find the 3rd side.
+537 So BC is \(\sqrt{19}\).
Then, by the law of sines, sin BCP is \(\frac{\sqrt3}{\sqrt{19}}\).
What now?
+118687 If
\(sin\angle{PCB}=\frac{\sqrt3}{\sqrt{19}}\\ then\\ \angle{PCB}=sin^{-1}\frac{\sqrt3}{\sqrt{19}}\\ \angle{PCB}\approx 23.413 ^{\circ}\)
Now, check you know how I did that on a calculator and then move onto step 3
sorry it has not displayed properly, I will see if I can fix it. It is fixed I hope.
