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).

MathCuber Jan 1, 2020

#1**+3 **

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.

Melody Jan 1, 2020

#3**+2 **

You can find BC using cosine rule. You have 2 sides and the included angle so you can find the 3rd side.

Melody
Jan 2, 2020

#4**+2 **

So BC is \(\sqrt{19}\).

Then, by the law of sines, sin BCP is \(\frac{\sqrt3}{\sqrt{19}}\).

What now?

MathCuber
Jan 2, 2020

#5**+2 **

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.

Melody
Jan 2, 2020