MathLabs

Problem 3

On the sides of an arbitrary triangle ABCABC, triangles ABRABR, BCPBCP, CAQCAQ are constructed externally with ∠CBP=∠CAQ=45∘\angle CBP = \angle CAQ = 45^\circ, ∠BCP=∠ACQ=30∘\angle BCP = \angle ACQ = 30^\circ, ∠ABR=∠BAR=15∘\angle ABR = \angle BAR = 15^\circ. Prove that ∠QRP=90∘\angle QRP = 90^\circ and QR=RPQR = RP.
Step 2 of 6: Find the six auxiliary side lengths by the sine rule
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PB=asin⁡30∘sin⁡105∘, PC=asin⁡45∘sin⁡105∘, QA=bsin⁡30∘sin⁡105∘, QC=bsin⁡45∘sin⁡105∘, RA=RB=csin⁡15∘sin⁡150∘PB=\frac{a\sin30^\circ}{\sin105^\circ},\ PC=\frac{a\sin45^\circ}{\sin105^\circ},\ QA=\frac{b\sin30^\circ}{\sin105^\circ},\ QC=\frac{b\sin45^\circ}{\sin105^\circ},\ RA=RB=\frac{c\sin15^\circ}{\sin150^\circ}
Detailed analysis

Apply the sine rule in BCPBCP, CAQCAQ, and ABRABR. Their angles are 45∘,30∘,105∘45^\circ,30^\circ,105^\circ; 45∘,30∘,105∘45^\circ,30^\circ,105^\circ; and 15∘,15∘,150∘15^\circ,15^\circ,150^\circ, respectively. The displayed formulas follow, with a=BCa=BC, b=CAb=CA, and c=ABc=AB.