In ΔPQR and ΔPST PQ = PT (Given) PR = PS (Given) ∠PTS=∠PQR (Angles opposite to equal sides of a triangle are equal) By SAS, ∆PQR≅∆PST ⇒∠QPR=∠TPS.........(i) Now,In ΔPQT ∠PTQ=62∘,∠PQT=62∘
∴∠QPT=180∘−(62∘+62∘)=56∘
Again ∠QPT=56∘ ⇒∠QPR+∠RPS+∠TPS=56∘ [∵∠QPR=∠mathrTPS from eq(i)] ⇒∠QPR+34∘+∠QPR=56∘