In triangle ΔPQR , ∠R = 70° and bisectors of angles Q and P meet at point S. Find ∠QSP
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given:In△PQR,Bisectors of ∠Q&∠R meet at point S.
ToProve:∠QSR=90
∘
+
2
1
∠P
Proof:Let∠PQR=∠Q;∠QRP=∠R;∠QPR=2P
In△PQR
∠P+∠Q+∠R=180
∘
(sum of angles of a triangle)
⟹∠Q+∠R=180−P−(1)
In△SQR
∠QSR+
2
∠Q
+
2
∠R
=180
∘
(sum of angles of a triangle)
2
∠Q+∠R
=180−∠QSR
from(1)
2
180−∠P
=180−∠QSR
⟹90−
2
∠P
=180−∠QSR
⟹∠QSR=90+
2
∠P
Hence,proved
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given:In△PQR,Bisectors of ∠Q&∠R meet at point S.
ToProve:∠QSR=90
∘
+
2
1
∠P
Proof:Let∠PQR=∠Q;∠QRP=∠R;∠QPR=2P
In△PQR
∠P+∠Q+∠R=180
∘
(sum of angles of a triangle)
⟹∠Q+∠R=180−P−(1)
In△SQR
∠QSR+
2
∠Q
+
2
∠R
=180
∘
(sum of angles of a triangle)
2
∠Q+∠R
=180−∠QSR
from(1)
2
180−∠P
=180−∠QSR
⟹90−
2
∠P
=180−∠QSR
⟹∠QSR=90+
2
∠P
if you don't understand goon G
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