CLASS 9 question.....
please help it's urgent....
in the given figure bisector of the exterior angles angles Q and R are formed by producing sides PQ and PR of triangle PQR intersect each other at the point is prove that angles QSR equals to 90-1/2 angle P.....
Answers
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
Answer:
Answer:
Explanation:
In the question,
The triangle PQR as shown in the figure,
The sides PQ and PR are extended to S and T.
QO and RO are bisectors.
To prove : ∠QOR = 90° - (1/2)(∠P)
Proof : Let us say,
∠RQO = ∠SQO = x°
and,
∠QRO = ∠TRO = y°
So, as PQS and PRT is a line.
∠PQR + 2x = 180°
So,
∠PQR = 180° - 2x ......(1)
And,
∠PRQ + 2y = 180°
So,
∠PRQ = 180° - 2y ..........(2)
In triangle PQR, as sum of the internal angles of the triangle is 180°.
So,
∠P + ∠PQR + ∠PRQ = 180°
On putting the values from the eqn. (1) and (2) we get,
∠P = 180° - (180° - 2x) - (180° - 2y)
∠P = 2x + 2y - 180° ........(3)
So,
Also,
In triangle QRO,
∠QOR = 180° - (x + y) =180° - x - y (Because, sum of the internal angles of the triangle is 180°.)
Now,
Taking half of eqn. (3) we get,
So,
Therefore,
Hence, Proved.