1.

If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is

Answer»

In ΔPQR

PQ=PR=R

1

and Q=R

where R

1

is circumradius

from sine RULE:

sinR

PQ

=2R

1

⇒sinR=

2

1

⇒R=

6

π

THEREFORE, P=π−Q−R=π−2R=

3

2π

Ans: D


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