Question

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

Answer 1

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

Answer 2

Answer:

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