Question

. In the given figure, PQR is a triangle in which

PQ = PR and ∠PQR = 75°, then show that QR is

equal to the radius of circumcircle of ∆PQR, whose

centre is O.​

Answer 1

Given- O is the centre of a circle in which an equilateral ΔPQR has been inscribed.

To find out- ∠QOR=?

Solution- The points P, Q & R are on the circumference of the circle since ΔPQR has been inscribed in the circle. Now the chord QR subtends ∠QOR to the centre O and ∠QPR to the circumference at P.

∴ By rule, ∠QOR=2∠QPR.........(i).

Now PQR is an equilateral Δ. Each of its angle is 75°

. i.e ∠QPR=75°

.

So, from (i), ∠QOR=2∠QPR=2×75=150°