In a non-right-angled triangle Δ P Q R , let p , q , r denote the lengths of the sides opposite to the angles at P , Q , R respectively. The median from R meets the side P Q at S , the perpendicular from P meets the side Q R at E , and R S and P E intersect at 0 . If p = 3 , q = 1 , and the radius of the circumcircle of the Δ P Q R equals 1 , then which of the following options is/are correct?

Step-by-step solution

By sine rule in P Q R p s i n P = q s i n Q = r s i n R = 2 R 3 s i n P = 1 s i n Q = 1 s i n R = 2 1 s i n P = 3 2 and s i n Q = s i n R = 1 2 P = 60 ° or 120 ° , Q = 30 ° and 150 ° ⇒ As P Q R is not a right angle triangle Only possibility is P = 120 ° , Q = R = 30 ° ⇒ P Q R is an isosceles triangle, hence, P E is also a median and ‘ O ’ will be centroid. ( ∆ P Q E and ∆ P R E are congruent and E is mid point of Q R ) ⇒ r = q = 1 , P = 3 Option A : Length of median from R R S = 1 2 2 p 2 + 2 q 2 - r 2 = 1 2 2 3 2