MHT CET202526 Apr 2025Evening ShiftMathematicsProperties of TrianglesActual
In a triangle P Q R with usual notations, R= 2 . If p 2 and q 2 are the roots of the equation. a x^2+b x+c=0 (a 0) , then
Options
- Aa+b=c
- Bb + c =a
- Ca+c=b
- Db = c
Correct answer
A. a+b=c
Step-by-step solution
Given that R = 2 in triangle PQR , the angle sum property implies P + Q = 2 , so P 2 + Q 2 = 4 . Taking tangent on both sides and applying the identity for ( P 2 + Q 2 ) yields: ( P 2 ) + ( Q 2 ) 1 - ( P 2 ) ( Q 2 ) = 1 Since ( P 2 ) and ( Q 2 ) are roots of ax^2 + bx + c = 0 , by Vieta’s formulas their sum is - b a and product is c a . Substituting into the trigonometric equation gives: - b a 1 - c a = 1 Simplifying leads to -b = a - c , or equivalently a + b = c . Final answer: A