JEE Mains · Maths · STD 11 - Trigonometrical equations
In a triangle, the sum of lengths of two sides is \(x\) and the product of the lengths of the same two sides is \(y.\) If \(x^2 - c^2 = y ,\) where \(c\) is the length of the third side of the triangle, then the circumradius of the triangle is
- A \(\frac {3}{2}\,y\)
- B \(\frac {c}{\sqrt 3}\)
- C \(\frac {c}{3}\)
- D \(\frac {y}{\sqrt 3}\)
Answer & Solution
Correct Answer
(B) \(\frac {c}{\sqrt 3}\)
Step-by-step Solution
Detailed explanation
\(x^{2}-c^{2}=y\) \((a+b)^{2}-c^{2}=a b\) \(a^{2}+b^{2}-c^{2}=-a b\) \(\frac{a^{2}+b^{2}-c^{2}}{2 a b}=-\frac{1}{2}\) \(\cos c=-\frac{1}{2}\) \(c=\frac{2 \pi}{3}\) \(\sin C=\frac{\sqrt{3}}{2}\) \(\frac{c}{\sin c}=2 R \Rightarrow R=\frac{c}{2 \sin c}=\frac{c}{\sqrt{3}}\)
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