MHT CET · Maths · Properties of Triangles
In a triangle \(P Q R\) with usual notations, \(\angle R=\frac{\pi}{2}\). If \(\tan \frac{p}{2}\) and \(\tan \frac{q}{2}\) are the roots of the equation. \(a x^2+b x+c=0 \quad(a \neq 0)\), then
- A \(a+b=c\)
- B \(\mathrm{b}+\mathrm{c}=a\)
- C \(a+c=b\)
- D \(\mathrm{b}=\mathrm{c}\)
Answer & Solution
Correct Answer
(A) \(a+b=c\)
Step-by-step Solution
Detailed explanation
\(P+Q+R=\pi \implies P+Q=\frac{\pi}{2}\) \(\frac{P}{2}+\frac{Q}{2}=\frac{\pi}{4}\)
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