MHT CET · Maths · Properties of Triangles
In \(\Delta \mathrm{ABC}\) with usual notations \(\mathrm{a}=4, \mathrm{~b}=3, \angle \mathrm{A}=60^{\circ}\), then \(\mathrm{c}\) is a root of the equation
- A \(c^{2}-3 c-7=0\)
- B \(c^{2}-3 c+7=0\)
- C \(c^{2}+3 c-7=0\)
- D \(c^{2}+3 c+7=0\)
Answer & Solution
Correct Answer
(A) \(c^{2}-3 c-7=0\)
Step-by-step Solution
Detailed explanation
\(\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}\)
\(\therefore \cos 60=\frac{1}{2}=\frac{9+c^{2}-16}{2 \times 3 c} \Rightarrow 1=\frac{c^{2}-7}{3 c} \Rightarrow 3 c=c^{2}-7\)
\(\therefore c^{2}-3 c-7=0\)
\(\therefore \cos 60=\frac{1}{2}=\frac{9+c^{2}-16}{2 \times 3 c} \Rightarrow 1=\frac{c^{2}-7}{3 c} \Rightarrow 3 c=c^{2}-7\)
\(\therefore c^{2}-3 c-7=0\)
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