JEE Mains · Maths · STD 11 - 9. straight line
Let \(A B C\) be an isosceles triangle in which \(A\) is at \((-1,0), \angle A=\frac{2 \pi}{3}, A B=A C\) and \(B\) is on the positive \(\mathrm{x}\)-axis. If \(\mathrm{BC}=4 \sqrt{3}\) and the line \(\mathrm{BC}\) intersects the line \(y=x+3\) at \((\alpha, \beta)\), then \(\frac{\beta^4}{\alpha^2}\) is :
- A \(85\)
- B \(36\)
- C \(45\)
- D \(75\)
Answer & Solution
Correct Answer
(B) \(36\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{c}}{\sin 30^{\circ}}=\frac{4 \sqrt{3}}{\sin 120^{\circ}}\) [By sine rule] \(2 c=8 \Rightarrow c=4\) \( \mathrm{AB}=|(\mathrm{b}+1)|=4 \) \( \mathrm{~b}=3, \mathrm{~m}_{\mathrm{AB}}=0 \) \( \mathrm{~m}_{\mathrm{BC}}=\frac{-1}{\sqrt{3}} \)…
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