Q.23. Let A, B be points on the two half-lines x-, a 0 at a distance of a from their point of inte

Q: What is the correct answer for JEE Main Mathematics Q23?

The correct answer is 9. Sol. $\triangle$PAB is equilateral [Figure: A coordinate plane with x and y axes. Point P is at the origin (0,0). Two lines are shown, symmetric about the x-axis, forming an angle. Points A and B are on these lines. The angle between the lines is $60^\circ$. The angle bisector is the x-axis. Point Q is on the x-axis. A is in the first quadrant, B is in the fourth quadrant. The coordinates of A are

Q: Q.23. Let A, B be points on the two half-lines x-, a > 0 at a distance of a from their point of intersection P. The line segment AB meets the angle bisector of the given half-lines at the point Q. If…

The correct answer is 9. Sol. $\triangle$PAB is equilateral [Figure: A coordinate plane with x and y axes. Point P is at the origin (0,0). Two lines are shown, symmetric about the x-axis, forming an angle. Points A and B are on these lines. The angle between the lines is $60^\circ$. The angle bisector is the x-axis. Point Q is on the x-axis. A is in the first quadrant, B is in the fourth quadrant. The coordinates of A are

Question

Q.23. Let A, B be points on the two half-lines x-, a > 0 at a distance of a from their point of intersection P. The line segment AB meets the angle bisector of the given half-lines at the point Q. If PQ =  and R is the radius of the circumcircle of PAB, then  is equal to ________.

FirstInTest · Accepted Answer

Sol. $	riangle$PAB is equilateral
[Figure: A coordinate plane with x and y axes. Point P is at the origin (0,0). Two lines are shown, symmetric about the x-axis, forming an angle. Points A and B are on these lines. The angle between the lines is $60^\circ$. The angle bisector is the x-axis. Point Q is on the x-axis. A is in the first quadrant, B is in the fourth quadrant. The coordinates of A are $(\alpha, \alpha/\sqrt{3})$ and B are $(\alpha, -\alpha/\sqrt{3})$. The distance PA = PB = a. The angle $\angle APQ = 30^\circ$.]

PQ = $\frac{\sqrt{3}\alpha}{2} = \frac{9}{2} \Rightarrow \alpha = 3\sqrt{3}$
$\frac{\alpha}{2r} = \cos30^\circ = \frac{\sqrt{3}}{2}$
$\alpha = \sqrt{3}R$
R = 3
$\frac{\alpha^2}{R} = \frac{(3\sqrt{3})^2}{3} = \frac{27}{3} = 9$

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