<p>Two lines \(y = -\sqrt{3}x\) and \(y = \sqrt{3}x\) intersect at the origin. A point \(P(2, 2)\) is given. If \(B\) is a point on the line \(y = \sqrt{3}x\) and \(K\) is on line \(y = \sqrt{3}x\) such that \(\angle AOB = 60°\) (where \(O\) is origin and \(\triangle OAB\) is equilateral), then \(\angle KBP\) equals:</p>
Step-by-Step Solution
Key Concept: The two given lines make angles of 120° and 60° with the positive x-axis respectively. Using the equilateral triangle condition (∠AOB = 60°), determine positions of A and B, then calculate ∠KBP using angle subtraction between lines through B.
<p><strong>Step 1:</strong> Identify the given lines. Line 1: y = -√3x makes angle 120° with positive x-axis. Line 2: y = √3x makes angle 60° with positive x-axis. These lines are perpendicular (120° - 60° = 60°).</p><p><strong>Step 2:</strong> Since A is on y = -√3x and B is on y = √3x with ∠AOB = 60° and △OAB equilateral, we have OA = OB. Let OA = OB = r. Then A and B are positioned such that they form an equilateral triangle with O.</p><p><strong>Step 3:</strong> For point A on y = -√3x at distance r from O: A = (-r/2, -r√3/2). For point B on y = √3x at distance r from O: B = (r/2, r√3/2).</p><p><strong>Step 4:</strong> Line BP connects B(r/2, r√3/2) to P(2, 2). Line BK lies along y = √3x (since K is on this line). The slope of BK is √3 (angle 60°).</p><p><strong>Step 5:</strong> Slope of BP = (2 - r√3/2)/(2 - r/2). For the equilateral triangle to exist with reasonable geometry, take r = 2. Then B = (1, √3).</p><p><strong>Step 6:</strong> Slope of BP = (2 - √3)/(2 - 1) = 2 - √3, which corresponds to angle arctan(2 - √3) = 15°. The line BK has slope √3, corresponding to 60°.</p><p><strong>Step 7:</strong> ∠KBP = 60° - 15° = 45°.</p><p>∴ Answer: C (45°)</p>
Correct Answer: C