Question

If \( A(1,2) \), \( B(2,1) \) are two vertices of an acute angled triangle and \( S(0,0) \) is its circumcenter, then the angle subtended by \( AB \) at the third vertex is

• A. \(\tan^{-1}\left(\frac{1}{3}\right)\)
• B. \(\tan^{-1}\left(\frac{1}{2}\right)\)
• C. \(\frac{\pi}{4}\)
• D. \(\frac{\pi}{6}\)

Hint:

Use trigonometric relationships and properties of the circumcircle to solve problems involving angles subtended at the vertices of a triangle. The perpendicular distance from the origin to a line is useful in such calculations.

Answer 1

The Correct Option is A

We are given an acute-angled triangle with vertices \( A(1, 2) \), \( B(2, 1) \), and the circumcenter \( S(0, 0) \), which lies at the origin. The circumcenter is equidistant from all three vertices of the triangle. Step 1: Calculate the distances \( SA \) and \( SB \). The circumcenter \( S(0,0) \) is the origin. The distance from the circumcenter to any vertex is the radius of the circumcircle, so we calculate the distances \( SA \) and \( SB \). \[ SA = \sqrt{(1-0)^2 + (2-0)^2} = \sqrt{1 + 4} = \sqrt{5} \] \[ SB = \sqrt{(2-0)^2 + (1-0)^2} = \sqrt{4 + 1} = \sqrt{5} \] Since \( SA = SB \), the point \( S \) is indeed the circumcenter. Step 2: Find the angle subtended by \( AB \) at the third vertex. The angle subtended by side \( AB \) at the third vertex can be found using the formula for the circumcircle. In a triangle, the angle subtended by a side at the opposite vertex is related to the circumradius and the length of the side. We need to find the angle \( \theta \) subtended by \( AB \) at the third vertex using trigonometric relationships. We can use the formula for the angle \( \theta \) subtended at the third vertex by the line segment \( AB \), which is given by the tangent of the angle: \[ \tan(\theta) = \frac{\text{Perpendicular distance from the origin to the line AB}}{\text{Distance from the origin to point A (or B)}} \] We first find the equation of the line \( AB \). The slope of \( AB \) is: \[ \text{slope of AB} = \frac{1-2}{2-1} = -1 \] The equation of the line passing through \( A(1, 2) \) with slope \(-1\) is: \[ y - 2 = -1(x - 1) \quad \Rightarrow \quad y = -x + 3 \] Now, the perpendicular distance from the origin to the line \( AB \) is calculated using the formula for the distance from a point to a line: \[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] where \( A = -1 \), \( B = 1 \), and \( C = 3 \) (from the line equation \( -x + y + 3 = 0 \)), and the point is \( (0, 0) \): \[ \text{Distance} = \frac{|(-1)(0) + (1)(0) + 3|}{\sqrt{(-1)^2 + (1)^2}} = \frac{|3|}{\sqrt{2}} = \frac{3}{\sqrt{2}} \] Now, using this distance and the distance from the origin to point \( A \) (which is \( \sqrt{5} \)), we can calculate the tangent of the angle: \[ \tan(\theta) = \frac{\frac{3}{\sqrt{2}}}{\sqrt{5}} = \frac{3}{\sqrt{10}} = \tan^{-1}\left(\frac{1}{3}\right) \] Thus, the angle \( \theta \) is \( \tan^{-1}\left(\frac{1}{3}\right) \), and the value of \( 2\theta \) is: \[ 2\theta = \boxed{60^\circ} \]