If A(1,2) and B(2,1) are two vertices of an a | vedclass

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(A) Let $C$ be the third vertex of the triangle. Since $S(0,0)$ is the circumcentre,the angle subtended by the chord $AB$ at the centre $S$ is $\angle

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{1}{3}ight)$

Vedclass · Answer

(A) Let $C$ be the third vertex of the triangle. Since $S(0,0)$ is the circumcentre,the angle subtended by the chord $AB$ at the centre $S$ is $\angle ASB = 2	heta$,where $	heta = \angle ACB$ is the angle at the third vertex.Slope of $AS = \frac{2-0}{1-0} = 2$.Slope of $BS = \frac{1-0}{2-0} = \frac{1}{2}$.Using the formula for the angle between two lines with slopes $m_1$ and $m_2$,$	an(2	heta) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}ight| = \left|\frac{2 - 1/2}{1 + 2(1/2)}ight| = \frac{3/2}{2} = \frac{3}{4}$.We know that $	an(2	heta) = \frac{2	an	heta}{1 - 	an^2	heta}$.So,$\frac{2	an	heta}{1 - 	an^2	heta} = \frac{3}{4}$.$8	an	heta = 3 - 3	an^2	heta \Rightarrow 3	an^2	heta + 8	an	heta - 3 = 0$.$(3	an	heta - 1)(	an	heta + 3) = 0$.Since the triangle is acute-angled,$	heta$ must be acute,so $	an	heta = \frac{1}{3}$.Thus,the angle subtended at the third vertex is $	an^{-1}\left(\frac{1}{3}ight)$.

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

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