Let A B C be a triangle such that A B=A C . Suppose the tangent to the circumcircle of A B C at B is perpendicular of AC . Find ABC measured in degrees.

Step-by-step solution

Consider the Isosceles triangle with vertex A, B, C such that A B=A C By symmetry circumcentre will lie on y - axis D (0, k ) circumcenter Radius of circumcircle = DA = DB = DC aligned & (a-k)^2 = k^2+1 & a^2+k^2-2 a k=k^2+1 & a^2-2 a k=1 aligned Now Circumcircle: x^2+(y-k)^2=k^2+1 Slope of tangent at B , aligned & 2 x+2(y-k) y^ =0 & . y^ |_B= . -x y-k |_B= 1 -k aligned Also, m _ AC =- a By given condition, (- a ) (- 1 k )=-1 a=-k From (1) & (2), a ^2+2 a ^2=1 a = 1 3 If a = 1 3 k =- 1 3 aligned & =m_ A B = 1 3 =30