The lengths of sides of triangle having vertices #A(7, 9)#, #B(8,8)# & #C(6, 6)# are given as
#AB=\sqrt{(7-8)^2+(9-8)^2}=\sqrt2#
#BC=\sqrt{(8-6)^2+(8-6)^2}=2\sqrt2#
#AC=\sqrt{(7-6)^2+(9-6)^2}=\sqrt{10}#
Now, using Cosine rule in #\Delta ABC# as follows
#\cos A=\frac{AB^2+AC^2-BC^2}{2(AB)(AC)}#
#\cos A=\frac{(\sqrt2)^2+(\sqrt10)^2-(2\sqrt2)^2}{2(\sqrt2)(\sqrt10)}#
#\cos A=1/\sqrt5#
#A=\cos^{-1}(1/\sqrt5)=63.435^\circ#
Similarly, we get
#\cos B=\frac{AB^2+BC^2-AC^2}{2(AB)(BC)}#
#\cos B=\frac{(\sqrt2)^2+(2\sqrt2)^2-(\sqrt10)^2}{2(\sqrt2)(2\sqrt2)}#
#\cos B=0#
#B=90^\circ#
#\cos C=\frac{AC^2+BC^2-AB^2}{2(AC)(BC)}#
#\cos C=\frac{(\sqrt10)^2+(2\sqrt2)^2-(\sqrt2)^2}{2(\sqrt10)(2\sqrt2)}#
#\cos C=2/\sqrt5#
#C=\cos^{-1}(2/\sqrt5)=26.565^\circ#
