A triangle is defined by the three points: A=(7,9) B=(8,8) C=(6,6). How to determine all three angles in the triangle (in radians) ?

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Answer 1 · 2018-06-28T22:47:57

#A=63.435^\circ#, #B=90^\circ#, #C=26.565^\circ#

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#