Question

If the co-ordinate of the vertices of a triangle ABC be $ A(-1,3,2),B(2,3,5) $ and $ C(3,5,-2), $ then $ \angle A $ is equal to:

• A. $ 45{}^\circ $
• B. $ 60{}^\circ $
• C. $ 90{}^\circ $
• D. $ 30{}^\circ $
• E. $ 135{}^\circ $

Answer 1

Answer: (C)

$ A(-1,3,2),B(2,3,5) $ and $ C(3,5,-2) $ $ \therefore $ $ AB=\sqrt{{{3}^{2}}+0+{{3}^{2}}}=\sqrt{18} $ $ CA=\sqrt{16+4+16}=6 $ and $ BC=\sqrt{1+4+49}=\sqrt{54} $ $ \because $ $ A{{B}^{2}}+C{{A}^{2}}=B{{C}^{2}} $ $ \therefore $ $ \Delta ABC $ is right angled triangle, right angled at A. $ \therefore $ $ \angle A=90{}^\circ $