Call Guide $ M $ is the midpoint of BC $ $ $ AM \ perp BC rightarrow AM \ perp (BCC

Call Guide $ M $ is the midpoint of BC $ $ $ AM \ perp BC \ rightarrow AM \ perp (BCC'B ') \ rightarrow AM \ perp $ in $ B'C utilitech (BCC'B') $ drawing $ MH \ perp B'C \ rightarrow B'C \ perp (AMH) $ be the angle between $ (AB'C) $ and $ (BB'C) $ is $ \ widehat {} = 60 ^ AHM I have $ 0 $ AM = \ dfrac {BC} {2} = a $. In a right triangle AMH $ $ we have: $ MH = \ dfrac {AM} {\ tan H} = \ dfrac {a} {\ sqrt {3}} $ In $ BCB triangle '$ $ BK $ drawing parallel with MH $ $ $ BK = 2MH = \ dfrac {2a} {\ sqrt {3}} $ $ where $ BK's high street of square triangle $ BCB '$ utilitech should utilitech be calculated: $ BB' = a \ sqrt { So volume 2} $ $ ABC.A 'B'C' $ is $ V = a ^ 3 \ sqrt {2} $
Write the equation for the line tangent create (d) an angle of $ 45 ^ 2 $ 0 asymptotic utilitech Tangential cut at A, B such that the radius of the inscribed circle biggest IAB. Plane contact with the sphere and two straight lines parallel to the sphere equation 2 line contact smallest radius. Find plane passing point, straight line parallel to Find M under (d) such that MA + MB minimum. Straight line (d) parallel to the horizontal axis and the graph cut (C) at two distinct points utilitech A, B such that triangle OAB $ $ O scales at the extreme point of the graph forms a triangle with the radius of the road circumscribed circle by 1