Do the bottom normal vector? A 1H？ ,? machine gun

Solution 1:

Known? AA 1？ =? 1? Can be pushed

AE？ =? 1? *? cos60？ =? 1/2

Huh? =? AE？ /? cos60？ =√3/3

Mg? =? A 1H？ =? √( 1-? Huh? )? =√6/3

Again? HG？ =? A 1M？ =? √3/2? Can be pushed

AG？ =? Huh? +? HG？ =? √3/3? +? √3/2? =? 5√3? /? six

therefore

Me? =? √(MG？ +? AG？ )? =√(? (√6/3)? +? (5√3? /? 6) )? =√ 1 1/2

Solution 1: Cosine Theorem

Ditto? Do you know? sin∠AA 1H？ =√3/3

cos(≈aa 1c 1)？ =? Because? (90 ? +∠AA 1H)？ =? cos∠AA 1H？ cos90？ -? sin∠AA 1H？ sin90？ =-√3/3

Me? =? √(AA 1+？ Me? -2 * aa 1 * AM * cos∠aa 1c 1)= √( 1+(√3/2)-？ 2? *? 1*? (√3/2)? *? (-√3/3)? )

=√ 1 1/2