PA, Pb and PC are three rays starting from point P. the angle between each two rays is 60 degrees. Then the cosine of the angle between PC and PAB is () A. 12B. 22C. 33D. 63

PA, Pb and PC are three rays starting from point P. the angle between each two rays is 60 degrees. Then the cosine of the angle between PC and PAB is () A. 12B. 22C. 33D. 63

Take any point D in PC and make do ⊥ plane APB, then ⊥ DPO is the angle formed by the line PC and plane PAB. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; through point O, make OE ⊥ PA, of ⊥ Pb, because do ⊥ plane APB, then de ⊥ PA, DF ⊥ Pb. △ dep ≌ DFP, ⊥ EP =

P is the point on the line AB, and AP = 25ab, M is the midpoint of AB, if PM = 2cm, then ab=______ cm．

∵ m is the middle point of AB, ∵ am = 12ab, ∵ P is the point on the line AB, and AP = 25ab, ∵ PM = am-ap = 12ab-25ab = 110ab = 2cm, ∵ AB = 20cm

As shown in the figure, P is the midpoint of line AB, and M is a point on Pb. Try to guess the size relationship between am-bm and 2pm, and briefly explain the reason
└───────────┴───────┴───┘
A P M B

The relationship between am-bm and 2pm is equal
∵ P is the midpoint of line ab
∴PB=AB/2,
∵ m is a point on Pb
∴PB=PM+BM
∴2PM=2PB-2BM=AB-2BM
∵AB=AM+BM
∴2PM=AM-BM

It is known that in △ PAB, the vertical bisector of AB side intersects Pa at point m, and the perpendicular foot is n. try to compare the size of PA and Pb

The relationship between PA and Pb is: PA > Pb
Connecting BM
∵ Mn vertical bisection ab
∴MA＝BM
∴PA＝PM＋MA＝PM＋BM
In triangular PMB, there are
PM＋BM＞PB
∵PA＝PM＋BM
∴PA＞PB