当AP与-|||-div 0 相切时,点B恰好落在 div 0 上,如图(2).-|||-请仅就图(2)的情形解答下列问题.-|||-(1)求证: angle PAO=2angle PBO;-|||-(2)若 div 0 的半径为5, =dfrac (20)(3), 求BP的长.-|||-P-|||-P-|||-M-|||-A 0 M A-|||-B B-|||-N-|||-图(1) 图(2)

(2angle(pi)/(12))^3=8angle(pi)/(4).A. 对B. 错

2.已知:如图,在 Delta ABC 中, angle BAE=angle CAE angle ADC=(90)^circ angle B=(40)^cir

三、判断题-|||-1.dfrac (2)(3)div 2-2div dfrac (2)(3)=0

如图所示, angle ABC=angle ABC, BD,B`D`分别是-|||-angle ABC , angle ABC 的平分线,求证: angle 1

设=sqrt ({x)^2+(y)^2+(z)^2} 则|div(grad)|(1,0,1)= () .-|||-(A) -sqrt (2) (B) sqrt

(湖北武汉江夏期中)已知: AD=AC =AE, AD交BC于点F.-|||-(1)如图1,若 angle BAD=angle CAE, 设DE交BC于点N,交

17.-|||-如图, Delta AOD 与 Delta BOC 存在对顶角 angle AOD=angle BOC=dfrac (pi )(4) ,AC=2

( d ^ 2 y ) div ( dx ^ 2 ) mid _ ( t = 0 ) =设函数y=y(x)由\cases{x=t^2-2t+1,e^y\sint

[单选题]过点（1，0，0）与（0，1，0），且与z＝x2＋y2相切的平面为（　　）。A.z＝0与x＋y?z＝1B.z＝0与2x＋2y?z＝2C.y＝x与x＋y

如图,在 Delta ABC 中, angle BAE=angle CAE ,-|||-angle ADC=(90)^circ angle B=(40)^ci