So AD= 1, AE = 2. In delta △ADE, ∠ DAE = 60,
From the cosine theorem DE= 12+22? 2× 1×2×cos60 =3。
Because AD2+DE2=AE2,
So advertising ⊥ Germany.
It is A 1D⊥DE after folding.
Because plane A 1DE⊥ plane BCED, and plane A 1DE∩ plane BCED=DE,
A 1D？ Aircraft A 1DE, A 1D⊥DE, so A 1D⊥ aircraft BCED.
Therefore, a 1d ⊥ EC.
(2) As shown in the figure, let PH⊥BD be at point H, and connect A 1H and a1p..
There is a 1 d ⊥ plane BCED from (1), PH? Aircraft BCED,
So A 1D⊥PH, and A 1D∩BD=D, so the PH⊥ plane A 1BD,
So ∠PA 1H is the angle formed by the straight line PA 1 and the plane A 1BD.
Let PB=x(0≤x≤3), then BH=x2, PH=3x2, DH=BD-BH=2-x2.
So A 1H=DH2+A 1D2=x2? 2x+54
So in Rt△PA 1H, tan ∠ pa1h = PHA1h = 3xx2? 8x+20
(1) if x=0, then tan ∠ pa1h = PHA1h = 3xx2? 8x+20=0，
② if x≠0, tan ∠ pa1h = PHA1h = 3xx2? 8x+20=3 1？ 8x+20x2
Let 1x=t(t≥ 13) and y=20t2-8t+ 1.
Because the function y=20t2-8t+ 1 monotonically increases over t≥ 13, ymin = 20×19-83+1= 59.
Therefore, the maximum value of tan∠PA 1H is 3. I've praised it all. I've stepped on it. What is your evaluation of this answer? Put away comments//high quality or satisfaction or special types or recommended answers dottimewindow.iperformance & window.iperformance.mark ('c _ best',+newdate); Lawyer recommendation service: If your problem has not been solved, please describe your problem in detail and consult other similar problems for free through Baidu Legal Professional Edition. In the equilateral triangle ABC 2022-07-30, D is a point on AC, E is on the extension line of AB, be = CD, and DE intersects BC at point P. Verification: DP=PE2022-06-06△ABC is an equilateral triangle. Point d is on AB, point e is on AC, AE=BD. Verification: Be = CD 2016-12-01In equilateral triangles AB and BC, DE is the point on AB and BC respectively, AD = BE, AE and CD intersect at point F, and Ag ⊥ CD. BD =/kloc-. In the known equilateral triangle ABC, DE is BC and AC respectively, AE=DC connecting AD and AD is at P.52065438 And CD=AE, which proves that DB = DE 652016-12-01equilateral triangle ABC has side lengths of 3, and points D and E are respectively. And satisfy ad db = CEEA =142013-07-21. In the equilateral triangle AB, AC with a side length of 2, D and E are the midpoint of AB and AC respectively, and the connection de (as shown in figure 12)2 is more similar? & gt recommended for you: F.context('cmsRight', [{'URL':'/d01373f082025aaf511aa256e9edab64034f1a07? x-BCE-process = image % 2f resize % 2Cm _ lfit % 2Cw _ 450% 2Ch _ 600% 2c limit _ 1% 2f quality % 2Cq _ 85% 2f format % 2Cf _ auto '，' contractId':'A24KA00562 '，}，{ ' URL ':/s？ word = % E6 % AC % A7 % E6 % B4 % B2 % E6 % 9D % AF & sa = searchpromo _ ozb _ zhidao _ tuijian '，' src ':'/3 BF 33 a 87 e 950352 AAF 5 df 4954 143 fbf2b 2 1 18 b6b？ x-BCE-process = image % 2f resize % 2Cm _ lfit % 2Cw _ 450% 2Ch _ 600% 2c limit _ 1% 2f quality % 2Cq _ 85% 2f format % 2Cf _ auto '，' contractId ':' '，}])； The price of electric cars has dropped many times. Is the quality guaranteed? What impact will the "network toilet" have? Is Huaqiang North's second-hand mobile phone reliable? Why is the cost of cancer treatment getting higher and higher? I recommend f.context ('recbrand', [{"img": "\/86D6277F9E2F07083523F69DFB 24B 899A901F20d? x-BCE-process = image % 2f resize % 2Cm _ lfit % 2Cw _ 450% 2Ch _ 600% 2c limit _ 1% 2f quality % 2Cq _ 85% 2f format % 2Cf _ auto "，" url":"/hm.js？ 6859 ce 5a af 00 FB 00387 e 6434 E4 FCC 925 "； var s = document . getelementsbytagname(" script ")[0]； s.parentNode.insertBefore(hm，s)； })(); window . TT = 1720 135282；