Drop a perpendicular from C to AB and label the intersection as point E. Let length AE = x and CE = y, therefore length BE = 10 - x. Consider ΔACE. By Pythagoras, x° + y° = (12)² = 144. Consider ΔBCE. (10-x)² + y² = 6² = 36. Expanding, 100 - 20x + x² + y² = 36. Replace x° + y° by 144 and 100 - 20x + 144 = 36 which solves for x = 10.4. Therefore, point E is to the right of point B, not between A and B, which would be implied by the diagram as drawn. So, ΔABC is an obtuse Δ, with 90°. What about length BE = 10 - 10.4 = -0.4? Length can not be negative. Since our calculations indicate the E may be to the right of B, BE = AE - AB = x - 10. Our equation for ΔBCE is now (x - 10)² + y² = 36. However, after expanding, we get the same equation as before and our calculations are still valid. We use the length AD = 8, as found in the video, so length DE = 10.4 - 8 = 2.4. Consider ΔCDE. (Length CD)² = (2.4)² + y² = (2.4)² + 144 -x² = (2.4)² + 144 - (10.4)² = 5.76 + 144 - 108.16 = 41.6 and length CD = √(41.6), as Math Booster also found. So, the diagram showed ΔABC as an acute Δ and our first cut at a solution assumed that it was acute. It turned out that Δ was actually obtuse with 90°, but we were able to recover and find the correct solution.
It is more normal to use as few letters to label your diagram as possible.You only needed two letters to mark the lengths on the whole diagram. If FC is a, so is EC and CG. AF is 12-a=AI. BG is 6-a = BH. ID then = DE =DH = b. AB then =12-a+b+b+6-a=10, so a=4+b. AD then =8, and BD =2. Then complete as you did.
Using CD = 4/5 * sqrt(65), you can continue the problem to find that the larger radius R = 16 * sqrt(14)/(25 + sqrt(65)) and the smaller radius r = 4 * sqrt(14)/(10 + sqrt(65)).
A me risulta 6,45..ma ci sono decimali,quindi non sono molto sicuro del risultato..ho usato le formule di Briggs per calcolare i 3 angoli,poi ho calcolato R e r,raggi dei due cerchi e infine il teorema dei seni .
Drop a perpendicular from C to AB and label the intersection as point E. Let length AE = x and CE = y, therefore length BE = 10 - x. Consider ΔACE. By Pythagoras, x° + y° = (12)² = 144. Consider ΔBCE. (10-x)² + y² = 6² = 36. Expanding, 100 - 20x + x² + y² = 36. Replace x° + y° by 144 and 100 - 20x + 144 = 36 which solves for x = 10.4. Therefore, point E is to the right of point B, not between A and B, which would be implied by the diagram as drawn. So, ΔABC is an obtuse Δ, with 90°. What about length BE = 10 - 10.4 = -0.4? Length can not be negative. Since our calculations indicate the E may be to the right of B, BE = AE - AB = x - 10. Our equation for ΔBCE is now (x - 10)² + y² = 36. However, after expanding, we get the same equation as before and our calculations are still valid. We use the length AD = 8, as found in the video, so length DE = 10.4 - 8 = 2.4. Consider ΔCDE. (Length CD)² = (2.4)² + y² = (2.4)² + 144 -x² = (2.4)² + 144 - (10.4)² = 5.76 + 144 - 108.16 = 41.6 and length CD = √(41.6), as Math Booster also found.
So, the diagram showed ΔABC as an acute Δ and our first cut at a solution assumed that it was acute. It turned out that Δ was actually obtuse with 90°, but we were able to recover and find the correct solution.
It is more normal to use as few letters to label your diagram as possible.You only needed two letters to mark the lengths on the whole diagram.
If FC is a, so is EC and CG. AF is 12-a=AI. BG is 6-a = BH. ID then = DE =DH = b.
AB then =12-a+b+b+6-a=10, so a=4+b. AD then =8, and BD =2. Then complete as you did.
I think he explains such extended way for mass better understanding
before i looked for the result, i wrote the graphics. without calculating angle bisectors, the number of unknown coordinates would be too much:
10 dim x(2),y(2):l1=12:l2=6:l3=10:@zoom%=1.4*@zoom%:if l3>l1+l2 then else 30
20 print "ungueltige eingabe":end
30 la=l1:lb=l2:lc=l3:lh=(la^2-lb^2+lc^2)/2/lc:hdr=sqr(la^2-lh^2)
40 x(0)=0:y(0)=0:x(1)=l3:y(1)=0:x(2)=lh:y(2)=hdr
50 adr=lc*hdr/2:dx1=lh:dy1=hdr:n1=sqr(lh^2+hdr^2)
60 xg11m=x(0):yg11m=y(0):dx1=lh:dy1=hdr:n1=sqr(hdr^2+lh^2):rem verschiedene variablen in haupt
70 x1=l1*dx1/n1:y1=l1*dy1/n1:x2=l1:y2=0:xg12m=(x1+x2)/2:yg12m=(y1+y2)/2:rem und subroutine verwenden
80 xg41=l3:yg41=0:goto 170
90 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12)
100 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22)
110 a13=a131+a132:a23=a231+a232
120 ngl1=a12*a21:ngl2=a22*a11
130 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end
140 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2
150 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2
160 xl=zx/ngl:yl=zy/ngl:return:rem print "x=";xl;"y=";yl
170 dxac=lh:dyac=hdr:n=sqr(dxac^2+dyac^2):dxac=l1*dxac/n:dyac=l1*dyac/n
180 dxab=l1:dyab=0:xg12m=(dxac+dxab)/2:yg12m=(dyac+dyab)/2:rem diese winkelhalbierende bleibt gleich ***
190 xg21m=x(2):yg21m=y(2):xg31m=xg21m:yg31m=yg21m:rem diese punkte bleiben gleich
200 dxbc=l3-lh:dybc=hdr:n=sqr(dxbc^2+dybc^2):dxbc=l1*dxbc/n:dybc=l1*dybc/n
210 xg41m=l3:yg41m=0:dxcb=l3-lh:dycb=hdr:n=sqr(dxcb^2+dycb^2):dxcb=l1*dxcb/n:dycb=l1*dycb/n
220 dxdb=l1:dydb=0:n=sqr(dxdb^2+dydb^2):dxdb=l1*dxdb/n:dydb=l1*dydb/n:rem print dydb:stop
230 xg421=xg41m-dxcb:yg421=yg41+dycb:xg422=xg41m-dxab:yg422=yg41m+dyab :rem stop
240 xg42m=(xg421+xg422)/2:yg42m=(yg421+yg422)/2:sw=sqr(l1^2+l2^2+l3^2)/1E3:lad=sw:yg1=0
250 xd=lh:yd=hdr:yg2=yd:yg31m=0:goto 410:rem die abstaende der punkte auf der tangente berechnen
260 dxdc=lh-lad:dydc=hdr:n=sqr(dxdc^2+dydc^2):dxdc=l1*dxdc/n:dydc=l1*dydc/n
270 xg221=xg21m-dxac:yg221=yg21m-dyac:xg222=xg21m-dxdc:yg222=yg21m-dydc
280 xg22m=(xg222+xg221)/2:yg22m=(yg222+yg221)/2:xg31m=lad
290 xg321=xg31m+dxdc:yg321=yg31m+dydc :rem stop
300 xg322=xg31m+dxab:yg322=yg31m+dyab:xg32m=(xg321+xg322)/2:yg32m=(yg321+yg322)/2 :rem stop
310 xg11=xg11m:yg11=yg11m:xg12=xg12m:yg12=yg12m:rem das lot faellen auf cd
320 xg11=xg11m:yg11=yg11m:xg12=xg12m:yg12=yg12m:xg21=xg21m:yg21=yg21m:xg22=xg22m:yg22=yg22m
330 gosub 90:xml=xl:yml=yl:xg11=xg31m:yg11=yg31m:xg12=xg32m:yg12=yg32m:xg21=xg41m:yg21=yg41m
340 xg22=xg42m:yg22=yg42m:gosub 90:xmr=xl:ymr=yl:rem print xml,"%",yml:print xmr,"%",ymr:stop
350 xg1=lad:xg2=lh:dx=xg2-xg1:dy=yg2-yg1:xp=xml:yp=yml:goto 380
360 zx=(xp-xg1)*dx:zy=(yp-yg1)*dy:k=(zx+zy)/(dx^2+dy^2):dxlo=k*dx:dylo=k*dy
370 xlo=xg1+dxlo:ylo=yg1+dylo:s=sqr((xp-xlo)^2+(yp-ylo)^2) :return
380 gosub 360:rl=s:xsl=xlo:ysl=ylo:xp=xmr:yp=ymr:gosub 360:rr=s:xsr=xlo:ysr=ylo
390 dgu1=(xsl-xd)^2/l1^2:dgu2=(ysl-yd)^2/l1^2:dgu3=(xsr-xd)^2/l1^2:dgu4=(ysr-yd)^2/l1^2
400 dg=dgu1+dgu2-dgu3-dgu4: return
410 gosub 260
420 lad1=lad:dg1=dg:lad=lad+sw:if lad>10*l1 then stop
430 lad2=lad:gosub 260:if dg1*dg>0 then 420
440 lad=(lad1+lad2)/2:gosub 260:if dg1*dg>0 then lad1=lad else lad2=lad
450 if abs(dg)>1E-10 then 440
460 print lad:mass=11E2/l1:goto 480
470 xb=x*mass:yb=y*mass:return
480 xba=0:yba=0:for a=1 to 3:ia=a:if ia=3 then ia=0
490 x=x(ia):y=y(ia):gosub 470:xbn=xb:ybn=yb:gosub 500:goto 510
500 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
510 next a:x=xd:y=yd:gosub 470:xba=xb:yba=yb:x=lad:y=0:gosub 470:xbn=xb:ybn=yb
520 gcol 4:gosub 500:gcol2:x=xml:y=yml:gosub 470:circle xb,yb,rl*mass
530 x=xmr:y=ymr:gosub 470:circle xb,yb,rr*mass
540 x=xg31m:y=yg31m:gosub 470:xba=xb:yba=yb:x=xg32m:y=yg32m:gosub 470:xbn=xb:ybn=yb:gosub 500
550 x=xg41m:y=yg41m:gosub 470:xba=xb:yba=yb:x=xg42m:y=yg42m:gosub 470:xbn=xb:ybn=yb:gosub 500
560 led=sqr((xd-lad)^2+yd^2):print "die gesuchte entfernung ist=";led
8
die gesuchte entfernung ist=6.4498062
>
run in bbc basic sdl and hit ctrl tab to copy from the results window
Using CD = 4/5 * sqrt(65), you can continue the problem to find that the larger radius R = 16 * sqrt(14)/(25 + sqrt(65)) and the smaller radius r = 4 * sqrt(14)/(10 + sqrt(65)).
Angle B>90' C1 is not equal to C2 The drawing is erroneous
Yes, badly so. This is very nearly a 30°60°90° triangle. 6, 12, 6sqr3=10.3923...
Why not simply subtract 2 from 10 to get the length of AD?
A me risulta 6,45..ma ci sono decimali,quindi non sono molto sicuro del risultato..ho usato le formule di Briggs per calcolare i 3 angoli,poi ho calcolato R e r,raggi dei due cerchi e infine il teorema dei seni .
Long and complicated and boring and what in the beginning seemed easy came out to be unnecessarily trivial.