Arc DEFB is twice the ∡C (i.e. 154°), then arc BCPD is 360°−154° = 206°. Arc EF is twice the ∡P (66°). x = ½(⌒BCPD−⌒EF) = 70° as the angle between two secants.
Draw DF. As both D and P are on the circumference and ∠EDF and ∠EPF both subtend arc EF, them ∠EDF = ∠EPF = 33°. As AB is a straight line, ∠DFB is an exterior angle to triangle ∆AFD at F, so ∠DFB = ∠FDA+∠DAF = 33°+x. As DFBC is a cyclic quadrilateral, its opposite vertices must sum to 180°. ∠DFB + ∠BCD = 180° 33° + x + 77° = 180° x + 110° = 180° x = 180° - 110° [ x = 70° ]
The angle P is on a circle so the arc EF is double the angle of P: EF = 2P = 2 * 33 = 66. The angle of E through F, back to E and stopping at F is a full circle + EF: Circle + EF = 360 + 66 = 426. That large arc can be split into two smaller arcs: EFBC and CPDE: EFBC + CPDE = 426. Angles B and D are on the circle and match the above arcs, there their angles are half of the arcs: B + D = 0.5 * (EFBC + CPDE) = 0.5 * 426 = 213. A quadrilateral has angles adding up to 360: B + C + D + x = 360. B + C + D + x = 360 x = 360 - (B + C + D) x = 360 - (B + D) - C x = 360 - 213 - 77 x = 70 Therefore, angle x is 70 degrees.
sometimes people must eat what's on the table. it is remarkable the angle x does neither depend on "w01=" in line 30 nor on "wkr=" in line 80. important is to consider that line fp and line bc are parallel: 10 print "premath-can you calculate the angle x" 20 dim x(1,3),y(1,3):n1=1:n2=3:yvi=850:xvi=1200 30 w01=15:w1=77:w2=33:r=1:xm=r:ym=r:x(0,0)=r*.8:y(0,0)=ym-sqr(r*r-(x(0,0)-xm)^2) 40 print y(0,0):wu=w01:xu=x(0,0):yu=y(0,0):gosub 50:goto 70 50 px=(xu-xm)*cos(rad(wu)):py=(yu-ym)*sin(rad(wu)):p=px+py:q=r*r-(xu-xm)^2-(yu-ym)^2 60 lu=-p+sqr(p*p+q):dx=lu*cos(rad(wu)):dy=lu*sin(rad(wu)):xn=xu+dx:yn=yu+dy:return 70 x(0,1)=xn:y(0,1)=yn:wu=180+wu-w1:xu=x(0,1):yu=y(0,1):gosub 50:x(0,2)=xn:y(0,2)=yn 80 wkr=200:dx=r*cos(rad(wkr)):dy=r*sin(rad(wkr)):x(1,0)=xm+dx:y(1,0)=ym+dy:xs1=x(1,0):ys1=y(1,0) 90 wu=w01:xu=xs1:yu=ys1:gosub 50:x(1,1)=xn:y(1,1)=yn:xu=xn:yu=yn:print xn,yn:wu=180+wu-33 100 gosub 50:xs2=xn:ys2=yn:x(1,2)=xn:y(1,2)=yn:x(0,3)=x(1,2):y(0,3)=y(1,2) 110 xg11=x(1,0):yg11=y(1,0):rem einen geradenschnittpunkt berechnen 120 xg12=x(0,0):yg12=y(0,0):xg21=xs2:yg21=ys2:xg22=x(0,2):yg22=y(0,2) 130 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12) 140 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22) 150 a13=a131+a132:a23=a231+a232 160 ngl1=a12*a21:ngl2=a22*a11 170 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end 180 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2 190 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2 200 xl=zx/ngl:yl=zy/ngl:print "x=";xl;"y=";yl:x(1,3)=xl:y(1,3)=yl 210 for a=0 to n1:for b=0 to n2:x=x(a,b):y=y(a,b) 220 sux=sux+x:suy=suy+y:anz=anz+1:next b:next a 230 mittelx=sux/anz:mittely=suy/anz:goto 250 240 xb=x-xmin:xb=xb*mass:yb=y-ymin:yb=yb*mass:return 250 xmin=mittelx:ymin=mittely:xmax=xmin:ymax=ymin 260 for a=0 to n1:for b=0 to n2:x=x(a,b):y=y(a,b) 270 if xxmax then xmax=x 290 if yymax then ymax=y 310 next b:next a:if xmin=xmax or ymin=ymax then else 350 320 if xmin=xmax then else 340 330 mass=yvi/ymin-ymax):goto 360 340 mass=xvi/(xmin-xmax):goto 360 350 masx=xvi/(xmax-xmin):masy=yvi/(ymax-ymin):if masx1 then stop else w=deg(acs(cw)):print "der winkel x=";w 440 a$=inkey$(0):if a$="" then 440 premath-can you calculate the angle x 0.0202041029 1.984807751.17364818 x=-0.589731019y=1.21845419 der winkel x=70 run in bbc basic sdl and hit ctrl tab to copy from the results window.
We may join EB. Exterior ang BED of 🔺 ABE= x + ABE --- (1) ang ABE= ang EPF =33 degs (both are inscribed on same arc) -(2) any BED = 180 -77=103 degs (as BECD is a cyclic quadrilateral) ---(3) From (1),(2)& (3) x = 103-33=70 degs
This is an example of "easier than it looks"!!!. I am kind of thinking that maybe you should make a playlist of problems solvable by the Insribed Angle Theorem or Cyclic Quadrilateral theorem. Also the x = 70°. I better use that for practice in order to autment to end my geometry training so that I can be an expert mathematician!!!
Ordinarily I give Professor PreMath's problems a good shake before watching the solution. I stared at this one all of three minutes and knew Id never get it. Some of these "theorems" Prof whips out are so obscure and esoteric...
Your methods are so unique!!! Beautiful
Thanks for the kind words! ❤️
Triangle ADF:
x + α + β = 180°
α = 77° (Vertex F)
β = 33° (Vertex D)
x = 180°-77°-33° = 70° ( Solved √ )
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Arc DEFB is twice the ∡C (i.e. 154°), then arc BCPD is 360°−154° = 206°. Arc EF is twice the ∡P (66°).
x = ½(⌒BCPD−⌒EF) = 70° as the angle between two secants.
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I do like it.
That's a great step-by-step explanation! It's clear and easy to follow.
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@PrithwirajSen-nj6qq and@@PreMath Thank you for the compliments!
Sir u r doing a very good job
I have never seen such a consistent maths questioner
Keep bringing quality questions pls❤
Draw DF. As both D and P are on the circumference and ∠EDF and ∠EPF both subtend arc EF, them ∠EDF = ∠EPF = 33°.
As AB is a straight line, ∠DFB is an exterior angle to triangle ∆AFD at F, so ∠DFB = ∠FDA+∠DAF = 33°+x.
As DFBC is a cyclic quadrilateral, its opposite vertices must sum to 180°.
∠DFB + ∠BCD = 180°
33° + x + 77° = 180°
x + 110° = 180°
x = 180° - 110°
[ x = 70° ]
I had no idea the solution would be so simple. Another satisfying video to review some principles of angle geometry. Thank you very much, PreMath.
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Bom dia Mestre
Muito Obrigado por mais uma aula
Grato
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The angle P is on a circle so the arc EF is double the angle of P: EF = 2P = 2 * 33 = 66.
The angle of E through F, back to E and stopping at F is a full circle + EF: Circle + EF = 360 + 66 = 426.
That large arc can be split into two smaller arcs: EFBC and CPDE: EFBC + CPDE = 426.
Angles B and D are on the circle and match the above arcs, there their angles are half of the arcs: B + D = 0.5 * (EFBC + CPDE) = 0.5 * 426 = 213.
A quadrilateral has angles adding up to 360: B + C + D + x = 360.
B + C + D + x = 360
x = 360 - (B + C + D)
x = 360 - (B + D) - C
x = 360 - 213 - 77
x = 70
Therefore, angle x is 70 degrees.
sometimes people must eat what's on the table.
it is remarkable the angle x does neither depend on "w01=" in line 30 nor on "wkr=" in line 80. important is to consider that line fp and line bc are parallel:
10 print "premath-can you calculate the angle x"
20 dim x(1,3),y(1,3):n1=1:n2=3:yvi=850:xvi=1200
30 w01=15:w1=77:w2=33:r=1:xm=r:ym=r:x(0,0)=r*.8:y(0,0)=ym-sqr(r*r-(x(0,0)-xm)^2)
40 print y(0,0):wu=w01:xu=x(0,0):yu=y(0,0):gosub 50:goto 70
50 px=(xu-xm)*cos(rad(wu)):py=(yu-ym)*sin(rad(wu)):p=px+py:q=r*r-(xu-xm)^2-(yu-ym)^2
60 lu=-p+sqr(p*p+q):dx=lu*cos(rad(wu)):dy=lu*sin(rad(wu)):xn=xu+dx:yn=yu+dy:return
70 x(0,1)=xn:y(0,1)=yn:wu=180+wu-w1:xu=x(0,1):yu=y(0,1):gosub 50:x(0,2)=xn:y(0,2)=yn
80 wkr=200:dx=r*cos(rad(wkr)):dy=r*sin(rad(wkr)):x(1,0)=xm+dx:y(1,0)=ym+dy:xs1=x(1,0):ys1=y(1,0)
90 wu=w01:xu=xs1:yu=ys1:gosub 50:x(1,1)=xn:y(1,1)=yn:xu=xn:yu=yn:print xn,yn:wu=180+wu-33
100 gosub 50:xs2=xn:ys2=yn:x(1,2)=xn:y(1,2)=yn:x(0,3)=x(1,2):y(0,3)=y(1,2)
110 xg11=x(1,0):yg11=y(1,0):rem einen geradenschnittpunkt berechnen
120 xg12=x(0,0):yg12=y(0,0):xg21=xs2:yg21=ys2:xg22=x(0,2):yg22=y(0,2)
130 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12)
140 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22)
150 a13=a131+a132:a23=a231+a232
160 ngl1=a12*a21:ngl2=a22*a11
170 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end
180 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2
190 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2
200 xl=zx/ngl:yl=zy/ngl:print "x=";xl;"y=";yl:x(1,3)=xl:y(1,3)=yl
210 for a=0 to n1:for b=0 to n2:x=x(a,b):y=y(a,b)
220 sux=sux+x:suy=suy+y:anz=anz+1:next b:next a
230 mittelx=sux/anz:mittely=suy/anz:goto 250
240 xb=x-xmin:xb=xb*mass:yb=y-ymin:yb=yb*mass:return
250 xmin=mittelx:ymin=mittely:xmax=xmin:ymax=ymin
260 for a=0 to n1:for b=0 to n2:x=x(a,b):y=y(a,b)
270 if xxmax then xmax=x
290 if yymax then ymax=y
310 next b:next a:if xmin=xmax or ymin=ymax then else 350
320 if xmin=xmax then else 340
330 mass=yvi/ymin-ymax):goto 360
340 mass=xvi/(xmin-xmax):goto 360
350 masx=xvi/(xmax-xmin):masy=yvi/(ymax-ymin):if masx1 then stop else w=deg(acs(cw)):print "der winkel x=";w
440 a$=inkey$(0):if a$="" then 440
premath-can you calculate the angle x
0.0202041029
1.984807751.17364818
x=-0.589731019y=1.21845419
der winkel x=70
run in bbc basic sdl and hit ctrl tab to copy from the results window.
We may join EB.
Exterior ang BED of 🔺 ABE= x + ABE --- (1)
ang ABE= ang EPF =33 degs (both are inscribed on same arc) -(2)
any BED = 180 -77=103 degs (as
BECD is a cyclic quadrilateral) ---(3)
From (1),(2)& (3)
x = 103-33=70 degs
Excellent!
Thanks for sharing ❤️
Solution is very simple. Minor EF is 66° and major DB is 206°. x is equal to 1/2(206 - 66) which is 70°.
This is an example of "easier than it looks"!!!. I am kind of thinking that maybe you should make a playlist of problems solvable by the Insribed Angle Theorem or Cyclic Quadrilateral theorem. Also the x = 70°. I better use that for practice in order to autment to end my geometry training so that I can be an expert mathematician!!!
Excellent!
Thanks for the feedback ❤️
شكرا لكم على المجهودات
يمكن استعمال
x+D+C+B=360
E+F+D+C+B=540
(E+F)-x=180
x=1/2(206+FB+206 +ED)-180
=206+44-180
=70
x=(marcDPC-marcEF) /2
=(206-66) /2=70
X=70°>❤❤❤
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Ordinarily I give Professor PreMath's problems a good shake before watching the solution. I stared at this one all of three minutes and knew Id never get it. Some of these "theorems" Prof whips out are so obscure and esoteric...
Thanks for the feedback ❤️
Gonna do the YMCA dance now! 🙂
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) Inscribed Angle (BCD) = 77º
02) Central Angle (BOD) = 77º * 2 = 154º
03) Arclength (BEFD) = 154º
04) Arclength (BCPD) = 360º - 154º = 206º
05) Inscribed Angle (EPF) = 33º
06) Central Angle (EOF) = 66º
07) Arclength (EF) = 66º
08) X = (206º - 66º) / 2
09) X = 140º / 2
10) X = 70º
Thus,
OUR BEST ANSWER :
Angle X equal to 70º.
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70 degrees without pen and paper
70 degrees.
*Solution:*
x = (bow BCPD - bow EF)/2
bow BCPD= 360° - bow BEFD
bow BCPD= 360° - 77°×2
bow BCPD= 360° - 154° = 206°
bow EF = 33° × 2 = 66°.
Therefore,
x = (206° - 66°)/2 → *x = 70°*
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asnwer=120 is it
asnwer=70 what what why