Can you calculate the angle X? | (Justify your answer) | #math #maths | #geometry

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Can you calculate the angle X? | (Justify your answer) | #math #maths | #geometry

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[apxmath]

Your methods are so unique!!! Beautiful

[marioalb]

Triangle ADF:
x + α + β = 180°
α = 77° (Vertex F)
β = 33° (Vertex D)
x = 180°-77°-33° = 70° ( Solved √ )

[jimlocke]

<EPF is an inscribed angle with measure 33°, so the arc it intercepts, EF, has twice that measure, or 66°. Let arc DE have measure a and BF have measure b. Then, arc BD has measure DE + EF + BF = a + 66° + b. However, arc BD is intercepted by inscribed angle <DCB, which has measure 77°, so has measure 154°. So, a + 66° + b = 154° and a + b = 88°. Construct BD. <ABD is an inscribed angle intercepting arc DF = DE + EF, so has measure (a + 66°)/2. <ADB is an inscribed angle intercepting arc EB = EF + BF, so has measure (66° + b)/2. <DAB is one angle of ΔABD, and the sum of interior angles is 180°, so <DAB + <ABD + <ADB = 180°. <DAB = x, <ABD = (a + 66°)/2, <ADB = (66° + b)/2, so x + (a + 66°)/2 + (66° + b)/2 = 180°, x + (a + b)/2 + 66° = 180°. Replace a + b by 88°. x + (88°)/2 + 66° = 180°, x + 44° + 66° = 180°, x + 110° = 180° and x = 70°, as PreMath also found.

[-wx--]

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.

[SkinnerRobot]

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.

[trishanuagarwal]

Sir u r doing a very good job
I have never seen such a consistent maths questioner
Keep bringing quality questions pls❤

[quigonkenny]

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° ]

[michaeldoerr]

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!!!

[alexundre]

Bom dia Mestre
Muito Obrigado por mais uma aula
Grato

[TurquoizeGoldscraper]

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.

[nandisaand]

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...

[zdrastvutye]

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
60
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
140
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
190
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 b:next a
230 250
240
250
260 for a=0 to n1:for b=0 to n2:x=x(a, b):y=y(a, b)
270 if x<xmin then xmin=x
280 if x>xmax then xmax=x
290 if y<ymin then ymin=y
300 if y>ymax 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<masy then mass=masx else mass=masy
360 for a=0 to n1:x=x(a, 0):y=y(a, 0):gosub 240:xba=xb:yba=yb
370 for b=1to n2:x=x(a, b):y=y(a, b):gosub 240:xbn=xb:ybn=yb:goto 390
380 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
390 gosub 380:next b:next a:x=xm:y=ym:gosub 240:circle xb, yb, r*mass
400 x=x(0, 0):y=y(0, 0):gosub 240:xba=xb:yba=yb:x=x(1, 3):y=y(1, 3):gosub 240:xbn=xb:ybn=yb
410 gosub 380:dx1=x(0, 3)-x(1, 3):dy1=y(0, 3)-y(1, 3):dx2=x(1, 0)-x(1, 3):dy2=y(1, 0)-y(1, 3)
420
430 cw=(zx+zy)/n:if abs(cw)>1 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.

[PrithwirajSen-njqq]

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

[imrannadafnadaf]

x=(marcDPC-marcEF) /2
=(206-66) /2=70

[LuisdeBritoCamacho]

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.

[DB-lgsq]

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

[imetroangola]

*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°*

[robinharwood]

My answer is completely unjustified, and I intend to leave it that way.