An easier way would be multiplying by -1 to solve for x first 9x+5y= 181 (8x+5y= 177)(-1) ============== x= 4 8(4)+5y= 177 32+5y= 177 5y= 45 y= 29 x+y= 33°
This is as far as I can go. 01) (10Y - 3)º + (17X + 5)º = 360º 02) (10Y)º + (17X)º = 358º 03) Y = - 17X/10 + 358/10 04) And the Solution is an Equation of a Straight LIne with Infinite Solutions. 05) I could say that (10Y - 3)º > 180º and (17X + 5)º < 180º. 06) Solving these 2 Inequations I can state that: 07) (10Y - 3)º > (17X + 5)º 08) Y > 17,7º and : 0º < X < 10,3º 09) This is a sort of Linear Programming Problem with a Solution Area. I am going to give a thought about this!!
Awesome video! Is it arbitrary which equations you add up? Or is the only reason for picking any two is that the arithmetic is clearer and easier for them?
My way of solution ▶ The sum of the opposite interior angles of a Cyclic Quadrilateral is 180°, therefore: ∠ADC + ∠CBA= 180° 8x+3+5y= 180° 8x+5y= 177° ∠BAD + ∠DCB= 180° 5y-3+9x+2= 180° 5y+9x= 181° ⇒ 5y+8x= 177° 5y+9x= 181° ⇒ 5y+9x= 181° -5y-8x= -177° ⇒ x= 4° y= (181°-9*4°)/5 y= 29° x+y= 4°+29° x+y= 33° is the answer !
Another way of solution ▶ ∠ ADC is the half of the circle segment ∠CA ∠CA= 2* ∠ ADC ∠CA= 2* (8x+3) ∠CA= 16x+6 ∠CBA is the half of the circle segment ∠AC ∠AC= 2* ∠CBA ∠AC= 2*5y ∠AC= 10y ⇒ the sum of the both circle segments is equal to 360° ∠CA+ ∠AC = 360° 16x+6 + 10y= 360° by dividing both sides of 2, we get: 8x+3+5y= 180° 8x+5y= 177° b) ∠BAD is the half of the circle segment ∠DB ∠DB= 2* ∠BAD ∠DB= 2*(5y-3) ∠DB= 10y-6 ∠DCB is the half of the circle segment ∠BD ∠BD= 2* ∠DCB ∠BD= 2*(9x+2) ∠BD= 18x+4 ⇒ the sum of the both circle segments is equal to 360° ∠DB+ ∠BD = 360° 10y-6+18x+4= 360° 10y+18x-2= 360° by dividing both sides of 2, we get: 5y+9x-1= 180° 5y+9x= 181° ⇒ 5y+8x= 177° 5y+9x= 181° ⇒ x= 4° y= (181°-9*4°)/5 y= 29° x+y= 4°+29° x+y= 33°
The answer is 33. Also I have to agree with some of the comments that you ckuldnhave just used the fact that it is a cyclic quadrilateral and just computed the opposite angles. I couls be wrong.
5y - 3° + 9x+ 2° = 8x+ 3° + 5y = 180° x = 4° 5y = 180° - 3° - 8x = 177°-8x y = 29° x+y = 33° ( Solved √ ) Where are those who everytime write "units" or " square units" ??? instead a recognized standard unit. Those who claim when we write "cm" or any unit of the international units system Why don't you write "units" now, in your comments ??? or "angle units" ??? if you pay attention, the statement of this exercise DOESN'T mention that are degrees , and in that case, the given data would be wrong !!
5y - 3 + 9x+ 2 = 8x+ 3 + 5y x = 4 radians 8x+ 3 + 5y = π 5y = π - 3 - 8x = 0,141592 -8x y = -6,37 radians, IMPOSIBLE !!! This exercise just works with sexagesimal degrees, and was not set in the input data !!! Where are those that everythime writes "units" ?? instead a recognized standard unit !!!
I had to look up the properties of a CQ first as it's been a while. It looks like x = 4 and y = 29, but I stand to be corrected. edit; Nearly forgot to answer the question LOL. x+y = 33
i have found a solution without any mathbooster theorem: 10 print "premath-can you find the value of x+y(cyclic quadrilateral)":dim x(3),y(3) 20 r=1:x(0)=r/3.9:wnu=rad(35):y(0)=r-sqr(r*r-(x(0)-r)^2):p1=(x(0)-r)*cos(wnu) 30 p2=(y(0)-r)*sin(wnu):p=p1+p2:q=(x(0)-r)^2+(y(0)-r)^2-r*r:xm=r:ym=r 40 l=-p+sqr(p*p-q):print l:dx=l*cos(wnu):dy=l*sin(wnu):sw=.000001 50 x(1)=x(0)+dx:y(1)=y(0)+dy:goto 440 60 xu1=x(0):yu1=y(0):w1=8*gx+rad(3):w2=9*gx+rad(2):dx03=r*cos(w1+wnu):xu2=xu1+dx03 70 dy03=r*sin(w1+wnu):yu2=yu1+dy03:gosub 80:goto 120 80 dx=xu2-xu1:dy=yu2-yu1:zx=dx*(xu1-xm):zy=dy*(yu1-ym):pk=(zx+zy)/(dx^2+dy^2) 90 qk=((xu1-xm)^2+(yu1-ym)^2-r*r)/(dx^2+dy^2):dis=pk^2-qk:k=-pk+sqr(dis):rem print k:stop 100 if abs(k)1 then return 200 wu=acs(cw):return 210 dx1=x(0)-x(3):dy1=y(0)-y(3):dx2=x(2)-x(3):dy2=y(2)-y(3):dxl3=x(2)-x(0):dyl3=y(2)-y(0) 220 gosub 150 230 if n=0 then return else w4=wu 240 dx1=x(1)-x(2):dy1=y(1)-y(2):dx2=-dx2:dy2=-dy2:dxl3=x(1)-x(3):dyl3=y(1)-y(3) 250 gosub 150 260 if n=0 then return else w3=wu 270 rem goto 270 280 rem if l1=0 or l2=0 then return else w3=wu 290 dgu1=w4*5*gy/4/pi^2:dgu2=(5*gy-rad(3))*w3/4/pi^2:dg=dgu1-dgu2:return 300 gx=sw:gosub 60:rem goto 310 310 if n=0 then else 330 320 gx=gx+sw:gosub 60:goto 310 330 dg1=dg:gx1=gx:gx=gx+sw:if gx>2*pi then stop 340 gx2=gx:gosub 60:rem goto 325 350 if n=0 then stop:300 else 360 360 if abs(dg)0 then 330 380 gx=(gx1+gx2)/2:gosub 60:if n=0 then 320 390 if abs(dg1-dg)0 then gx1=gx else gx2=gx 420 if abs(dg)>1E-10 then 380 430 return 440 gosub 300:mass=1200/2/r:gosub 450:goto 470 450 print "gx=",gx,"%","gy=",gy:print deg(gx),"%",deg(gy),"grad":print "x+y=";deg(gx)+deg(gy);"grad" 460 gosub 510:cls:return 470 gx=gx+sw:goto 490 480 gosub 60:gosub 310:return 490 gosub 480:gosub 450:goto 470 500 xbu=x*mass:ybu=y*mass:return 510 gcol5:x=x(0):y=y(0):gosub 500:xba=xbu:yba=ybu:for a=1 to 4:ia=a:if ia=4 then ia=0 520 x=x(ia):y=y(ia):gosub 500:xbn=xbu:ybn=ybu:goto 540 530 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 540 gosub 530:next a:x=xm:y=ym:gcol8:gosub 500:circle xbu,ybu,r*mass 550 print "eine taste druecken" 560 a$=inkey$(0):if a$="" then 560 else return gx= 0.2585845% gy= 0.185234222 14.8158005% 10.6131392grad x+y=25.4289397grad eine taste druecken because w4/w3=(5*y-3)/5y as written in line 290. run in bbc basic sdl and hit ctrl tab to copy from the results window. you may add "@zoom%=@zoom%*1.4" at the beginning
Thank you
You are very welcome!
Thanks for the feedback ❤️
Opposite angles adding up to 180 degrees.
(5y - 3) + (9x + 2) = 180..........(1)
Also, 5y + (8x + 3) = 180.........(2)
Subtracting (2) from (1).
(5y - 3) + (9x + 2) - 5y - (8x + 3) = 0.
5y - 3 + 9x + 2 - 5y - 8x - 3 = 0.
x - 4 = 0.
x = 4.
Substituting in (2).
5y + 35 = 180.
5y = 145.
y = 29.
x + y = 4 + 29 = 33.
Excellent!
Thanks for sharing ❤️
You only needed two equations. You didn’t need equation 3. Two equations and two unknowns will get you the solution. No need to do extra work.
I would do that, too, if only to "check my work". 🧠
Do you think a tenured math professor doesn’t know that?
Obviously two equations and two unknowns are needed, however, I think Mister wanted to detail another resource for the solution
Você tem toda razão!!!
@@joeschmo622 não sei se ele é professor de matemática, pois professor com a matemática apurada não faz isso!
Interesting and easy too!
Thanks for the feedback ❤️
✨Magic!✨
Excellent!
Glad to hear that!
Thanks for the feedback ❤️
An easier way would be multiplying by -1 to solve for x first
9x+5y= 181
(8x+5y= 177)(-1)
==============
x= 4
8(4)+5y= 177
32+5y= 177
5y= 45
y= 29
x+y= 33°
Excellent!
Thanks for sharing ❤️
Very nice
Good job
Thanks Sir
With my respects
❤❤❤❤❤❤
So nice of you dear
You are very welcome!
Thanks for the feedback ❤️
Thank you!
You are very welcome!
Thanks for the feedback ❤️
8x+3+5y=180°
5y-3+9x+2=180°
8x+3+5y=5y-3+9x+2
-x=-4
x=4
y=29
x+y=4+29=33
Answer: 33
Excellent!
Thanks for sharing ❤️
9x+5y-3+2=180
9x+5y=181 (1)
8x+5y+3=180
8x+5y=177 (2)
(1)&(2)
9x-8x=181-177=4
So x=4°
9(4)+5y=181
5y=181-36=145
So y=45°
So x+y=4+29=33°.❤❤❤
Excellent!
Thanks for sharing ❤️
29 + 4 = 33 Answer
Super very easy
In a cyclic quadrilateral, the sum of the opposite sides =180
Hence, (5y -3) + 9x + 2 = 180 or
5y + 9x - 1 = 180 equation 1
and 5y + (8x +3) = 180 equation 2
-------------------------------------------- subtract
0 + x - 4 = 0
x = 4
5y + (8*4 + 3) = 180 substitute into equation 2
5y + 35 = 180
5y = 145
y = 29
x+ y = 4+ 29 = 33
Excellent!
Thanks for sharing ❤️
Solution:
It's a Cyclic Quadrilateral, therefore, the opposite interior angles must add up to 180°
(9x + 2) + (5y - 3) = 180°
(8x + 3) + 5y = 180°
9x + 2 + 5y - 3 = 180°
9x + 5y = 181° ... ¹
8x + 3 + 5y = 180°
8x + 5y = 177° ... ²
¹ - ²
x = 4
8 (4) + 5y = 177°
32 + 5y = 177
5y = 177 - 32
y = 145/5
y = 29
x + y = 4 + 29
x + y = 33
Excellent!
Thanks for sharing ❤️
Let's face this challenge:
.
..
...
....
.....
A convex quadrilateral ABCD is cyclic if and only if its opposite angles add up to 180°:
∠BAD + ∠DCB = ∠CBA + ∠ADC = 180°
(5y − 3°) + (9x + 2°) = (5y) + (8x + 3°)
5y − 3° + 9x + 2° = 5y + 8x + 3°
⇒ x = 4°
5y − 3° + 9x + 2° = 180°
5y + 9x = 181°
5y + 9*4° = 181°
5y + 36° = 181°
5y = 145°
⇒ y = 29°
5y + 8x + 3° = 180°
5y + 8x = 177°
5y + 8*4° = 177°
5y + 32° = 177°
5y = 145°
⇒ y = 29° ✓
Now we are able to solve this problem:
x + y = 4° + 29° = 33°
Best regards from Germany
Excellent!
Thanks for sharing ❤️
This is as far as I can go.
01) (10Y - 3)º + (17X + 5)º = 360º
02) (10Y)º + (17X)º = 358º
03) Y = - 17X/10 + 358/10
04) And the Solution is an Equation of a Straight LIne with Infinite Solutions.
05) I could say that (10Y - 3)º > 180º and (17X + 5)º < 180º.
06) Solving these 2 Inequations I can state that:
07) (10Y - 3)º > (17X + 5)º
08) Y > 17,7º and : 0º < X < 10,3º
09) This is a sort of Linear Programming Problem with a Solution Area.
I am going to give a thought about this!!
5y + 8x + 3 = 9x + 2 + 5y - 3 → x = 4; 10y + 17x = 358 → y = 29 → x + y = 33
Excellent!
Thanks for sharing ❤️
Solution:
In a Cyclic Quadrilateral the opposite angles are together 180°. Therefore is:
(1) 5y-3+9x+2 = 180
(2) 5y+8x+3 = 180
(1) - (2) = (3) x-4 = 0 |+4 ⟹ (3a) x = 4 |in (2) ⟹
(2a) 5y+8*4+3 = 180 ⟹
(2b) 5y+35 = 180 |-35 ⟹
(2c) 5y = 145 |/5 ⟹
(2d) y = 29 ⟹ x+y = 4+29 = 33
Awesome video! Is it arbitrary which equations you add up? Or is the only reason for picking any two is that the arithmetic is clearer and easier for them?
Equation 1 is essential! Thanks for asking😀❤️
My way of solution ▶
The sum of the opposite interior angles of a Cyclic Quadrilateral is 180°,
therefore:
∠ADC + ∠CBA= 180°
8x+3+5y= 180°
8x+5y= 177°
∠BAD + ∠DCB= 180°
5y-3+9x+2= 180°
5y+9x= 181°
⇒
5y+8x= 177°
5y+9x= 181°
⇒
5y+9x= 181°
-5y-8x= -177°
⇒
x= 4°
y= (181°-9*4°)/5
y= 29°
x+y= 4°+29°
x+y= 33° is the answer !
Another way of solution ▶
∠ ADC is the half of the circle segment ∠CA
∠CA= 2* ∠ ADC
∠CA= 2* (8x+3)
∠CA= 16x+6
∠CBA is the half of the circle segment ∠AC
∠AC= 2* ∠CBA
∠AC= 2*5y
∠AC= 10y
⇒
the sum of the both circle segments is equal to 360°
∠CA+ ∠AC = 360°
16x+6 + 10y= 360°
by dividing both sides of 2, we get:
8x+3+5y= 180°
8x+5y= 177°
b) ∠BAD is the half of the circle segment ∠DB
∠DB= 2* ∠BAD
∠DB= 2*(5y-3)
∠DB= 10y-6
∠DCB is the half of the circle segment ∠BD
∠BD= 2* ∠DCB
∠BD= 2*(9x+2)
∠BD= 18x+4
⇒
the sum of the both circle segments is equal to 360°
∠DB+ ∠BD = 360°
10y-6+18x+4= 360°
10y+18x-2= 360°
by dividing both sides of 2, we get:
5y+9x-1= 180°
5y+9x= 181°
⇒
5y+8x= 177°
5y+9x= 181°
⇒
x= 4°
y= (181°-9*4°)/5
y= 29°
x+y= 4°+29°
x+y= 33°
Excellent!
Thanks for sharing ❤️
The answer is 33. Also I have to agree with some of the comments that you ckuldnhave just used the fact that it is a cyclic quadrilateral and just computed the opposite angles. I couls be wrong.
Thanks for the feedback ❤️
5 y - 3 + 9 x + 2
= 5 y + 8 x + 3
= 180°
x = 4
& 5 y + 9 * 4 = 180 + 1 i.e y = 29 °
Hereby x + y = 33 °
Excellent!
Thanks for sharing ❤️
5y - 3° + 9x+ 2° = 8x+ 3° + 5y = 180°
x = 4°
5y = 180° - 3° - 8x = 177°-8x
y = 29°
x+y = 33° ( Solved √ )
Where are those who everytime write "units" or " square units" ??? instead a recognized standard unit.
Those who claim when we write "cm" or any unit of the international units system
Why don't you write "units" now, in your comments ??? or "angle units" ???
if you pay attention, the statement of this exercise DOESN'T mention that are degrees , and in that case, the given data would be wrong !!
5y - 3 + 9x+ 2 = 8x+ 3 + 5y
x = 4 radians
8x+ 3 + 5y = π
5y = π - 3 - 8x = 0,141592 -8x
y = -6,37 radians, IMPOSIBLE !!!
This exercise just works with sexagesimal degrees,
and was not set in the input data !!!
Where are those that everythime writes "units" ??
instead a recognized standard unit !!!
😀
Thanks for the feedback ❤️
Also 2A+2C=360 (or 2B+2D=360) gives the second needed equation.
Thanks for the feedback ❤️
I had to look up the properties of a CQ first as it's been a while.
It looks like x = 4 and y = 29, but I stand to be corrected.
edit; Nearly forgot to answer the question LOL.
x+y = 33
Excellent!
Thanks for sharing ❤️
i have found a solution without any mathbooster theorem:
10 print "premath-can you find the value of x+y(cyclic quadrilateral)":dim x(3),y(3)
20 r=1:x(0)=r/3.9:wnu=rad(35):y(0)=r-sqr(r*r-(x(0)-r)^2):p1=(x(0)-r)*cos(wnu)
30 p2=(y(0)-r)*sin(wnu):p=p1+p2:q=(x(0)-r)^2+(y(0)-r)^2-r*r:xm=r:ym=r
40 l=-p+sqr(p*p-q):print l:dx=l*cos(wnu):dy=l*sin(wnu):sw=.000001
50 x(1)=x(0)+dx:y(1)=y(0)+dy:goto 440
60 xu1=x(0):yu1=y(0):w1=8*gx+rad(3):w2=9*gx+rad(2):dx03=r*cos(w1+wnu):xu2=xu1+dx03
70 dy03=r*sin(w1+wnu):yu2=yu1+dy03:gosub 80:goto 120
80 dx=xu2-xu1:dy=yu2-yu1:zx=dx*(xu1-xm):zy=dy*(yu1-ym):pk=(zx+zy)/(dx^2+dy^2)
90 qk=((xu1-xm)^2+(yu1-ym)^2-r*r)/(dx^2+dy^2):dis=pk^2-qk:k=-pk+sqr(dis):rem print k:stop
100 if abs(k)1 then return
200 wu=acs(cw):return
210 dx1=x(0)-x(3):dy1=y(0)-y(3):dx2=x(2)-x(3):dy2=y(2)-y(3):dxl3=x(2)-x(0):dyl3=y(2)-y(0)
220 gosub 150
230 if n=0 then return else w4=wu
240 dx1=x(1)-x(2):dy1=y(1)-y(2):dx2=-dx2:dy2=-dy2:dxl3=x(1)-x(3):dyl3=y(1)-y(3)
250 gosub 150
260 if n=0 then return else w3=wu
270 rem goto 270
280 rem if l1=0 or l2=0 then return else w3=wu
290 dgu1=w4*5*gy/4/pi^2:dgu2=(5*gy-rad(3))*w3/4/pi^2:dg=dgu1-dgu2:return
300 gx=sw:gosub 60:rem goto 310
310 if n=0 then else 330
320 gx=gx+sw:gosub 60:goto 310
330 dg1=dg:gx1=gx:gx=gx+sw:if gx>2*pi then stop
340 gx2=gx:gosub 60:rem goto 325
350 if n=0 then stop:300 else 360
360 if abs(dg)0 then 330
380 gx=(gx1+gx2)/2:gosub 60:if n=0 then 320
390 if abs(dg1-dg)0 then gx1=gx else gx2=gx
420 if abs(dg)>1E-10 then 380
430 return
440 gosub 300:mass=1200/2/r:gosub 450:goto 470
450 print "gx=",gx,"%","gy=",gy:print deg(gx),"%",deg(gy),"grad":print "x+y=";deg(gx)+deg(gy);"grad"
460 gosub 510:cls:return
470 gx=gx+sw:goto 490
480 gosub 60:gosub 310:return
490 gosub 480:gosub 450:goto 470
500 xbu=x*mass:ybu=y*mass:return
510 gcol5:x=x(0):y=y(0):gosub 500:xba=xbu:yba=ybu:for a=1 to 4:ia=a:if ia=4 then ia=0
520 x=x(ia):y=y(ia):gosub 500:xbn=xbu:ybn=ybu:goto 540
530 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
540 gosub 530:next a:x=xm:y=ym:gcol8:gosub 500:circle xbu,ybu,r*mass
550 print "eine taste druecken"
560 a$=inkey$(0):if a$="" then 560 else return
gx= 0.2585845% gy= 0.185234222
14.8158005% 10.6131392grad
x+y=25.4289397grad
eine taste druecken
because w4/w3=(5*y-3)/5y as written in line 290.
run in bbc basic sdl and hit ctrl tab to copy from the results window. you may add "@zoom%=@zoom%*1.4" at the beginning
The problem is easy to solve .... but only once you know that the angles are in degrees !
I love this problem because I have the guilty pleasure of falling down the Rabbit hole to the orthocenter too meet the Grand Poobah of Rabbits. 🙂
Wow😀
Thanks for the feedback ❤️
Linear equations are also tested
Thanks for the feedback ❤️
33 is the correct answer
trivial
Thanks for the feedback ❤️
Too routine.😢
Thanks for the feedback ❤️
ไม่จำเป็นต้องเอา(-2)× แต่ให้เอา (1)-(2)จะได้9X+5Y-8X-5Y=181-177 ; X=4
Thanks for sharing ❤️