Well, by quick inspection we have the following equations: r*sin(A1) = 7.5 r*sin(A2) = 3.5 4*A1 + 2*A2 = 180 So right away we have three equations in three unknowns. Manipulate the third equation: 4*A1 + 2*A2 = 180 2*A1 + A2 = 90 A2 = 90 - 2*A1 Substitute into the first to equations: r*sin(A1) = 7.5 r*sin(90-2*A1) = 3.5 --> r*cos(2*A1) = 3.5 --> r*(1-2*sin^2(A1)) = 3.5 Now let u = sin(A1). Then r*u = 7.5 --> u = 7.5/r r*(1-2*u^2) = 3.5 --> r*(1-2*(7.5/r)^2) = 3.5 r - 2*7.5^2/r = 3.5 r^2 - 3.5*r - 2*7.5^2 = 0 Roots are 12.5 and -9. A negative radius won't do, so r=12.5, d = 25. Q.E.D.
the following algorithm works, unless l10 then xs11=xs1 else xs12=xs1 90 if abs(dg)>1E-10 then 80 100 print "der radius=";r:mass=1E3/2/r:goto 120 110 xbu=x*mass:ybu=y*mass:return 120 x=r*2:y=0:gosub 110:xba=xbu:yba=ybu:for a=1 to nu:wa=a/nu*pi 130 x=r*cos(wa):x=x+r:y=r*sin(wa):gosub 110:xbn=xbu:ybn=ybu:goto 150 140 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 150 gosub 140:next a:x=0:y=0:gosub 110:gcol4:xba=xbu:yba=ybu:x=xs1:y=ys1:gosub 110 160 xbn=xbu:ybn=ybu:gosub 140:x=xs2:y=ys2:gosub 110:gcol5:xbn=xbu:ybn=ybu:gosub 140 170 x=2*r:y=0:gosub 110:xbn=xbu:ybn=ybu:gcol6:gosub 140 der radius=12.5 > run in bbc basic sdl and hit ctrl tab to copy from the results window
Let AB=d, BC=CD=15, DA=7, E be the point of intersection of the lines AD and BC. Since DC=BC and AD are the diameter, AC is the bisector and height of the BAE triangle. Therefore, AE=AB=d, BC=CE=15. From the equality EA * ED = EB * EC we get the equation d*(d-7) = 30 * 15. It follows that d=25. The root d=-18 does not satisfy.
A more simple way:put the 7cord between the two 15 cords ; now the spalier chord îs paralel with the diameter.From the center of the circle draw a line perpendicular on the smaler chord. Draw another line between the center of the circle and the point between the small and large chord. Now apply the Pitagorian teorema în the two triangles!
HOW TO FIND THE EXACT EXACT EXACT CIRCUMFERENCE OF CIRCLE . Procedure Step 1. Draw a square, 2 diagonals and inscribe a circle in the square. Thereby side and diameter will be the same Step 2 : Subtract 2 diagonals from the perimeter of square Step3. Divide step2 with 8 Step4. Add 3 times of the side. to Step 3 At the end we get the EXACT circumference of the inscribed circle
I'm 𝗰𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗶𝗻𝗴 you to solve this 𝗺𝗮𝘁𝗵 𝗽𝗿𝗼𝗯𝗹𝗲𝗺 and if you will be able to solve this i'll give you money as a reward. 𝗩𝗶𝗱𝗲𝗼 𝗹𝗶𝗻𝗸-:ua-cam.com/video/iFN4DMh8Wto/v-deo.htmlfeature=shared
Well, by quick inspection we have the following equations:
r*sin(A1) = 7.5
r*sin(A2) = 3.5
4*A1 + 2*A2 = 180
So right away we have three equations in three unknowns. Manipulate the third equation:
4*A1 + 2*A2 = 180
2*A1 + A2 = 90
A2 = 90 - 2*A1
Substitute into the first to equations:
r*sin(A1) = 7.5
r*sin(90-2*A1) = 3.5 --> r*cos(2*A1) = 3.5 --> r*(1-2*sin^2(A1)) = 3.5
Now let u = sin(A1). Then
r*u = 7.5 --> u = 7.5/r
r*(1-2*u^2) = 3.5 --> r*(1-2*(7.5/r)^2) = 3.5
r - 2*7.5^2/r = 3.5
r^2 - 3.5*r - 2*7.5^2 = 0
Roots are 12.5 and -9. A negative radius won't do, so r=12.5, d = 25.
Q.E.D.
the following algorithm works, unless l10 then xs11=xs1 else xs12=xs1
90 if abs(dg)>1E-10 then 80
100 print "der radius=";r:mass=1E3/2/r:goto 120
110 xbu=x*mass:ybu=y*mass:return
120 x=r*2:y=0:gosub 110:xba=xbu:yba=ybu:for a=1 to nu:wa=a/nu*pi
130 x=r*cos(wa):x=x+r:y=r*sin(wa):gosub 110:xbn=xbu:ybn=ybu:goto 150
140 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
150 gosub 140:next a:x=0:y=0:gosub 110:gcol4:xba=xbu:yba=ybu:x=xs1:y=ys1:gosub 110
160 xbn=xbu:ybn=ybu:gosub 140:x=xs2:y=ys2:gosub 110:gcol5:xbn=xbu:ybn=ybu:gosub 140
170 x=2*r:y=0:gosub 110:xbn=xbu:ybn=ybu:gcol6:gosub 140
der radius=12.5
>
run in bbc basic sdl and hit ctrl tab to copy from the results window
I did enjoy the video, your straightforward style is awesome; I suscribe right now.
Let AB=d, BC=CD=15, DA=7, E be the point of intersection of the lines AD and BC. Since DC=BC and AD are the diameter, AC is the bisector and height of the BAE triangle. Therefore, AE=AB=d, BC=CE=15.
From the equality EA * ED = EB * EC we get the equation d*(d-7) = 30 * 15. It follows that d=25. The root d=-18 does not satisfy.
Well done dude, only mistake AB is diameter instead of AD
A more simple way:put the 7cord between the two 15 cords ; now the spalier chord îs paralel with the diameter.From the center of the circle draw a line perpendicular on the smaler chord. Draw another line between the center of the circle and the point between the small and large chord. Now apply the Pitagorian teorema în the two triangles!
This is clever, turning a non-symmetric problem to a symmetric one.
HOW TO FIND THE EXACT EXACT EXACT CIRCUMFERENCE OF CIRCLE
. Procedure
Step 1. Draw a square, 2 diagonals and inscribe a circle in the square. Thereby side and diameter will be the same Step 2 : Subtract 2 diagonals from the perimeter of square Step3. Divide step2 with 8 Step4. Add 3 times of the side. to Step 3
At the end we get the EXACT circumference of the inscribed circle
I did it with trig; your way is much better. Nice!
beautiful!
I'm 𝗰𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗶𝗻𝗴 you to solve this 𝗺𝗮𝘁𝗵 𝗽𝗿𝗼𝗯𝗹𝗲𝗺 and if you will be able to solve this i'll give you money as a reward. 𝗩𝗶𝗱𝗲𝗼 𝗹𝗶𝗻𝗸-:ua-cam.com/video/iFN4DMh8Wto/v-deo.htmlfeature=shared
Would you rather vibes 💀💀💀
Very good teacher
25
Oui tres beau