

* If angle x is in radian, curved side BKC=ax & curved side DKE=ax1.
..........(1) * BC=2a*sin(x/2) and DE=2a*sin(x1/2) .....(1a) Since
BC^{2}=a^{2}+a^{2}2a^{2}cos x
* BkC/BC=(x/2)/sin (x/2) and DkE/DE=(x1/2)/sin (x1/2)
.........(1b)
As x > 0, the RHS tends to 1 and Bkc=BC and
DkE=DE
* AL/KL= a*cos (x/2)/[a*(1cos (x/2)]=1/[1/cos(x/2) 1]
and AT/KT= 1/[1/cos(x1/2) 1] ......(1c)
* Perimeter of the curved triangle ABkC is a(2+x) & curved triangle
ADkE=a(2+x1) ........(2)
* Area of the curved triangle ABkC is (π*a^{2}/2π)*x =
a^{2}x/2=(1/2)*ax*a .........(3) * Area of the curved triangle ADkE is (π*a^{2}/2π)*x1 =
a^{2}x1/2=(1/2)*ax1*a ........(4) * Analogous to a
triangle ABC where a/(b+ca) +b/(a+cb)+c/(a+bc) =z which has a
minimum value 3, here z=2/ x +x/(2x) for
Curved triangle ABkC ........(5) z1=2/ x1 +x1/(2x1)
for Curved triangle ADkE ......(6) * Triangular inequality : a+ax
>a => ax > 0 which is true .....(7) *
Triangular
inequality : a+a >ax may or may not hold good unlike a st.
line triangle where it invariably holds good. * Putting the cosine
law, a^{2}x^{2}=a^{2}+a^{2}2a^{2}cos Y
where Y is a hypothetical angle, we get Y=2*sin^{1}(x/2) *
Similarly, Putting the cosine
law, a^{2}=a^{2}x^{2}+a^{2}2x^{2}a^{2}cos
U where U is a hypothetical angle, we get U=cos^{1}(x/2)
* In curved triangle ABkC, sum of angles Y+2U= 2*sin^{1}(x/2)
+2*cos^{1}(x/2) = 2*π/2 = π where Y,U are
hypothetical angles. Same for the other curved triangle.
While analysing the equations of curved triangles, one finds that
the area formula is equivalent to that of straight triangle.The
triangular inequality partly holds good. Cosine formula can be
fitted in with hypothetical angles etc.
Explanation: Suppose slope of AE is m2, Coordinate of E is (x2,y2),
then m2=(y2h1)/(x2g1) or y2=m2x2+(h1g1m2)=m2x2+k where k=h1g1m2
or y2=m2x2+k.......(5)
Now, (g1x2)^{2} +(h1y2)^{2} =a^{2} or
(g1x2)^{2} +(h1km2x2)^{2} =a^{2} or
(g1x2)^{2} +(g1m2m2x2)^{2} =a^{2} or
on simplification, we get
(1+m2^{2} )x2^{2} + (2g1)(1+m2^{2}
)x2 + (g1^{2}m2^{2} +g1^{2}  a^{2} ) =0
which is a quadratic equn. which gives 2 value of x2 for each value
of m and hence we get in total 4 values of x2.
* If the slope of AE is m2', then the x & y coordinate of E (m2,n2)
are related by formula m2m2' g1m2'+h1=n2
; and m2 can be found out by solving the quadratic equation
(1+m2'*m2')m2*m2 +2g1(m2'*m2' 1)m2 +(g1*g1+g1*g1*m2'*m2'a*a)=0.
since m2' has 2 values and for each value of m2', m2 has 2 values =>
m2 has total 4 values .
* TZ=a*sin(x1r/2)*tan(x1r/2) and AZ=AT+TZ=a / Cos(x1r/2).
Previously, TZ was written as [(a*cosx1)/(2*cos(x1/2))] which is
wrong. pl. check.
* Value of AT, TZ found out from length and from coordinates tally
for angle > 0 and up to 162.54 degree and at 270 degree. From
coordinates , we take AT1 and AT8 for AT and TZ1 & TZ4 for TZ.
Reasons not yet explored. 
Value of a 

Value of x in Degree 

Value of x1 in Degree 

Data below is independent of data above 

xcoordinate of center at A(g1) 

ycoordinate of center at A(h1) 

xcoordinate of Dput any arbitrary value (m1) 




curved1b 
x in radian 

Comment on x 

x1 in radian 

Comment on x1 

BC 

DE 

DZ=ZE [a1=a/(2*cos (x1/2))] 

a1/a [1/(2*cos (x1/2)] 

AT (a*cos[x1/2]) 

TZ [a*sin(x1/2)*tan(x1/2)] 

AZ [a/( cos (x1/2)] 

AT/TZ 

area of Δ DAE = a^{2} * sinx1 / 2 

area of Δ DZE = a1^{2} * sin(180°x1) / 2 =(a1^{2} *sinx1
)/ 2 

area DAEZ =( a^{2}+a1^{2} )*(sinx1) / 2 

area of Δ DAE / area of Δ DZE 

BkC arc 

DkE arc 

area of DBkCZ (red area)=[sinx1*(a^{2}+a1^{2}) /2]  a^{2}x1/2 



BkC/BC 

DkE/DE 

AL/KL 

AT/TK 

sum of 2 sides/3rd side(2a/ax) in ABkC 

sum of 2 sides/3rd side(2a/ax1) in ADkE 

Perimeter of ABkC:P1 [a(2+x)] 

Perimeter of ADkE:P2 [a(2+x1)] 

P1/P2 

Area of ABkC:A1 

Area of ADkE:A2 

Area of DkET:A3 [(a^{2}/2)*(x1sinx1)] 

Area of BkCL:A4 [(a^{2}/2)*(xsinx)] 

A1/A2 

A3/A4 [(x1sinx1)/(xsinx)], x,x1 in radian 

Angle Y 

Angle U 

Y+2U in degree 

Angle Y1 

Angle U1 

Y1+2U1 in degree 

If BC becomes the new arc, corresponding angle
x2 is 
(radian)
AND
(degree) 
If BC becomes the new arc of the circle, % change in angle x
is(Ve) 
%
(this is independent of magnitude of radius) 
If angle between AB & AC are reduced to zero,
maximum stretch of chord beyond k is DE/2 = asin(x1/2) or BC/2=a*sin(x/2)
as the case may be. Taking start point as BC, stretch is asin(x/2) 

Look to fig1 and fig. 2
and find perimeter , area of fig 2. Green lines are 
tangent to the circle with radius a and join
at the center of circle 
Perimeter (perimeterfactor*a) 

Area
(areafactor*a^{2}) 

Area factor 

Perimeter factor 



General Converter 

Put angle in Radians 

Find angle in Degree 

sin 

cos 

tan 



Equation of the circle 
(x
)^{2} +(y)^{2}
=
^{2 }= 
or simplified equation of circle 
x^{2 }+ y^{2 } () x  () y + () = 0 
ycoordinate of D(n1a) & (n1b) =± √[a^{2}(m1g1)^{2}] +h1 
and

Slope of AD w.r.t. xaxis 
and

Slope of DZ w.r.t. xaxis (DZ is perpendicular to AD) 
and

YIntercept of DZ 
,
,, 
Equn. of st. line DZ......(1) 
y=x
+

do
.....(2) 
y=x
+

do
.....(3) 
y=x
+

do
.....(4) 
y=x
+

Slope of AE w.r.t. xaxis
((slopeADtanx1)/(1+slopeAD*tanx1)) 
and

Slope of EZ w.r.t. xaxis (EZ is perpendicular to AE) 
and

YIntercept of EZ for slope

,

do for slope

, 
xcoordinate of E (m2) for slope

,

xcoordinate of E (m2) for slope

and

ycoordinate of E (n2) for slope

,

ycoordinate of E (n2) for slope

,

Equn. of st. line EZ......(1) 
y=x
+

do
.....(2) 
y=x
+

do
.....(3) 
y=x
+

do
.....(4) 
y=x
+

CoOrdinates of Z set 1(z1x,z1y) (AZ=
) 
&

CoOrdinates of Z set 2(z2x,z2y) (AZ=
) 
&

CoOrdinates of Z set 3(z3x,z3y) (AZ=
) 
&

CoOrdinates of Z set 4(z4x,z4y) (AZ=
) 
&

coordinate of T, midpoint of DE (x,y) T1* lengthAT1

&

Slope of AT1 & AT8 (length of
AT1=length of AT8) 
&

do T2 * lengthAT2

&

doT3 * lengthAT3

&

doT4 * lengthAT4

&

do T5 * lengthAT5

&

doT6 * lengthAT6

&

doT7 * lengthAT7

&

do T8 * lengthAT8

&

Slope of TZ1 & TZ4 
&

Length of TZ 1 (t1ax,t1ay) (z1x,z1y) 

Length of TZ 2 (t2ax,t2ay) (z2x,z2y) 

Length of TZ 3 (t3bx,t3by) (z3x,z3y) 

Length of TZ 4 (t4bx,t4by)(z4x,z4y) 

Length of TZ 5 (t1ax,t1ay)
(z2x,z2y) 

Length of TZ 6 (t2ax,t2ay)
(z1x,z1y) 

Length of TZ 7 (t3bx,t3by)
(z4x,z4y) 

Length of TZ 8 (t4bx,t4by)
(z3x,z3y) 

Length of TZ 9 (t1ax,t1ay)
(z3x,z3y) 

Length of TZ 10 (t2ax,t2ay)
(z4x,z4y) 

Length of TZ 11 (t1ax,t1ay)
(z4x,z4y) 

Length of TZ 12 (t2ax,t2ay)
(z3x,z3y) 

Length of TZ 13 (t3bx,t3by)
(z1x,z1y) 

Length of TZ 14 (t4bx,t4by)
(z2x,z2y) 

Length of TZ 15 (t3bx,t3by)
(z2x,z2y) 

Length of TZ 16 (t4bx,t4by)
(z1x,z1y) 

Length of TZ 17 (t3ax,t3ay)
(z3x,z3y) 

Length of TZ 18 (t4ax,t4ay)
(z4x,z4y) 

Length of TZ 19 (t1bx,t1by)
(z1x,z1y) 

Length of TZ 20 (t2bx,t2by)
(z2x,z2y) 

Length of TZ 21 (t4ax,t4ay)
(z3x,z3y) 

Length of TZ 22 (t3ax,t3ay)
(z4x,z4y) 

Length of TZ 23 (t2bx,t2by)
(z1x,z1y) 

Length of TZ 24 (t1bx,t1by)
(z2x,z2y) 

Length of TZ 25 (t3ax,t3ay)
(z1x,z1y) 

Length of TZ 26 (t4ax,t4ay)
(z2x,z2y) 

Length of TZ 27 (t3ax,t3ay)
(z2x,z2y) 

Length of TZ 28 (t4ax,t4ay)
(z1x,z1y) 

Length of TZ 29 (t1bx,t1by)
(z3x,z3y) 

Length of TZ 30 (t2bx,t2by)
(z4x,z4y) 

Length of TZ 31 (t1bx,t1by)
(z4x,z4y) 

Length of TZ 32 (t2bx,t2by)
(z3x,z3y) 


