of
the
inscribed circle
l
=
1,
m =
1/2,
n
=
1/3.
PROBLEM
1
a
sin
2
=
£
=
P
sin
/3
_=
P
m
~
=
2p
sin
^
=

=
3p
2
n
Since for each
4
in
general:
sin2
^ +
sin2
~
+
sin2
~
+ 2
sin
~
sin
^
sin
^
=
1,
after the
substitution
of the above values the
equation
will read:
1614 p2
+
12 p3

1
0
P
=
x
p
x
16p
+
1614
x
12
x
1
=
0
1614x
+
12

x3
=
0
or
:
x3

16
14x

12
=
0
Now
we
have
to
apply
Cardan's formula
x
=
i
I+
tir
+
fPl
J
[2] [3]
+
J
7
2
4
fSL
1
12J
+
fPl
[3]
where
p
=
1614,
q
=
121412.
The discriminant
J
f7
I2
fPl3
\fql
[2]
+.
L3J
[2]
/'
,4r\
+
!£.
fPl'
L3J
is
negative,
consequently its root is irrational.
Hence the trigonometric
method
should be
applied.
Then
22