Isosceles triangle pt.2
A
B
C
is an isosceles
triangle
with
A
C
=
B
C
and has an angle
A
C
B
=
1
0
8
∘
.
If the ratio of
B
C
A
B
can be represented as
c
a
+
b
, where
a
,
b
and
c
positive integers
with
b
square-free, find
a
+
b
+
c
.
Bonus : Try to do this without trigonometry.
The answer is 8.
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2 solutions
I've submitted another solution (without using trigonometry) to above my first.
@ahmad saad nice solution ;;))
how did you get it as 2cos36
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AB/BC=sin{ACB}/sin{CAB}=sin{108}/sin{36}=sin{180-72}/sin{36}=sin{72}/sin{36}
= 2*sin{36}.cos{36}/sin{36}=2*cos{36} .... (sinse, sin{36} =/= 0)
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sin (180-72)=cos 72
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@Abhishek Alva – No, the correct identity is : sin(180-72)=sin72 & cos72=sin(90-72)
U
s
i
n
g
S
i
n
L
a
w
:
−
S
i
n
3
6
B
C
=
S
i
n
1
0
8
A
B
=
S
i
n
(
3
∗
3
6
)
A
B
=
S
i
n
3
6
∗
(
3
−
4
∗
S
i
n
2
3
6
)
A
B
.
B
u
t
S
i
n
3
6
=
8
5
−
5
∴
B
C
A
B
=
3
−
2
5
−
5
=
2
1
+
5
=
c
a
+
b
.
⟹
a
+
b
+
c
=
1
+
5
+
2
=
8
.
O
R
L
e
t
C
F
⊥
A
B
.
∠
C
A
B
=
3
6
o
,
a
n
d
A
F
=
2
1
∗
A
B
.
C
A
2
1
∗
A
B
=
C
A
A
F
=
C
A
∗
C
o
s
C
A
B
=
C
A
∗
C
o
s
3
6
=
C
A
∗
4
1
+
5
.
⟹
C
A
A
B
=
2
1
+
5
=
c
a
+
b
.
⟹
a
+
b
+
c
=
1
+
5
+
2
=
8
.
<ACB=108 ---> <CAB=<CBA=36 , sin108 = sin72 = 2sin36.cos36
AB/BC = sin108/sin36 = 2cos36 , cos36=(1+sqrt5)/4 .... it is easy to prove that.
AB/BC = (1+sqrt5)/2 ---> a=1 , b=5 & c=2 ---> a+b+c = 8
Anther solution without using Trigonometry