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National and Regional Contests
Netherlands Contests
Dutch Mathematical Olympiad
1973 Dutch Mathematical Olympiad
3
3
Part of
1973 Dutch Mathematical Olympiad
Problems
(1)
<PCA=? <PAB = 30^o, <PBA = 20^o in 100-40-40 triangle
Source: Netherlands - Dutch NMO 1973 p3
1/27/2023
The angles
A
A
A
and
B
B
B
of base of the isosceles triangle
A
B
C
ABC
A
BC
are equal to
4
0
o
40^o
4
0
o
. Inside
△
A
B
C
\vartriangle ABC
△
A
BC
,
P
P
P
is such that
∠
P
A
B
=
3
0
o
\angle PAB = 30^o
∠
P
A
B
=
3
0
o
and
∠
P
B
A
=
2
0
o
\angle PBA = 20^o
∠
PB
A
=
2
0
o
. Calculate, without table,
∠
P
C
A
\angle PCA
∠
PC
A
.
angles
geometry
isosceles