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Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
1951 Poland - Second Round
5
5
Part of
1951 Poland - Second Round
Problems
(1)
(a^2 + b^2) \sin (A - B) = (a^2 - b^2) \sin (A + B)
Source: Polish MO second round 1951 p5
8/28/2024
Prove that if the relationship between the sides and opposite angles
A
A
A
and
B
B
B
of the triangle
A
B
C
ABC
A
BC
is
(
a
2
+
b
2
)
sin
(
A
−
B
)
=
(
a
2
−
b
2
)
sin
(
A
+
B
)
(a^2 + b^2) \sin (A - B) = (a^2 - b^2) \sin (A + B)
(
a
2
+
b
2
)
sin
(
A
−
B
)
=
(
a
2
−
b
2
)
sin
(
A
+
B
)
then such a triangle is right-angled or isosceles.
geometry
trigonometry