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The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60^{\circ}. Find the angles of the parallelogram.

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Answer: 60^{\circ},120^{\circ}

Solution.
Given that the angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60^{\circ}.

In this figure
\angle ADC and  \angle ABC are obtuse angles and DE and DF are altitudes.
In quadrilateral BEDF,
\angle BED= \angle BFD = 90^{\circ}
 \angle FBE=360^{\circ}-(\angle FDE+\angle BED+\angle BFD)
=360^{\circ}-(60^{\circ}+90^{\circ}+90^{\circ})

=360^{\circ}-240^{\circ}
=120^{\circ}
It is given that ABCD is a parallelogram
ADC = 120^{\circ}
\angle A+\angle B=180^{\circ}  {Sum of interior angle =180^{\circ}}

\angle A=180^{\circ}-B
\angle A=180^{\circ}-120^{\circ}
\angle A=60^{\circ}
\Rightarrow \angle C=\angle A=60^{\circ}
Hence angles of the parallelogram are 60^{\circ}, 120^{\circ}

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