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Question
The angle between the two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 45°. Find the angles of the parallelogram.
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Solution
Let ABCD be a parallelogram, where BE and BF are the perpendiculars through the vertex B to the sides DC and AD, respectively.
Let ∠A = ∠C = x, ∠B = ∠D = y ...[Opposite angles are equal in parallelogram]
Now, ∠A + ∠B = 180° ...[Adjacent sides of a parallelogram are supplementary]
In triangle ABF;
∠ABF = 90° – x
And in triangle BEC,
∠EBC = 90° – x
So, x + 90° – x + 45° + 90° – x = 180°
⇒ – x = 180° – 225°
⇒ x = 45°
So, ∠A = ∠C = 45°
∠B = 45° + 45° + 45° = 135°
⇒ ∠D = 135°
Hence, the angles are 45°, 135°, 45° and 135°.
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