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.
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.
Answer:
:
Let ABCD be a parallelogram, where BE and BF are the perpendicular 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]
⇒x+∠ABF+∠FBE+∠EBC=180∘
⇒x+90∘−x+45∘+90∘−x=180∘
[∵InΔABF,∠ABF=90∘−xandinΔBEC,∠EBC=90∘−x]
⇒−x=180∘−225∘⇒x=45∘∴∠A=∠C=45∘∠B=45∘+45∘+45∘=135∘⇒∠D=135∘
Hence the angles are 45∘,135∘,45∘,135∘.
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:
Let ABCD be a parallelogram, where BE and BF are the perpendicular 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]
⇒x+∠ABF+∠FBE+∠EBC=180∘
⇒x+90∘−x+45∘+90∘−x=180∘
[∵InΔABF,∠ABF=90∘−xandinΔBEC,∠EBC=90∘−x]
⇒−x=180∘−225∘⇒x=45∘∴∠A=∠C=45∘∠B=45∘+45∘+45∘=135∘⇒∠D=135∘
Hence the angles are 45∘,135∘,45∘,135∘.
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