The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of parallelogram is 60 degrees. find angles of the parallelogram?? please help day after is my final examination.
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In quad. DPBQ, by angle sum property we have
∠PDQ + ∠DPB + ∠B + ∠BQD = 360°
60° + 90° + ∠B + 90° = 360°
∠B = 360° – 240°
Therefore, ∠B = 120°
But ∠B = ∠D = 120° opposite angles of parallelogram
As, AB || CD opposite sides of a parallelogram
∠B + ∠C = 180° sum of adjacent interior angles is 180°
120° + ∠C = 180°
∠C = 180° – 120° = 60°
Hence ∠A = ∠C = 60° Opposite angles of parallelogram are equal
∠PDQ + ∠DPB + ∠B + ∠BQD = 360°
60° + 90° + ∠B + 90° = 360°
∠B = 360° – 240°
Therefore, ∠B = 120°
But ∠B = ∠D = 120° opposite angles of parallelogram
As, AB || CD opposite sides of a parallelogram
∠B + ∠C = 180° sum of adjacent interior angles is 180°
120° + ∠C = 180°
∠C = 180° – 120° = 60°
Hence ∠A = ∠C = 60° Opposite angles of parallelogram are equal
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Step-by-step explanation:
n quad. DPBQ, by angle sum property we have
∠PDQ + ∠DPB + ∠B + ∠BQD = 360°
60° + 90° + ∠B + 90° = 360°
∠B = 360° – 240°
Therefore, ∠B = 120°
But ∠B = ∠D = 120° opposite angles of parallelogram
As, AB || CD opposite sides of a parallelogram
∠B + ∠C = 180° sum of adjacent interior angles is 180°
120° + ∠C = 180°
∠C = 180° – 120° = 60°
Hence ∠A = ∠C = 60° Opposite angles of parallelogram are equal
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