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Question
In Fig. below, ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute
∠CDB and ∠ADB.
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Solution
To find `∠`CDB and `∠`ADB
`∠`CBD = `∠`ABD = 60° [Alternative interior angle AD || BC and BD is the transversal]
In a parallelogram ABCD
`∠`A = `∠`C = 75° [ ∵ Opposite side angles of a parallelogram are equal]
In `∠`BDC
`∠`CBD + `∠`C + `∠`CDB =180° [Angle sum property]
⇒ 60° + 75° + `∠`CDB = 180°
⇒`∠`CDB = 180° - (60° + 75°)
⇒ `∠`CDB = 45°
Hence `∠`CDB = 45°, `∠`ADB = 60°
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