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In the given figure, ABCD and BDCE are parallelograms with common base DC. If BC ⊥ BD, then ∠BEC =(a) 60o (b) 30o (c) 150o (d) 120o |
Answer» (a) 60o From the given figure, ∠BAD = 30o ∠BCD = 30o … [∵opposite angles of parallelogram are equal] Now, let us consider the triangle CBD From angle sum property, ∠DBC + ∠BCD + ∠CDB = 180o 90o + 30o + ∠CDB = 180o 120o + ∠CDB = 180o ∠CDB = 180o – 120o ∠CDB = 60o ∴∠BEC = 60o, because opposite angles of parallelogram are equal. |
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