Question

in the adjoining figure,ABCD is a parallelogram in which angle BAD 75° and angle DBC =60° calculate angle
CDB​

Answer 1

Explanation:

In ▲DBC,

→ $\bf\angle{DBC + \angle BCD + \angle CDB = 180°}$

→  $\bf\angle{60° + 75° + \angle CDB = 180°}$

Note that angle BAD is equal to angle BCD as opposite sides of parallelogram are equal.

→  $\bf\angle{135° + \angle CDB = 180°}$

→  $\bf{\angle CDB = 180-135}$

→  $\bf{\angle CDB =45° }$

Final Answer:

$\bf{\angle CDB =45° }$