Given : BC = CE, ∠CBE = 84° and ∠ADC = 78° To find : ∠DEC = $$ heta$$ = ? Solution : Adjacent angles of a rhombus are supplementary => ∠ADC + ∠BCD = $$180^circ$$ => ∠BCD = $$180-78=102^circ$$ ---------------(i) $$ riangle$$ BCE is an isosceles triangle with BC = CE, => ∠CBE = ∠CEB = 84° Thus, in $$ riangle$$ BCE, => ∠CBE + ∠CEB + ∠BCE = $$180^circ$$ => ∠BCE = $$180-84-84=12^circ$$ ---------------(ii) Adding equations (i) and (ii), => ∠BCD + ∠BCE = $$102+12$$ => ∠DCE = $$114^circ$$ Now, ABCD is a rhombus and BC = CE, => CD = CE and thus CDE is an isosceles triangle with ∠CDE = ∠DEC = $$ heta$$ In $$ riangle$$CDE => $$ heta+ heta+angle DCE = 180^circ$$ => $$2 heta = 180-114=66^circ$$ => $$ heta = frac{66}{2}=33^circ$$ => Ans - (C)