In the given figure chord $E D$ is parallel to diameter $A C$ of the circle given $∠ C B E = 65^{\circ} c a l c u l a t e ∠ D E C .$

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Solution

$\begin{matrix} F r o m t h e g i v e n f i g u r e J o i n O E a r c E C s u b t e n d s ∠ E O C a t t h e c e n t r e a n d ∠ E B C a t t h e r e m a i n i n g p a r t s o f t h e c i r c l e ∠ E O C = 2 ∠ E B C = 2 \times 65^{\circ} = 130^{\circ} N o w, I n Δ O E C, O E = O C (r a d i i o f t h e s a m e c i r c l e) ∴ ∠ O E C = ∠ O C E B u t i n Δ E O C ∠ O E C + ∠ O C E + ∠ E O C = 180^{\circ} \Rightarrow 2 ∠ O C E = 180^{\circ} - 130^{\circ} ∴ ∠ O C E = 25^{\circ} A C ∥ E D [g i v e n] ∴ ∠ D E C = ∠ O C E \Rightarrow ∠ D E C = 25^{\circ} \end{matrix}$

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