In the given figure 'O' is the centre of the circle and AB , CD are equal to chords .If $< A O B = 70^{o} k$ .Find the angles of $△ O C D$ .

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Solution

From the given figure
$\begin{matrix} I n Δ A O B a n d Δ C O D O A = O D [b o t h a r e r a d i u s] O B = O C [b o t h a r e r a d i u s] A B = D C [C h o r d a r e e q u a l] ∴ Δ A O B ≅ Δ C O D (b y s - s - s c o n g r u e n t) ∴ ∠ A O B = ∠ C O D = 70^{\circ} N o w I n Δ O C D O C = O D (b o t h a r e r a d i u s) t h e n, ∠ O D C = ∠ O C D = x (L e t) ∴ s u m o f a n g l e o f Δ = 180^{\circ} ∴ x + x + 70^{\circ} = 180^{\circ} \Rightarrow 2 x = 180^{\circ} - 70^{\circ} ∴ x = \frac{110^{\circ}}{2} H e n c e ∠ O D C = 55^{\circ} ∠ O C D = 55^{\circ} ∠ C O D = 70^{\circ} \end{matrix}$

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Similar questions

∠ A O B = 70^{o}

△ O C D

In the figure below, O is the centre of the circle and AB, CD are chords, with ∠OAB = ∠OCD,

Prove that AB = CD.

A B

C D

O

P

A B, Q

C D

∠ P O Q = 150^{o}

∠ A P Q

∠ A O B = 80^{\circ}

∠ C O D

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