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# In the figure given below, if O is the centre of the circle and

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In the figure given below, if O is the centre of the circle and  [#permalink]

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27 Oct 2018, 00:53
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In the figure given below, if O is the center of the circle and $$\angle BAO=54^{\circ}$$, then what is $$\angle OAD+ \angle OCD?$$
Attachment:

Image.jpg [ 18.36 KiB | Viewed 249 times ]

A. $$108^{\circ}$$
B. $$126^{\circ}$$
C. $$112^{\circ}$$
D. $$120^{\circ}$$
E. $$128^{\circ}$$

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Re: In the figure given below, if O is the centre of the circle and  [#permalink]

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27 Oct 2018, 01:41
eswarchethu135 wrote:
In the figure given below, if O is the center of the circle and $$\angle BAO=54^{\circ}$$, then what is $$\angle OAD+ \angle OCD?$$
Attachment:
The attachment Image.jpg is no longer available

A. $$108^{\circ}$$
B. $$126^{\circ}$$
C. $$112^{\circ}$$
D. $$120^{\circ}$$
E. $$128^{\circ}$$

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Re: In the figure given below, if O is the centre of the circle and  [#permalink]

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27 Oct 2018, 02:06
ABCD is an isosceles trapezoid. (AB = DC)

OB = OA
∠B = ∠ C = 54

∠AOB = 180-(54+54)=72
∠AOC = 180-72 = 108
∠A + ∠D = 360 - ∠B - ∠C = 252
∠A = ∠D ; ∠D = 126

∠OAD + ∠OCD = 360- ∠AOC - ∠D = 360 - 108 - 126 = 126
Re: In the figure given below, if O is the centre of the circle and   [#permalink] 27 Oct 2018, 02:06
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