A) \[\angle D\]
B) \[\angle AOD\]
C) \[{{45}^{o}}\]
D) indeterminate
Correct Answer: C
Solution :
First, we draw the diagonal AC. Now, in \[{{30}^{o}}\] and in \[\angle B={{90}^{o}},\angle C={{30}^{o}}\], we have \[\angle PXR=\angle UXS\]and \[{{180}^{o}}\](common) \[{{0}^{o}}\] \[{{360}^{o}}\] \[{{90}^{o}}\] \[\angle DAB={{75}^{o}}\] coincides with \[\angle DBC={{60}^{o}}\] Since, ABCD is a cyclic quadrilateral in which sum of opposite angles is \[\angle CDB=....\] i.e., \[{{60}^{o}}\] \[{{75}^{o}}\] Each angle = \[{{45}^{o}}\]\[{{135}^{o}}\] \[\sqrt{407}\] .You need to login to perform this action.
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