A) \[55{}^\circ \]
B) \[35{}^\circ \]
C) \[145{}^\circ \]
D) \[125{}^\circ \]
Correct Answer: C
Solution :
[c] In \[\Delta ACD\] \[\angle DAC=55{}^\circ \] [given] \[\angle ACD=90{}^\circ =\] angle in a semicircle \[\angle ADC=180{}^\circ -90{}^\circ -55{}^\circ \] \[=180{}^\circ -145{}^\circ =35{}^\circ \] Now, in a cyclic quadrilateral Sum of opposite angles \[=\text{ }180{}^\circ \] \[\angle ABC+\angle ADC=180{}^\circ \] \[\angle ABC=180{}^\circ -\angle ADC\] \[=180{}^\circ -35{}^\circ \] \[=145{}^\circ \] |
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