The side AC of a\[\Delta ABC\]is extended to D such That\[BC=CD.\] If \[\angle ACB\]is \[70{}^\circ ,\]then \[\angle ADB\]is equal to |
A) \[35{}^\circ \]
B) \[45{}^\circ \]
C) \[70{}^\circ \]
D) \[110{}^\circ \]
Correct Answer: A
Solution :
\[\angle ACB+\angle BCD=180{}^\circ \] [linear pair] |
\[\angle BCD=180{}^\circ -70{}^\circ =110{}^\circ \] |
In \[\Delta BCD,\] \[BC=CD\] |
\[\angle CBD=\angle CDB\] ... (i) |
[angles opposite to equal sides] |
Also, \[\angle BCD+\angle CBD+\angle CDB=180{}^\circ \] |
\[2\angle CDB=180{}^\circ -\angle BCD\] |
\[=180{}^\circ -110{}^\circ =70{}^\circ \] |
\[\therefore \] \[\angle CDB=\angle ADB=\frac{70{}^\circ }{2}=35{}^\circ \] |
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