A) \[35{}^\circ \]
B) \[45{}^\circ \]
C) \[70{}^\circ \]
D) \[110{}^\circ \]
Correct Answer: A
Solution :
\[\angle \]ACB + \[\angle \]BCD = \[180{}^\circ \left( linear pair \right)\] \[\angle \]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° 2\[\angle \]CDB = 180°-\[\angle \]BCD \[= 180{}^\circ - 110{}^\circ = 70{}^\circ \] \[\therefore \]\[\angle \]CBD = \[\angle \]ADB =\[\frac{{{70}^{{}^\circ }}}{2}\]= \[35{}^\circ \]You need to login to perform this action.
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