question_answer

The side AC of a triangle ABC is produced to D such that BC = CD. If \[\angle ACB\]is, then what is \[\angle ADB\]equal to?

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 =\[\frac{{{70}^{{}^\circ }}}{2}\]= \[35{}^\circ \]