In a triangle ABC, `/_ABC=45^@` and a point D is on BC such that `2BD=CD` and `/_DAB=15^@` then `/_ACB=`

Answer

3

Explanation

`F` in segment `BC` choose point `D` on segment `BC` such that `2|BD|=|CD|` & construct ray `BP` such that `/_PBC=45^(@)` & let `A` be point on ray `BP` that moves from `B` in the direction of the ray. Applying sine rule in `/_\ACD` & `/_\ABC` Let `alpha=/_CAD` & `/_CDA=/_CBA+/_DAB=60^(@)` wer have `(CD)/(sinalpha)=(CA)/(sin60^(@))` & `(BC)/(sin(alpha+15^(@)))=(CA)/(sin45^(@))` Dividing the first equation by second equation gives
`(|CD|sin(alpha+15^(@)))/(|BC|sinalpha)=(sin45^(@))/(sin60^(@))`
`|(CD)/(BC)|=2/3=((sin45^(@))/(sin60^(@)))^(2)`

`((sin45^(@))/(sin60^(@)))^(2)=(sinalpha)/(sin(alpha+15^(@))).(sin45^(@))/(sin60^(@))`
Clearly `alpha=45^(@)` satisfies the equation `/_ABC=45^(@),/_CAB=60^(@)` & `/_ACB=75^(@)`