A) 15,130
B) 15,125
C) 35, 40
D) 30,150
Correct Answer: B
Solution :
[b] In the adjoining figure, \[\angle ACB=30{}^\circ \]and \[\angle ACB=2x{}^\circ \](Given \[\Rightarrow \]\[30{}^\circ =2x{}^\circ \]) \[\Rightarrow \]\[x=15{}^\circ \]and \[\angle BDC=y{}^\circ \] (Given) From figure, \[\angle BDC=\left[ 180{}^\circ -\left( \frac{\angle ABC}{2}+\frac{\angle ACB}{2} \right) \right]\] \[y{}^\circ =\left[ 180{}^\circ -\left( \frac{80{}^\circ }{2}+\frac{30{}^\circ }{2} \right) \right]=[180{}^\circ -(40{}^\circ +15{}^\circ )]\] \[y=(180{}^\circ -55{}^\circ )\]\[\Rightarrow \]\[y=125{}^\circ \]\[\therefore \]x and y = 15 and 125 |
You need to login to perform this action.
You will be redirected in
3 sec