A) \[40{}^\circ ,\text{ }50{}^\circ ,\text{ }60{}^\circ \]
B) \[60{}^\circ ,\text{ }60{}^\circ ,\text{ }60{}^\circ \]
C) \[50{}^\circ ,\text{ }50{}^\circ ,\text{ }60{}^\circ \]
D) \[60{}^\circ ,\text{ }70{}^\circ ,\text{ }70{}^\circ \]
Correct Answer: C
Solution :
Since, ABCD is a parallelogram \[\therefore \angle ADC=\angle ABC\Rightarrow x=z~~~~~~...(i)\] Now in right triangle EBC, we have \[\angle \text{BEC+}\angle \text{EBC+}\angle \text{ECB=180 }\!\!{}^\circ\!\!\text{ }\] \[\Rightarrow 90{}^\circ +x+40{}^\circ =180{}^\circ \] \[\Rightarrow x=180{}^\circ -130{}^\circ =50{}^\circ \] From (i), \[x\text{ }=\text{ }z\text{ }=\text{ }50{}^\circ \] Now, in\[\Delta FCD\], we have \[\angle CFD+\angle FDC+\angle FCD=180{}^\circ \] [Angle sum property] \[\Rightarrow 90{}^\circ +50{}^\circ +\angle FCD=180{}^\circ \] \[\Rightarrow 140{}^\circ +\angle FCD=180{}^\circ \] \[\Rightarrow \angle FCD=180{}^\circ -140{}^\circ =40{}^\circ \] Also, in parallelogram ABCD, we have \[\angle ADC+\angle DCB=180{}^\circ \] \[\Rightarrow 50{}^\circ +(\angle FCD+y+40{}^\circ )=180{}^\circ \] \[\Rightarrow 50{}^\circ +40{}^\circ +y+40{}^\circ =180{}^\circ \] \[\Rightarrow y=180{}^\circ -130{}^\circ \Rightarrow y=50{}^\circ \] \[\therefore x=y=z=50{}^\circ \]
You need to login to perform this action.
You will be redirected in
3 sec
