A) \[56{}^\circ \]
B) \[112{}^\circ \]
C) \[28{}^\circ \]
D) \[65{}^\circ \]
Correct Answer: A
Solution :
[a] \[X=\angle PQB=28{}^\circ \] (angles in alternate segment) also, \[OB\bot TB\] we get, \[\angle POB=90{}^\circ -x=90{}^\circ -28{}^\circ \] \[=62{}^\circ \] Now, in BPQ \[\angle BPQ+\angle PBQ+\angle PQB=180{}^\circ \] \[(62+y)+(62+z)+x=180\] \[x+y+z=180-2\times 62\] \[=180-124=56{}^\circ \] |
You need to login to perform this action.
You will be redirected in
3 sec