A) \[\angle ALC\]
B) \[\angle ACB\]
C) \[\angle BAC\]
D) \[\angle B-\angle BAL\]
Correct Answer: B
Solution :
\[\angle BAL+\angle B+90{}^\circ =180{}^\circ \] \[\Rightarrow \]\[\angle BAL+\angle B=90{}^\circ \]\[\angle BAL=90{}^\circ -\angle B\] ?(i) Now in \[\Delta ABC,\]\[\angle ACB+\angle B+\angle A=180{}^\circ \] \[\Rightarrow \]\[\angle ACB+\angle B=180{}^\circ -90{}^\circ \] (Given) \[\angle ACB+\angle B=90{}^\circ \] \[\angle ACB=90{}^\circ -\angle B\] From Eqs. (i) and (ii) \[\angle BAL=\angle ACB\]
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