# A triangle ABC is right angled at A. AL is drawn perpendicular to BC. Then ∠BAL is equal to::

A. ∠ALC B. ∠ACB C. ∠BAC D. ∠BLA Answer: Option B
$\begin{array}{l}\text{In }\triangle\ \text{ABL}\\ \angle\mathrm{BAL}+\angle\mathrm{ALB}+\angle\mathrm{B}=180^{\circ}\\ \angle\mathrm{BAL}+90^{\circ}+\angle\mathrm{B}=180^{\circ}\\ \angle\mathrm{BAL}=90^{\circ}-\angle\mathrm{B}\quad\ldots..(1)\\ \text{In}\ \triangle\ \text{ABC},\\ \angle\mathrm{A}+\angle\mathrm{B}+\angle\mathrm{C}=180^{\circ}\\ \angle\mathrm{B}+\angle\mathrm{C}=90^{\circ}\\ \angle\mathrm{C}=90^{\circ}-\angle\mathrm{B}\\ \angle\mathrm{ACB}=90^{\circ}-\angle\mathrm{B}\ ….\left(2\right)\\ \text{From }\left(1\right)\ \&\ \ \left(2\right),\ \\ \text{We get,}\\ \angle BAC=\angle ACB\end{array}$