Question Bank - P2 — Maths STD 10 — Question

$\angle B=90^{\circ}, \angle Q=30^{\circ} \quad \ldots [ Given ]$

$\therefore \angle A=60^{\circ} \quad \ldots[\text { Remaining angle of a triangle ] }$

$\therefore \triangle A B Q \text { is a } 30^{\circ}-60^{\circ}-90^{\circ} \text { triangle. }$

$\therefore A B=\frac{1}{2} A Q$ $\quad$ $\ldots$[ Side opposite to $30^{\circ}$ ]

$\therefore A B=\frac{1}{2} \times 8$

$\therefore A B=4 \mathrm{~cm}$