Question

The altitude AD of $\Delta ABC$ in which $\angle A$ obtuse and AD = 10 cm. If BD = 10 cm and CD = $10\sqrt{3}$ cm, determine $\angle A$.

Answer 1

$\Delta ABC$ is a right triangle , right angled at D such that = 10 cm and BD = 10 cm

$\therefore tan\angle BAD=\frac{BD}{AD}$
$\Rightarrow tan\angle BAD=\frac{10}{10}=1$
$\Rightarrow tan\angle BAD=tan{45}^{\circ }$
TRIGONOMETRICRATIOS
$\Rightarrow \angle BAD={45}^{\circ }$
$\Delta ACD$ is a right triangle right angled at D such that AD = 10 cm and DC = 10$\sqrt{3}$ cm.
$\therefore tan\angle CAD=\frac{CD}{AD}$
$\Rightarrow tan\angle CAD=\frac{10\sqrt{3}}{10}=\sqrt{3}$
$\Rightarrow tan\angle CAD=tan{60}^{\circ }$
$\therefore \angle CAD={60}^{\circ }$
From (i) and (ii) , we have

$\angle BAC=\angle BAD+\angle CAD={45}^{\circ }+{60}^{\circ }={105}^{\circ }$