Question

In a triangle ABC, the perpendicular AD from point A, to the side BC meets BC at point D. If BD = 8 cm, DC = 2 cm and AD = 4 cm. Then find the value of angle BAC. $\left[3\text{Marks}\right]$

Answer 1

In right angled triangle ACD
$A{C}^2=C{D}^2+A{D}^2$
$\Rightarrow A{C}^2=2^2+4^2=20$

In right angled triangle ADB
$A{B}^2=A{D}^2+D{B}^2$
$\Rightarrow A{B}^2=4^2+8^2=80$ $\left[1\text{Mark}\right]$

$\text{Now,}BC=BD+DC=8+2=10cm$
$\text{We can observe that}B{C}^2={10}^2=100$
$\text{and}A{B}^2+A{C}^2=80+20=100$
$\text{Hence,}B{C}^2=A{C}^2+A{B}^2$
$\text{Hence, from converse of Pythagoras theoren we have}$
$\angle BAC={90}^{\circ }$ $\left[2\text{Marks}\right]$