Question

In a △ABC, the perpendicular 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 measure of ∠BAC.

Answer 1

△ADC is a right angled triangle.
${\left(AD\right)}^2+{\left(DC\right)}^2={\left(AC\right)}^2{\left(4cm\right)}^2+{\left(2cm\right)}^2={\left(AC\right)}^2{16cm}^2+{4cm}^2={\left(AC\right)}^2{\left(AC\right)}^2=20c{m}^2$

△ADB is a right angled triangle.
${\left(AD\right)}^2+{\left(DB\right)}^2={\left(AB\right)}^2{\left(4cm\right)}^2+{\left(8cm\right)}^2={\left(AB\right)}^2{16cm}^2+{64cm}^2={\left(AB\right)}^2{\left(AB\right)}^2=80c{m}^2$

In △ABC
${\left(AB\right)}^2=80c{m}^2{\left(AC\right)}^2=20c{m}^2{\left(BC\right)}^2={(BD+DC)}^2={(8cm+2cm)}^2=100c{m}^2{\left(AB\right)}^2+{\left(AC\right)}^2=80c{m}^2+20c{m}^2=100c{m}^2={\left(BC\right)}^2$
So triangle ABC is a right angled triangle with right angle at A
So, $\angle BAC=90°$