Question

In a triangle ABC; $A=\frac{\pi }{3},b=50,c=30$. If AD is a median through A, then $A{D}^2$ is equal to

Answer 1

In triangle ABC, we have $AB=c,BC=a,AC=b$
Given AD is median of triangle $ABC$
$\Rightarrow BD=DC=\frac{a}{2}$
$\cos A=\frac{b^2+c^2-a^2}{2bc}$
$\cos \frac{\pi }{3}=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow b^2+c^2-a^2=bc$ (1)
In triangle ABD,
$\cos B=\frac{c^2+{\left(\frac{a}{2}\right)}^2-{AD}^2}{ac}$
$A{D}^2=c^2+\frac{a^2}{4}-ac\cos B$
$\Rightarrow 4A{D}^2=4c^2+a^2-4ca\frac{(c^2+a^2-b^2)}{2ca}$
$=2c^2+2b^2-a^2$
$=2c^2+2b^2-(b^2+c^2-bc)$
[(using (1))]
$=b^2+c^2+bc$
$4A{D}^2=2500+900+1500=4900$
$\Rightarrow A{D}^2=1225$