In a triangle $ABC, \angle{A}=\dfrac{\pi}{3},b=40,c=30,AD$ is the median through $A,$ then $4{\left(AD\right)}^{2}$ must be:
• A
$3007$
• B
$3070$
• C
$3700$
• D
$7003$

Solution

$\cos{A}=\dfrac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}$
Given $A=\dfrac{\pi}{3}$
$\Rightarrow \cos{\dfrac{\pi}{3}}=\dfrac{1}{2}=\dfrac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}$
$\Rightarrow bc={b}^{2}+{c}^{2}-{a}^{2}$              ..............$\left(1\right)$
In $\triangle ABD,$
${AD}^{2}={c}^{2}+\dfrac{{a}^{2}}{4}-2c\times\dfrac{a}{2}\cos{B}$
$\Rightarrow 4{AD}^{2}=4{c}^{2}+{a}^{2}-4ca\left(\dfrac{{c}^{2}+{a}^{2}-{b}^{2}}{2ca}\right)$
$\Rightarrow 4{AD}^{2}=4{c}^{2}+{a}^{2}-2\left({c}^{2}+{a}^{2}-{b}^{2}\right)$
$\Rightarrow 4{AD}^{2}=4{c}^{2}+{a}^{2}-2{c}^{2}-2{a}^{2}+2{b}^{2}$
$\Rightarrow 4{AD}^{2}=2{c}^{2}-{a}^{2}+2{b}^{2}$
$\Rightarrow 4{AD}^{2}={c}^{2}+{b}^{2}+\left({c}^{2}+{b}^{2}-{a}^{2}\right)$
$\Rightarrow 4{AD}^{2}={c}^{2}+{b}^{2}+bc$ from $\left(1\right)$
Substituting for $b=40,c=30$ we get
$\Rightarrow 4{AD}^{2}={30}^{2}+{40}^{2}+40\times 30$
$=900+1600+1200=3700$