Question
In $\triangle\text{ABC},\ \angle\text{A}=60^\circ.$ Prove that $BC^2 = AB^2 + AC^2 - AB \times AC.$
✓
Answer
In $\triangle\text{ABC}$ in which A is an acute angle with $60^\circ.$
$\sin60^\circ=\frac{\text{CD}}{\text{AC}}=\frac{\sqrt{3}}{2}$
$\Rightarrow\text{CD}=\frac{\sqrt{3}}{2}\text{AC}\ \ \ ....(1)$
$\cos60^\circ=\frac{\text{AD}}{\text{AC}}=\frac{1}{2}$
$\Rightarrow\text{AD}=\frac{1}{2}\text{AC}\ \ \ ....(2)$
Now apply Pythagoras theorem in triangle BCD
$BC^2 = CD^2 + BD^2$
$= CD^2 + (AB - AD)^2$
$=\Big(\frac{\sqrt{3}}{2}\text{AC}\Big)+\text{AB}^2+\Big(\frac{1}{2}\text{AC}\Big)^2-2\text{AB}\frac{1}{2}\text{AC}$
$= AC^2 + AB^2 - AB \times AC$
Hence $BC^2 = AB^2 + AC^2 - AB \times AC$
$\sin60^\circ=\frac{\text{CD}}{\text{AC}}=\frac{\sqrt{3}}{2}$
$\Rightarrow\text{CD}=\frac{\sqrt{3}}{2}\text{AC}\ \ \ ....(1)$
$\cos60^\circ=\frac{\text{AD}}{\text{AC}}=\frac{1}{2}$
$\Rightarrow\text{AD}=\frac{1}{2}\text{AC}\ \ \ ....(2)$
Now apply Pythagoras theorem in triangle BCD
$BC^2 = CD^2 + BD^2$
$= CD^2 + (AB - AD)^2$
$=\Big(\frac{\sqrt{3}}{2}\text{AC}\Big)+\text{AB}^2+\Big(\frac{1}{2}\text{AC}\Big)^2-2\text{AB}\frac{1}{2}\text{AC}$
$= AC^2 + AB^2 - AB \times AC$
Hence $BC^2 = AB^2 + AC^2 - AB \times AC$
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