Question

In the adjoining figure, if $BC=a,AC=b,AB=c$ and $\angle CAB={120}^{\circ }$, then the correct relation is

• A. a² = b² + c² + 2 bc
• B. a² = b² + c² – 2 bc
• C. a² = b² + c² + bc
• D. a² = b² + c² – bc

Answer 1

The correct option is C

a² = b² + c² + bc

In $△BDC,$
${BC}^2={BD}^2+{CD}^2$
As, $BD=AB+AD,$ upon simplification, we can write

$B{C}^2=(AB+AD{)}^2+C{D}^2$

${BC}^2={AB}^2+{AD}^2+2AB.AD+C{D}^2$

$\Rightarrow B{C}^2={AB}^2+{AC}^2+2AB.AD$ $[\because \text{in}△ADC,A{D}^2+C{D}^2=A{C}^2]$

$\Rightarrow B{C}^2={AB}^2+{AC}^2+2AB.\frac{1}{2}AC$ $[\because \angle CAD=180-120={60}^{\circ }\text{and}AD=AC\cos {60}^{\circ }=\frac{1}{2}AC]$

$\Rightarrow B{C}^2={AB}^2+{AC}^2+AB.AC$

$\Rightarrow a^2=b^2+c^2+bc$