Question

In $\triangle ABC$, $\angle CAB=60^\circ$, $BC=a$, $AC=b$, $AB=c$. Which relation is correct? \includegraphics[width=0.5\linewidth]{image83.png}

• A. $a^{2}=b^{2}+c^{2}-bc$
• B. $a^{2}=b^{2}+c^{2}-2bc$
• C. $a^{2}=b^{2}+c^{2}+bc$
• D. $a^{2}=b^{2}+c^{2}+2bc$

Hint:

Remember the cosine rule: $a^2=b^2+c^2-2bc\cos A$. For $60^\circ$, $\cos 60^\circ=\tfrac12$, which simply replaces the $-2bc$ term with $-bc$.

Answer 1

The Correct Option is A

By the Law of Cosines for side $a$ opposite angle $A$: \[ a^{2}=b^{2}+c^{2}-2bc\cos A. \] Here $A=\angle CAB=60^\circ$ and $\cos 60^\circ=\tfrac12$. Substituting, \[ a^{2}=b^{2}+c^{2}-2bc\left(\tfrac12\right)=b^{2}+c^{2}-bc. \] \[ \boxed{a^{2}=b^{2}+c^{2}-bc} \]