In a $\Delta ABC$, if ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=ac+\sqrt{3}ab$ , then the triangle is
A. Equilateral
B. Right-angled and isosceles
C. Right angled and not isosceles
D. None of the above

Answer Verified Verified
Hint: We have to see whether the triangle is of which type. So use ${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos
A$, ${{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cos B$ ,${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C$and compare it with ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=ac+\sqrt{3}ab$. You will get the answer.

Complete step by step solution:

In $\Delta ABC$, we have given ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=ac+\sqrt{3}ab$.
But we know that in $\Delta ABC$,
& {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A \\
& {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cos B \\
& {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C \\
Now adding above three equations we get,
${{a}^{2}}+{{b}^{2}}+{{c}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A+{{a}^{2}}+{{c}^{2}}-2ac\cos
B+{{a}^{2}}+{{b}^{2}}-2ab\cos C$
Simplifying we get,
${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=2bc\cos A+2ac\cos B+2ab\cos C$
Now in question we are given ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=ac+\sqrt{3}ab$.
So comparing we get,
$2\cos A=0$ , $\cos B=\dfrac{1}{2}$ and $\cos C=\dfrac{\sqrt{3}}{2}$.
$\cos A=\cos \dfrac{\pi }{2}$, $\cos B=\cos \dfrac{\pi }{3}$ and $\cos C=\cos \dfrac{\pi }{6}$.
So from above we get that the triangle is a right angled triangle and not isosceles.
The correct answer is option (C).

Note: Read the question carefully. Also, you must know the concept regarding all types of triangles.
Also, take care while comparing. Do not miss any term while subtracting. Take care that no terms are
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