Question

In a triangle $a^2+b^2+c^2=ca+ab\sqrt{3}$. Then the triangle is

• A. Equilateral
• B. Right angle and isosceles
• C. Right angled with $A={90}^o,B={60}^o,C={30}^o$
• D. $noneoftheabove$

Answer 1

The correct option is C Right angled with $A={90}^o,B={60}^o,C={30}^o$

Given,

$a^2+b^2+c^2=ca+ab\sqrt{3}$

$⟹a^2+b^2+c^2=2a(\frac{c}{2}+b\frac{\sqrt{3}}{2})$

According to sine formula

$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=\frac{1}{k}$

so

$k^{2}si{n}^{2}A+k^{2}si{n}^{2}B+k^{2}si{n}^{2}C=2.k.sinA(\frac{1}{2}.ksinC+\frac{\sqrt{3}}{2}.ksinB)$

$⟹si{n}^{2}A+si{n}^{2}B+si{n}^{2}C=2sinA(\frac{1}{2}sinC+\frac{\sqrt{3}}{2}sinB)$

On comparing it with standard trignnometric identity

we get

cosB=$\frac{1}{2}$

$⟹B={60}^{\circ }$

also cosC=$\frac{\sqrt{3}}{2}=cos{30}^{\circ }$

$C={30}^{\circ }$

As we know In triangle

$A+B+C=180$

$A+60+30=180$

$A+90=180$

$A={90}^{\circ }$