Question

For a $\Delta ABC$ it is given that $CosA+CosB+CosC=\frac{3}{2}$ then the triangle is

• A. isosceles
• B. equilateral
• C. right angled
• D. scalene

Answer 1

Answer: (B)

equilateral

$cosA+cosB+cosC=1+4sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}$
$sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=\frac{1}{8}$
$sin\frac{A}{2}=sin\frac{B}{2}=sin\frac{C}{2}$
$(sin\frac{A}{2}{)}^3=\frac{1}{2^3}$
$sin\frac{A}{2}=\frac{1}{2}$
$\frac{A}{2}={30}^0$
$A={60}^0$
$\therefore A=B=C={60}^0$