Question

In $△ABC,sinA+sinB+sinC=1+\sqrt{2}$ and $cosA+cosB+cosC=\sqrt{2}$ if the triangle is

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

Answer 1

The correct option is D right-angled isosceles

Given, $\sin A+\sin B+\sin C$$=1+\sqrt{2}$
Consider $B={90}^0$
Hence we get
$\sin A+\cos A=\sqrt{2}$.
$1+2\sin A.\cos A=2$
$\sin 2A=1$
$2A=\frac{\pi }{2}$
$A=C=\frac{\pi }{4}$.
Hence an isosceles right angled triangle.