Question

In $△$ ABC,$\sin A+\sin B+\sin C=1+\sqrt{2}$ and $\cos A+\cos +\cos C=\sqrt{2}$ if the triangle is

• A. Equilateral
• B. Isosceles
• C. Right angled
• D. Right-angled isosceles

Answer 1

Answer: (D)

Right-angled isosceles

Given:

$A+B+C=180$

$\Rightarrow$$\sin A+\sin B+\sin C$$=1+\sqrt{2}$

$\Rightarrow$Consider $B={90}^0$

$\therefore$ $A+C=90$

$\therefore$ $C=90-A$

$\Rightarrow$$\sin A+\sin C$$=\sqrt{2}$

Hence we get

$\Rightarrow$$\sin A+\cos A=\sqrt{2}$.

Upon squaring both sides we get

$\Rightarrow$$1+2\sin A.\cos A=2$

$\Rightarrow$$\sin 2A=1$

$\Rightarrow$$2A=\frac{\pi }{2}$

$\Rightarrow$$A=C=\frac{\pi }{4}$.

Hence an isosceles right angled triangle.