Question

In a $\Delta ABC$, if cos A cos B+ sin A sin B sin C=1 then a:b:c is equal to

• A. $1:2:3$
• B. $1:\sqrt{3}:2$
• C. $1:1:1$
• D. $1:1:\sqrt{2}$

Answer 1

Answer: (D)

$1:1:\sqrt{2}$

$cosAcosB+sinAsinBsinC=1$

$cosAcosB+sinAsinB-sinAsinB+sinAsinBsinC=1-\left(1\right)$

using

$cos(A-B)=cosAcosB+sinAsinB,from\left(1\right)$

$cos(A-B)-sinAsinB(1-sinC)=1$

$1-cos(A-B)+sinAsinB(1-sinC)=0$

$2si{n}^2\left(\frac{A-B}{2}\right)+sinAsinB(1-sinC)=0$

so,

$2si{n}^2\left(\frac{A-B}{2}\right)=0$

$\frac{A-B}{2}=0$

$A=B$

and $sinAsinB(1-sinC)=0$

$A\ne 0$ and $B\ne 0$

$1-sinC=0$

$sinC=1$

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

$\therefore$ ABC is right angled isosceles triangle

so

$a:b:c=1:1:\sqrt{2}$