Q66 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics
In a triangle $A B C$. If $\cos A \cos B+\sin A \sin B \sin C=1$, then $\sin A+\sin B+\sin C=$
- A$\frac{2+\sqrt{3}}{2}$
- Correct answerB$1+\sqrt{2}$
- C$\frac{2 \sqrt{3}-1}{2}$
- D$\frac{3+\sqrt{3}}{2}$
Solution
We have, In $\triangle A B C$,
$$
\cos A \cos B+\sin A \sin B \sin C=1
$$
$$
\cos A \cos B+\sin A \sin B \sin C=1
$$
is possible of $A=45^{\circ}, B=45^{\circ}, C=90^{\circ}$
$$
\begin{aligned}
& \text { i.e. } \cos 45^{\circ} \cos 45^{\circ}+\sin 45^{\circ} \sin 45^{\circ} \sin 90^{\circ}=\frac{1}{2}+\frac{1}{2}=1 \\
& \therefore \quad \sin A+\sin B+\sin C \\
& =\sin 45^{\circ}+\sin 45^{\circ}+\sin 90^{\circ} \\
& =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+1=\sqrt{2}+1 \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'B'.