Q68 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics
In a triangle $A B C$, if $\cos A \cos B+\sin A \sin B \sin C=1$, then $a: b: c=$
- Correct answerA$1: 1: \sqrt{2}$
- B$1: 1: 1$
- C$\sqrt{2}: 1: 1$
- D$1: \sqrt{2}: 1$
Solution
We have,
$\cos A \cos B+\sin A \sin B \sin C=1$
$\begin{aligned} & \begin{array}{r}\Rightarrow \quad \cos A \cos B+\sin A \sin B-\sin A \sin B \\ +\sin A \sin B \sin C=1\end{array} \\ & \Rightarrow \quad \cos (A-B)-\sin A \sin B(1-\sin C)=1 \\ & \Rightarrow 1-\cos (A-B)+\sin A \sin B(1-\sin C)=0 \\ & \Rightarrow 2 \sin ^2\left(\frac{A-B}{2}\right)+\sin A \sin B(1-\sin C)=0\end{aligned}$
$\begin{aligned} & \therefore \sin \left(\frac{A-B}{2}\right)=0 \text { and } \sin A \sin B(1-\sin C)=0 \\ & \Rightarrow A-B=0 \text { and } \sin C=1[\because \sin A, \sin B \neq 0] \\ & \Rightarrow \quad A=B \text { and } C=90^{\circ} \\ & \therefore \quad A=B=45^{\circ}, C=90^{\circ}\end{aligned}$
$\begin{array}{ll}\text { Now, } & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ \Rightarrow & \frac{a}{\sin 45^{\circ}}=\frac{b}{\sin 45^{\circ}}=\frac{c}{\sin 90^{\circ}} \\ \Rightarrow & a: b: c=\frac{1}{\sqrt{2}}: \frac{1}{\sqrt{2}}: 1=1: 1: \sqrt{2}\end{array}$
Hence, the correct answer is option: 'A'.