Properties of Triangles 2 Question 13
13. If in a $\triangle A B C, \cos A \cos B+\sin A \sin B \sin C=1$, then show that $a: b: c=1: 1: \sqrt{2}$.
$(1986,5$ M)
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Solution:
- Given, $\cos A \cos B+\sin A \sin B \sin C=1$
$$ \begin{array}{lc} \Rightarrow & \sin C=\frac{1-\cos A \cos B}{\sin A \sin B} \\ \Rightarrow & \frac{1-\cos A \cos B}{\sin A \sin B} \leq 1 \quad[\because \sin C \leq 1] \\ \Rightarrow & 1-\cos A \cos B \leq \sin A \sin B \\ \Rightarrow & \cos (A-B) \geq 1 \\ \Rightarrow & \cos (A-B)=1 \\ \Rightarrow & A-B=0 \end{array} \quad[\because \operatorname{ascos}(\theta) \leq 1] $$
On putting $A=B$ in Eq. (i), we get
$$ \begin{array}{lc} & \sin C=\frac{1-\cos ^{2} A}{\sin ^{2} A} \\ \Rightarrow & \sin C=1 \\ \Rightarrow & C=\pi / 2 \\ \text { Now, } & A+B+C=\pi \end{array} $$
$$ \begin{aligned} & \Rightarrow \quad A+B=\frac{\pi}{2} \Rightarrow A=\frac{\pi}{4} \quad \because A=B \text { and } C=\frac{\pi}{2} \\ & \therefore \quad \sin A: \sin B: \sin C=\sin \frac{\pi}{4}: \sin \frac{\pi}{4}: \sin \frac{\pi}{2} \\ & \Rightarrow \quad a: b: c=\frac{1}{\sqrt{2}}: \frac{1}{\sqrt{2}}: 1 \\ & =1: 1: \sqrt{2} \end{aligned} $$
