Question

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

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

Answer 1

The correct option is C

$1:1:\sqrt{2}$

Determine the ratio of $a:b:c$.

Step 1: Determine the angles of the triangle.

Given, $\cos A\cos B+\sin A\sin B\sin C=1$

Assume $\sin C=1$, then $C=90°[\because \sin 90°=1]$

$$\begin{gathered} \therefore \cos A\cos B+\sin A\sin B=1 \\ \Rightarrow \cos (A-B)=1[\because \cos (A-B)=\cos A\cos B+\sin A\sin B] \\ \Rightarrow \cos (A-B)=\cos 0°[\because \cos 0°=1] \\ \Rightarrow A-B=0 \\ \Rightarrow A=B \end{gathered}$$

Thus, $A=B=45°[\because C=90°,A=B\mathrm{and}\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°]$

Step 2: Determine the ratio

Since two angles are equal, the triangle is an isosceles right angle triangle and hence two sides will be equal.

Therefore let sides $a=k,b=k.$

Thus we have, by the converse of Pythagoras theorem

$$\begin{gathered} c=\sqrt{a^2+b^2} \\ =\sqrt{k^2+k^2} \\ =\sqrt{2k^2} \\ c=\sqrt{2}k \end{gathered}$$

The ratio of $a:b:c$ is $1k:1k:\sqrt{2}kor1:1:\sqrt{2}$.

Hence, option (C).is correct .