In $\triangle \mathrm{ABC}$, if $\cos ^2 \mathrm{~A}+\cos ^2 \mathrm{~B}+\cos ^2 \mathrm{C}=1$, then…

In $\triangle \mathrm{ABC}$, if $\cos ^2 \mathrm{~A}+\cos ^2 \mathrm{~B}+\cos ^2 \mathrm{C}=1$, then $\triangle \mathrm{ABC}$ is
  1. an equilateral Triangle
  2. an isosceles triangle
  3. a right angled triangle
  4. a scalene triangle

Solution

$\because \cos ^2 \mathrm{~A}+\cos ^2 \mathrm{~B}+\cos ^2 \mathrm{C}=1$ $\begin{aligned} & \Rightarrow \cos ^2 A+\cos ^2 B-\sin ^2 B=0 \\ & \Rightarrow \cos ^2 A+\cos (B+C) \cos (B-C)=0 \\ & \Rightarrow \cos ^2 A+\cos (180-A) \cos (B+C)=0 \\ & \Rightarrow \cos A(\cos A-\cos (B-C))=0 \\ & \Rightarrow \cos A(-\cos (B+C)-\cos (B-C))=0 \\ & \Rightarrow \cos A \cdot \cos B \cdot \cos C=0\end{aligned}$ $\therefore$ Either $\cos A=90^{\circ}$ or $\cos B=90^{\circ}$ or $\cos C=90^{\circ}$ So, its triangle with on angle $=90^{\circ}$ Here $\triangle \mathrm{ABC}$ is right angle triangle.

Asked in: AP EAMCET 2023 (17 May Shift 2)

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