मराठी

If in δ a B C , Cos 2 a + Cos 2 B + Cos 2 C = 1 Prove that the Triangle is Right-angled. - Mathematics

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प्रश्न

If in \[∆ ABC, \cos^2 A + \cos^2 B + \cos^2 C = 1\] prove that the triangle is right-angled. 

 

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उत्तर

Let ABC be any triangle.
In \[∆ ABC\]

\[\cos^2 A + \cos^2 B + \cos^2 C = 1\]

\[ \Rightarrow \cos^2 A + \cos^2 B + \cos^2 \left[ \pi - \left( B + A \right) \right] = 1 \left( \because A + B + C = \pi \right)\]

\[ \Rightarrow \cos^2 A + \cos^2 B + \cos^2 \left( B + A \right) = 1\]

\[ \Rightarrow \cos^2 A + \cos^2 B = 1 - \cos^2 \left( B + A \right)\]

\[ \Rightarrow \cos^2 A + \cos^2 B = \sin^2 \left( B + A \right)\]

\[ \Rightarrow \cos^2 A + \cos^2 B = \left( \sin A\cos B + \cos A\sin B \right)^2 \]

\[ \Rightarrow \cos^2 A + \cos^2 B = \sin^2 A \cos^2 B + \cos^2 A \sin^2 B + 2\sin A\sin B\cos A\cos B\]

\[ \Rightarrow \cos^2 A\left( 1 - \sin^2 B \right) + \cos^2 B\left( 1 - \sin^2 A \right) = 2\sin A\sin B\cos A\cos B\]

\[ \Rightarrow 2 \cos^2 A \cos^2 B = 2\sin A\sin B\cos A\cos B\]

\[ \Rightarrow \cos A\cos B = \sin A\sin B\]

\[ \Rightarrow \cos A\cos B - \sin A\sin B = 0\]

\[ \Rightarrow \cos \left( A + B \right) = 0\]

\[ \Rightarrow \cos \left( A + B \right) = \cos {90}^° \] 

\[ \Rightarrow A + B = {90}^°\] 

\[ \Rightarrow C = {90}^° \left( \because A + B + C = {180}^° \right)\] 

Hence, \[∆\]ABC is right angled

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Sine and Cosine Formulae and Their Applications
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पाठ 10: Sine and cosine formulae and their applications - Exercise 10.2 [पृष्ठ २५]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 10 Sine and cosine formulae and their applications
Exercise 10.2 | Q 17 | पृष्ठ २५

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