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  • Prove that in any triangle ABC, cos A equals b^2 +c^2 -a^2 by 2bc where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.

Prove that in any triangle ABC, \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}  where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.

Answers (1)

Given:

a, b, c are magnitudes of the sides opposite to the vertices A, B, C respectively.

⇒ AB = c, BC = a and CA = b

To Prove:

In triangle ABC,

\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}

Construction: We have constructed a triangle ABC and named the vertices according to the question.

Note the height of the triangle, BD.

If ∠BAD = A

Then, BD = c sin A

\\ \qquad \sin \mathrm{A}=\frac{\text { perpendicular }}{\text { hypotenuse }} \text { in } \Delta \mathrm{BAD} \\ \qquad \because \sin \mathrm{A}=\frac{\mathrm{BD}}{\mathrm{c}} \\ \Rightarrow \mathrm{BD}=\mathrm{c} \sin \mathrm{A}

\\ \text { And, } A D=c \cos A \\ \qquad \cos A=\frac{\text { base }}{\text { hypotenuse }} \text { in } \Delta B A D \\ \because \\ \Rightarrow \cos A=\frac{A D}{C} \\ \Rightarrow A D=c \cos A

Proof:

Here, components of c which are:

c sin A

c cos A

are drawn on the diagram.

Using Pythagoras theorem which says that,

(hypotenuse)2 =(perpendicular)2 + (base)2

Take triangle BDC, which is a right-angled triangle.

Here,

Hypotenuse = BC

Base = CD

Perpendicular = BD

We get,\\ (BC)\textsuperscript{2} = (BD)\textsuperscript{2} + (CD)\textsuperscript{2}\\ \\ $ \Rightarrow $ a\textsuperscript{2} = (c sin A)\textsuperscript{2} + (CD)\textsuperscript{2} [\because $ from the diagram, BD = c sin A]\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + (b - c cos A)\textsuperscript{2}\\ $\because$ from the diagram, AC = CD + AD\\
\\ \\$ \Rightarrow $ CD = AC - AD\\ $ \Rightarrow $ CD = b - c cos A]\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + (b\textsuperscript{2} + (-c cos A)\textsuperscript{2} - 2bc cos A) [\because$ from algebraic identity, (a -b)\textsuperscript{2} = a\textsuperscript{2} + b\textsuperscript{2} - 2ab]\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + b\textsuperscript{2} + c\textsuperscript{2} cos\textsuperscript{2} A - 2bc cos A\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + c\textsuperscript{2} cos\textsuperscript{2} A + b\textsuperscript{2} - 2bc cos A\\

 

 \\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} (sin\textsuperscript{2} A + cos\textsuperscript{2} A) + b\textsuperscript{2} - 2bc cos A\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} + b\textsuperscript{2} - 2bc cos A [\because$ from trigonometric identity, sin\textsuperscript{2} $ \theta $ + cos\textsuperscript{2} $ \theta $ = 1]\\ \\ $ \Rightarrow $ 2bc cos A = c\textsuperscript{2} + b\textsuperscript{2} - a\textsuperscript{2}\\ \\ $ \Rightarrow $ 2bc cos A = b\textsuperscript{2} + c\textsuperscript{2} - a\textsuperscript{2}\\.

\\\Rightarrow \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}

Hence proved

Posted by

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