A2-b-c2-4 b c cos 2A-2

We know that by cosine rule cos A=b^2+c^2 a^2/2bc ⇒ 2bc cos A=b^2+c^2 a^2 ⇒ a^2=b^2+c^2 2bc cos A ⇒ a^2=b^2+c^2 2bc (2 cos^2 A/2 1 ) ⇒ a^2=b^2+c^2+2bc 4bc cos ...

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Standard XII

Mathematics

Cosine Rule

a2=b+c2-4 b c...

Question

$a^{2} = {(b + c)}^{2} - 4 b c c o s^{2} \frac{A}{2}$

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Solution

$We know that by cosine rule c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \Rightarrow 2 b c c o s A = b^{2} + c^{2} - a^{2} \Rightarrow a^{2} = b^{2} + c^{2} - 2 b c c o s A \Rightarrow a^{2} = b^{2} + c^{2} - 2 b c (2 c o s^{2} \frac{A}{2} - 1) \Rightarrow a^{2} = b^{2} + c^{2} + 2 b c - 4 b c c o s^{2} \frac{A}{2} \Rightarrow a^{2} = {(b + c)}^{2} - 4 b c c o s^{2} \frac{A}{2} a^{2} = {(b + c)}^{2} - 4 b c c o s^{2} \frac{A}{2}$

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a2=(b+c)2−4 bc cos2 A2

We know that by cosine rulecos A=b2+c2−a22bc⇒ 2bc cos A=b2+c2−a2⇒ a2=b2+c2−2bc cos A⇒ a2=b2+c2−2bc (2 cos2 A2−1)⇒ a2=b2+c2+2bc−4bc cos2 A2⇒ a2=(b+c)2−4bc cos2 A2a2=(b+c)2−4 bc cos2 A2

$a^{2} = {(b + c)}^{2} - 4 b c c o s^{2} \frac{A}{2}$