Question
Prove that in any triangle ABC, $\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}},$ where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.
✓
Answer
Let, $\overrightarrow{\text{AB}}=\text{c},\overrightarrow{\text{BC}}=\text{a}$ and $\overrightarrow{\text{AC}}=\text{b}$
Hare, components of c are ccos A and csin A is drawn.
Since, $\overrightarrow{\text{CD}}=\text{b}-\text{c}\cos\text{A}$
In $\triangle\text{BDC},$
$\text{a}^2=(\text{b}-\text{c}\cos\text{A})^2+(\text{c}\sin\text{A})^2$
$\Rightarrow\text{a}^2=\text{b}^2+\text{c}^2\cos^2\text{A}-2\text{bc}\cos\text{A}+\text{c}^2\sin^2\text{A}$
$\Rightarrow2\text{b}\text{c}\cos\text{A}=\text{b}^2-\text{a}^2+\text{c}^2(\cos^2\text{A}+\sin^2\text{A})$
$\therefore\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
Hare, components of c are ccos A and csin A is drawn.
Since, $\overrightarrow{\text{CD}}=\text{b}-\text{c}\cos\text{A}$
In $\triangle\text{BDC},$
$\text{a}^2=(\text{b}-\text{c}\cos\text{A})^2+(\text{c}\sin\text{A})^2$
$\Rightarrow\text{a}^2=\text{b}^2+\text{c}^2\cos^2\text{A}-2\text{bc}\cos\text{A}+\text{c}^2\sin^2\text{A}$
$\Rightarrow2\text{b}\text{c}\cos\text{A}=\text{b}^2-\text{a}^2+\text{c}^2(\cos^2\text{A}+\sin^2\text{A})$
$\therefore\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
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