Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions3 Marks
Question
In any triangle $\text{ABC}$ with usual notations prove $c=a \cos B +b \cos A$.
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Answer
By cosine rule, we have $b^2 = c^2 + a^2 - 2ca \cos B$
$\therefore \cos B=\frac{c^2+a^2-b^2}{2} a c$
Similarly $\cos A=\frac{b^2+c^2-a^2}{2 b c}$
$\text{R.H.S.} = a \cos B + b \cos A$
$=a \cdot \frac{c^2+a^2-b^2}{2 a c}+b \cdot \frac{b^2+c^2-a^2}{2 b c}$
$=\frac{c^2+a^2-b^2}{2 c}+\frac{b^2+c^2-a^2}{2 c}$
$=\frac{c^2+a^2-b^2+b^2+c^2-a^2}{2 c}$
$=\frac{2 c^2}{2 c}$
$=c$
$=\text{L.H.s}$
$\therefore c = a \cos B + b \cos A$
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