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Question Number 21430 by Tinkutara last updated on 23/Sep/17

In any ΔABC, the value of  2ac sin (((A − B + C)/2)) is

$$\mathrm{In}\:\mathrm{any}\:\Delta{ABC},\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{2}{ac}\:\mathrm{sin}\:\left(\frac{{A}\:−\:{B}\:+\:{C}}{\mathrm{2}}\right)\:\mathrm{is} \\ $$

Answered by $@ty@m last updated on 23/Sep/17

=2acsin (((π−2B)/2))  =2accos B  =c^2 +a^2 −b^2

$$=\mathrm{2}{ac}\mathrm{sin}\:\left(\frac{\pi−\mathrm{2}{B}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}{ac}\mathrm{cos}\:{B} \\ $$$$={c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$

Commented by Tinkutara last updated on 23/Sep/17

But book says b^2 −a^2 −c^2 . I also think  your answer true because I also got the  same answer.

$$\mathrm{But}\:\mathrm{book}\:\mathrm{says}\:{b}^{\mathrm{2}} −{a}^{\mathrm{2}} −{c}^{\mathrm{2}} .\:\mathrm{I}\:\mathrm{also}\:\mathrm{think} \\ $$$$\mathrm{your}\:\mathrm{answer}\:\mathrm{true}\:\mathrm{because}\:\mathrm{I}\:\mathrm{also}\:\mathrm{got}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{answer}. \\ $$

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