Question Number 196625 by sniper237 last updated on 28/Aug/23
$${Let}\:\:{a}={BC}\:\:{b}={CA}\:\:{c}={AB} \\ $$$${ABC}\:\:{is}\:{rectangle}\:{iff}\:\:{tan}\left(\frac{\hat {{A}}}{\mathrm{2}}\right)=\frac{{a}}{{b}+{c}} \\ $$
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Question Number 196625 by sniper237 last updated on 28/Aug/23 | ||
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$${Let}\:\:{a}={BC}\:\:{b}={CA}\:\:{c}={AB} \\ $$$${ABC}\:\:{is}\:{rectangle}\:{iff}\:\:{tan}\left(\frac{\hat {{A}}}{\mathrm{2}}\right)=\frac{{a}}{{b}+{c}} \\ $$ | ||