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Category: Geometry

Does-sin-A-sin-B-imply-sin-A-2-sin-B-2-For-example-A-pi-and-B-pi-

Question Number 13305 by prakash jain last updated on 18/May/17 $$\mathrm{Does} \\ $$$$\mathrm{sin}\:\mathrm{A}=\mathrm{sin}\:\mathrm{B} \\ $$$$\mathrm{imply}\:\mathrm{sin}\:\frac{\mathrm{A}}{\mathrm{2}}=\mathrm{sin}\frac{\mathrm{B}}{\mathrm{2}}? \\ $$$$\mathrm{For}\:\mathrm{example}\:\mathrm{A}=\pi\:\mathrm{and}\:\mathrm{B}=−\pi? \\ $$ Answered by ajfour last updated on…

For-positive-a-b-c-such-that-a-b-c-1-show-that-a-b-c-b-c-a-c-a-b-1-solution-a-b-c-b-c-a-c-a-b-a-b-a-c-b-c-b-a-c-a-c-b-b-c-a-a-c-b-a-b-c-a-0-

Question Number 13236 by Abbas-Nahi last updated on 17/May/17 $${For}\:{positive}\:\:{a},{b},{c}\:\:{such}\:{that}\:\:{a}\:{b}\:{c}=\mathrm{1} \\ $$$${show}\:{that}\:\:{a}^{{b}+{c}} \:\:{b}^{{c}+{a}} \:\:{c}^{{a}+{b}} \:\leqslant\mathrm{1} \\ $$$${solution}: \\ $$$${a}^{{b}+{c}} \:\:{b}^{{c}+{a}} \:\:{c}^{{a}+{b}} \:=\left({a}^{{b}} {a}^{{c}} \:{b}^{{c}} {b}^{{a}}…

Question-13228

Question Number 13228 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17 $${in}\:{triangle}\:{ABC},{show}\:{that}: \\ $$$$\boldsymbol{{r}}=\sqrt{\boldsymbol{{r}}_{\boldsymbol{{a}}} .\boldsymbol{{r}}_{\boldsymbol{{b}}} }+\sqrt{\boldsymbol{{r}}_{\boldsymbol{{b}}} .\boldsymbol{{r}}_{\boldsymbol{{c}}} }+\sqrt{\boldsymbol{{r}}_{\boldsymbol{{c}}} .\boldsymbol{{r}}_{\boldsymbol{{a}}} }\:.…

Question-144268

Question Number 144268 by aliibrahim1 last updated on 23/Jun/21 Answered by imjagoll last updated on 24/Jun/21 $$\mathrm{consider}\:\measuredangle\mathrm{CDA}\:=\:\mathrm{90}°−\alpha\:\mathrm{and}\: \\ $$$$\measuredangle\mathrm{CAD}=\mathrm{90}°−\alpha\:\mathrm{so}\:\mathrm{x}\:=\:\mathrm{CD}=\mathrm{3}\:\mathrm{cm} \\ $$ Commented by som(math1967) last…

Question-13067

Question Number 13067 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/May/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/May/17 $${in}\:{triangle}\:{ABC}: \\ $$$$\left.{i}\right)\:\:\:\:\:\:\:\measuredangle{BIA}=\measuredangle{CIB}=\measuredangle{AIC} \\ $$$$\left.{ii}\right)\:\:\:\:{DE}\bot{BC},{DF}\bot{AC},{DH}\bot{AB} \\ $$$$\left.{iii}\right)\:\:\:{S}_{{ABC}} =\:\:\mathrm{1}\:\:\left({area}\:{of}\:\Delta{ABC}\right) \\…