Question Number 182407 by Acem last updated on 09/Dec/22 Answered by a.lgnaoui last updated on 09/Dec/22 $${Area}=\mathrm{AC}×\mathrm{AB}\frac{\mathrm{sin}\:\beta}{\mathrm{2}} \\ $$$$\beta=\frac{\mathrm{12}}{\mathrm{5}}\:\mathrm{x}\:\:\:\gamma=\frac{\mathrm{3x}}{\mathrm{2}}\:\:\:\alpha=\frac{\mathrm{3}}{\mathrm{5}}\mathrm{x}\:\:\:\left({degre}\right) \\ $$$$\alpha+\beta+\gamma=\mathrm{180}\:\:\:\left(\:\frac{\mathrm{45}}{\mathrm{10}}\right)×\mathrm{x}=\mathrm{180}\:\:\mathrm{x}=\mathrm{40} \\ $$$$\beta=\frac{\mathrm{12}×\mathrm{40}}{\mathrm{5}}=\mathrm{96}° \\ $$$${Area}=\frac{\mathrm{2}×\mathrm{AB}}{\mathrm{2}}×\mathrm{sin}\:\mathrm{96}…
Question Number 182392 by Acem last updated on 08/Dec/22 Commented by Acem last updated on 08/Dec/22 $$\:{Prove}:\:\frac{\mathrm{1}}{\:\sqrt{{a}}}=\:\frac{\mathrm{1}}{\:\sqrt{{b}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{{c}}} \\ $$ Answered by HeferH last updated on…
Question Number 182375 by Acem last updated on 08/Dec/22 Commented by Acem last updated on 08/Dec/22 $$\:{I}\:{just}\:{solved}\:{it},\:\mathrm{2}\:{sec}\:{ago},\: \\ $$$$\left.\:\phi\:\in\:\right]\mathrm{20},\:\mathrm{27}\left[\right. \\ $$$$\:{Btw}\:{extend}\:{of}\:{NA}\:{intersects}\:{with}\:{extend}\:{of}\:{PC} \\ $$$$\:{at}\:{point}\:\in\:{this}\:{circle} \\ $$$$\:{Hope}\:{you}\:{success}\:{to}\:{find}\:\phi\:!…
Question Number 116800 by mr W last updated on 06/Oct/20 Answered by bobhans last updated on 07/Oct/20 $$\Rightarrow\frac{\mathrm{sin}\:\mathrm{8x}}{\mathrm{sin}\:\mathrm{4x}}\:=\:\frac{\mathrm{sin}\:\mathrm{5x}}{\mathrm{sin}\:\mathrm{3x}}\:;\:\frac{\mathrm{2sin}\:\mathrm{4x}.\mathrm{cos}\:\mathrm{4x}}{\mathrm{sin}\:\mathrm{4x}}\:=\:\frac{\mathrm{sin}\:\mathrm{5x}}{\mathrm{sin}\:\mathrm{3x}} \\ $$$$\Rightarrow\mathrm{2sin}\:\mathrm{3x}.\mathrm{cos}\:\mathrm{4x}\:=\:\mathrm{sin}\:\mathrm{5x} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{7x}+\mathrm{sin}\:\left(−\mathrm{x}\right)=\mathrm{sin}\:\mathrm{5x} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{7x}−\mathrm{sin}\:\mathrm{5x}\:+\mathrm{sin}\:\left(−\mathrm{x}\right)=\mathrm{0} \\…
Question Number 51245 by Tawa1 last updated on 25/Dec/18 Answered by afachri last updated on 25/Dec/18 $$\left(\mathrm{2}\right)\:\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{evaluate}}\:\bigtriangleup{H}_{\mathrm{Reaction}} \:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{bond}}\:\boldsymbol{\mathrm{energy}}\:: \\ $$$$\bigtriangleup{H}_{\mathrm{Reaction}} \:=\:\bigtriangleup{H}_{\mathrm{r}} \:−\:\bigtriangleup{H}_{\mathrm{p}} \\ $$$$\bigtriangleup{H}_{\mathrm{p}} \:;\:\left(\:\mathrm{N}−\mathrm{N}\right)\:+\:\mathrm{4}\left(\mathrm{N}−\mathrm{H}\right)\:=\:\left(+\mathrm{167}\right)\:\:+\:\:\mathrm{4}\left(\mathrm{385}.\mathrm{9}\right)…
Question Number 182310 by mnjuly1970 last updated on 07/Dec/22 Answered by Acem last updated on 08/Dec/22 Commented by mnjuly1970 last updated on 08/Dec/22 $${superexcellen}\:{sir}\:{Acem}..{thanks}\:{alot} \\…
Question Number 182311 by mnjuly1970 last updated on 07/Dec/22 Commented by Frix last updated on 07/Dec/22 $$\mathrm{Another}\:\mathrm{question}: \\ $$$$\mathrm{At}\:\mathrm{which}\:\mathrm{angle}\:\alpha\:\mathrm{the}\:\mathrm{areas}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangles} \\ $$$${ABY}\:\mathrm{and}\:{BCY}\:\mathrm{are}\:\mathrm{equal}? \\ $$ Commented by…
Question Number 182292 by HeferH last updated on 07/Dec/22 Answered by Acem last updated on 07/Dec/22 $$\:{x}=\:\cancel{\mathrm{90}°}\:\:\:\:“{I}\:{doupt}'' \\ $$$$\:{from}\:{the}\:{common}\:{side}\:{and}\:{the}\:\mathrm{2}\:{equals}\:{bases} \\ $$$$\:\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{sin}\:{x}}=\:\frac{\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{120}−{x}}\:\:\Leftrightarrow\:\:\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{20}}{\mathrm{sin}\:{x}}=\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{120}−{x}} \\ $$$$\:\mathrm{2}\:\mathrm{cos}\:\mathrm{20}\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{cos}\:{x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{sin}\:{x}\right)=\:\mathrm{sin}\:{x} \\ $$$$\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{20}\:\mathrm{cos}\:{x}=\:\mathrm{sin}\:{x}\:\left(\mathrm{1}−\:\mathrm{cos}\:\mathrm{20}\right)…
Question Number 51141 by Tawa1 last updated on 24/Dec/18 Answered by ajfour last updated on 24/Dec/18 Commented by hassentimol last updated on 24/Dec/18 $$\mathrm{Excuse}\:\mathrm{me}\:\mathrm{sir}… \\…
Question Number 182192 by mr W last updated on 05/Dec/22 Commented by mr W last updated on 05/Dec/22 $${find}\:{the}\:{smallest}\:{area}\:{of}\:{inscribed} \\ $$$${right}−{angle}\:{triangle}\:{in}\:{a}\:{given} \\ $$$${triangle}\:{with}\:{sides}\:{a},{b},{c}.\:\left({a}\geqslant{b}\geqslant{c}\right) \\ $$…