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Question Number 219093 by hardmath last updated on 19/Apr/25
Commented by hardmath last updated on 19/Apr/25
ABC = △ B - acute angle ∠B = 2 ∙ ∠C AB = 10 BC = 22 S_(△ABC) = ?
$$\mathrm{ABC}\:=\:\bigtriangleup \\ $$$$\mathrm{B}\:-\:\mathrm{acute}\:\mathrm{angle} \\ $$$$\angle\mathrm{B}\:=\:\mathrm{2}\:\centerdot\:\angle\mathrm{C} \\ $$$$\mathrm{AB}\:=\:\mathrm{10} \\ $$$$\mathrm{BC}\:=\:\mathrm{22} \\ $$$$\mathrm{S}_{\bigtriangleup\boldsymbol{\mathrm{ABC}}} \:=\:? \\ $$
Answered by mr W last updated on 19/Apr/25
Commented by mr W last updated on 19/Apr/25
(y/(10))=(x/(22))=k ⇒x=22k, y=10k x×10^2 +y×22^2 =(x+y)(x^2 +xy) 22k×10^2 +10k×22^2 =22k(22+10)^2 k^2 10=(22+10)k^2 ⇒k^2 =((10)/(22+10)) ⇒k=((√5)/4) ⇒x=22×((√5)/4)=((11(√5))/2) cos α=((11×2)/(11(√5)))=(2/( (√5))) ⇒sin α=(1/( (√5))) S_(ABC) =((10×22 sin 2α)/2) =10×22×(1/( (√5)))×(2/( (√5)))=88
$$\frac{{y}}{\mathrm{10}}=\frac{{x}}{\mathrm{22}}={k}\:\Rightarrow{x}=\mathrm{22}{k},\:{y}=\mathrm{10}{k} \\ $$$${x}×\mathrm{10}^{\mathrm{2}} +{y}×\mathrm{22}^{\mathrm{2}} =\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{xy}\right) \\ $$$$\mathrm{22}{k}×\mathrm{10}^{\mathrm{2}} +\mathrm{10}{k}×\mathrm{22}^{\mathrm{2}} =\mathrm{22}{k}\left(\mathrm{22}+\mathrm{10}\right)^{\mathrm{2}} {k}^{\mathrm{2}} \\ $$$$\mathrm{10}=\left(\mathrm{22}+\mathrm{10}\right){k}^{\mathrm{2}} \\ $$$$\Rightarrow{k}^{\mathrm{2}} =\frac{\mathrm{10}}{\mathrm{22}+\mathrm{10}}\:\Rightarrow{k}=\frac{\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\Rightarrow{x}=\mathrm{22}×\frac{\sqrt{\mathrm{5}}}{\mathrm{4}}=\frac{\mathrm{11}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{11}×\mathrm{2}}{\mathrm{11}\sqrt{\mathrm{5}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$${S}_{{ABC}} =\frac{\mathrm{10}×\mathrm{22}\:\mathrm{sin}\:\mathrm{2}\alpha}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{10}×\mathrm{22}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}=\mathrm{88} \\ $$
Commented by hardmath last updated on 19/Apr/25
dear professor, answer: 60, 66, 72, 80, 88
$$\mathrm{dear}\:\mathrm{professor},\:\mathrm{answer}:\:\:\mathrm{60},\:\mathrm{66},\:\mathrm{72},\:\mathrm{80},\:\mathrm{88} \\ $$
Answered by mr W last updated on 19/Apr/25
Commented by mr W last updated on 19/Apr/25
10b^2 +22b^2 =32(10^2 +10×22) ⇒b^2 =320 ⇒b=8(√5) cos α=((16)/( 8(√5)))=(2/( (√5))) ⇒sin α=(1/( (√5))) ⇒sin 2α=(4/5) S_(ABC) =(1/2)×10×22×(4/5)=88
$$\mathrm{10}{b}^{\mathrm{2}} +\mathrm{22}{b}^{\mathrm{2}} =\mathrm{32}\left(\mathrm{10}^{\mathrm{2}} +\mathrm{10}×\mathrm{22}\right) \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\mathrm{320}\:\Rightarrow{b}=\mathrm{8}\sqrt{\mathrm{5}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{16}}{\:\mathrm{8}\sqrt{\mathrm{5}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\: \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{sin}\:\mathrm{2}\alpha=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${S}_{{ABC}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{22}×\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{88} \\ $$
Commented by hardmath last updated on 19/Apr/25
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by Nicholas666 last updated on 20/Apr/25
applying the law of sinus; ((22)/(sin (180°−3α))) = ((10)/(sin α)) ⇒ ((22)/(sin 3α)) = ((10)/(sin α)) → 22 sin α = 10(sin 3α −4 sin^3 α) →22 = 10(3 − 4 sin^2 α) ⇒ sin^2 α = (1/5) →sin α = ((√5)/5) = α = arcsin(((√5)/5)) = 26.57° →AC = 10 ((sin 2 α)/(sin α)) = 20 cos α = 20 (((2(√5))/5)) = 8(√5) = 17.89 → (1/2)(10)(22) sin 2α =110 ((4/5)) = 88 α = 26.57° AC = 17.89 S_(ABC) = 88 sequare units
$${applying}\:{the}\:{law}\:{of}\:{sinus}; \\ $$$$\:\:\frac{\mathrm{22}}{{sin}\:\left(\mathrm{180}°−\mathrm{3}\alpha\right)}\:=\:\frac{\mathrm{10}}{{sin}\:\alpha}\:\Rightarrow\:\frac{\mathrm{22}}{{sin}\:\mathrm{3}\alpha}\:=\:\frac{\mathrm{10}}{{sin}\:\alpha}\:\: \\ $$$$\:\rightarrow\:\mathrm{22}\:{sin}\:\alpha\:=\:\mathrm{10}\left({sin}\:\mathrm{3}\alpha\:−\mathrm{4}\:{sin}^{\mathrm{3}} \:\alpha\right) \\ $$$$\:\rightarrow\mathrm{22}\:=\:\mathrm{10}\left(\mathrm{3}\:−\:\mathrm{4}\:{sin}^{\mathrm{2}} \:\alpha\right)\:\Rightarrow\:{sin}^{\mathrm{2}} \:\alpha\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\:\: \\ $$$$\rightarrow{sin}\:\alpha\:=\:\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\:\:=\:\alpha\:=\:{arcsin}\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\right)\:=\:\mathrm{26}.\mathrm{57}° \\ $$$$\rightarrow{AC}\:=\:\mathrm{10}\:\:\frac{{sin}\:\mathrm{2}\:\alpha}{{sin}\:\alpha}\:=\:\mathrm{20}\:{cos}\:\alpha\:=\:\mathrm{20}\:\left(\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}\right)\:=\:\:\mathrm{8}\sqrt{\mathrm{5}}\:=\:\mathrm{17}.\mathrm{89} \\ $$$$\rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}\right)\left(\mathrm{22}\right)\:{sin}\:\mathrm{2}\alpha\:=\mathrm{110}\:\left(\frac{\mathrm{4}}{\mathrm{5}}\right)\:=\:\mathrm{88}\:\: \\ $$$$ \\ $$$$\alpha\:=\:\mathrm{26}.\mathrm{57}° \\ $$$${AC}\:=\:\mathrm{17}.\mathrm{89} \\ $$$${S}_{{ABC}} \:=\:\mathrm{88}\:{sequare}\:{units} \\ $$$$ \\ $$

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