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Question Number 222176 by cherokeesay last updated on 19/Jun/25
Answered by mr W last updated on 19/Jun/25
∠B=β ∠A=2β ∠C=π−3β (a/(sin 2β))=(c/(sin 3β))=((25)/(sin β)) ⇒a=50 cos β ⇒c=25(3−4 sin^2 β) r=15 sin β =radius of incircle Δ=(((a+b+c)r)/2)=((15 sin β (25+50 cos β+75−100 sin^2 β))/2) Δ=((ac sin β)/2)=((50 cos β×25 (3−4 sin^2 β) sin β)/2) ((50 cos β×25 (3−4 sin^2 β) sin β)/2)=((15 sin β (25+50 cos β+75−100 sin^2 β))/2) 5 cos β (3−4 sin^2 β)=3(2+cos β−2 sin^2 β) 5(4 cos^2 β−1)=3(2 cos β+1) 10 cos^2 β−3 cos β−4=0 (5 cos β−4)(2 cos β+1)=0 ⇒cos β=(4/5) ⇒sin β=(3/5) Δ=((50×0.8×25 (3−4×0.6^2 )×0.6)/2) =468 ✓
$$\angle{B}=\beta \\ $$$$\angle{A}=\mathrm{2}\beta \\ $$$$\angle{C}=\pi−\mathrm{3}\beta \\ $$$$\frac{{a}}{\mathrm{sin}\:\mathrm{2}\beta}=\frac{{c}}{\mathrm{sin}\:\mathrm{3}\beta}=\frac{\mathrm{25}}{\mathrm{sin}\:\beta} \\ $$$$\Rightarrow{a}=\mathrm{50}\:\mathrm{cos}\:\beta \\ $$$$\Rightarrow{c}=\mathrm{25}\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\beta\right) \\ $$$${r}=\mathrm{15}\:\mathrm{sin}\:\beta\:={radius}\:{of}\:{incircle} \\ $$$$\Delta=\frac{\left({a}+{b}+{c}\right){r}}{\mathrm{2}}=\frac{\mathrm{15}\:\mathrm{sin}\:\beta\:\left(\mathrm{25}+\mathrm{50}\:\mathrm{cos}\:\beta+\mathrm{75}−\mathrm{100}\:\mathrm{sin}^{\mathrm{2}} \:\beta\right)}{\mathrm{2}} \\ $$$$\Delta=\frac{{ac}\:\mathrm{sin}\:\beta}{\mathrm{2}}=\frac{\mathrm{50}\:\mathrm{cos}\:\beta×\mathrm{25}\:\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\beta\right)\:\mathrm{sin}\:\beta}{\mathrm{2}} \\ $$$$\frac{\mathrm{50}\:\mathrm{cos}\:\beta×\mathrm{25}\:\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\beta\right)\:\mathrm{sin}\:\beta}{\mathrm{2}}=\frac{\mathrm{15}\:\mathrm{sin}\:\beta\:\left(\mathrm{25}+\mathrm{50}\:\mathrm{cos}\:\beta+\mathrm{75}−\mathrm{100}\:\mathrm{sin}^{\mathrm{2}} \:\beta\right)}{\mathrm{2}} \\ $$$$\mathrm{5}\:\mathrm{cos}\:\beta\:\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\beta\right)=\mathrm{3}\left(\mathrm{2}+\mathrm{cos}\:\beta−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\beta\right) \\ $$$$\mathrm{5}\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\beta−\mathrm{1}\right)=\mathrm{3}\left(\mathrm{2}\:\mathrm{cos}\:\beta+\mathrm{1}\right) \\ $$$$\mathrm{10}\:\mathrm{cos}^{\mathrm{2}} \:\beta−\mathrm{3}\:\mathrm{cos}\:\beta−\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{5}\:\mathrm{cos}\:\beta−\mathrm{4}\right)\left(\mathrm{2}\:\mathrm{cos}\:\beta+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\beta=\frac{\mathrm{4}}{\mathrm{5}}\:\Rightarrow\mathrm{sin}\:\beta=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\Delta=\frac{\mathrm{50}×\mathrm{0}.\mathrm{8}×\mathrm{25}\:\left(\mathrm{3}−\mathrm{4}×\mathrm{0}.\mathrm{6}^{\mathrm{2}} \right)×\mathrm{0}.\mathrm{6}}{\mathrm{2}} \\ $$$$\:\:\:\:=\mathrm{468}\:\checkmark \\ $$
Commented by cherokeesay last updated on 19/Jun/25
thank you master !
$${thank}\:{you}\:{master}\:! \\ $$
Commented by mr W last updated on 20/Jun/25

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