Question Number 224570 by Abdulazim last updated on 19/Sep/25
Answered by fantastic last updated on 19/Sep/25
![30^0 AB=2Rsin α CH=2Rcos α[R=circumradius] CH=(√3)AB 2Rcos α=(√3)×2Rsin α (√3) tan α=1 α=tan^(−1) ((1/( (√3))))=30^0](https://www.tinkutara.com/question/Q224575.png)
$$\mathrm{30}^{\mathrm{0}} \\ $$$${AB}=\mathrm{2}{R}\mathrm{sin}\:\alpha \\ $$$${CH}=\mathrm{2}{R}\mathrm{cos}\:\alpha\left[{R}={circumradius}\right] \\ $$$${CH}=\sqrt{\mathrm{3}}{AB} \\ $$$$\mathrm{2}{R}\mathrm{cos}\:\alpha=\sqrt{\mathrm{3}}×\mathrm{2}{R}\mathrm{sin}\:\alpha \\ $$$$\sqrt{\mathrm{3}}\:\mathrm{tan}\:\alpha=\mathrm{1} \\ $$$$\alpha=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\mathrm{30}^{\mathrm{0}} \\ $$
Commented by mr W last updated on 19/Sep/25
$${right}! \\ $$
Answered by mr W last updated on 19/Sep/25
Commented by mr W last updated on 19/Sep/25
$$\angle{EAD}=\angle{EBC}=\angle{HAD} \\ $$$$\Rightarrow{AH}={AE}=\mathrm{2}{R}\:\mathrm{sin}\:\angle{ABE} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{R}\:\mathrm{sin}\:\left(\mathrm{90}°−\angle{BAC}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{R}\:\mathrm{cos}\:\angle{BAC} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{R}\:\mathrm{cos}\:\angle{A} \\ $$$${BC}=\mathrm{2}{R}\:\mathrm{sin}\:\angle{A} \\ $$$$\Rightarrow\frac{{BC}}{{AH}}=\frac{\mathrm{2}{R}\:\mathrm{sin}\:\angle{A}}{\mathrm{2}{R}\:\mathrm{cos}\:\angle{A}}=\mathrm{tan}\:\angle{A} \\ $$$${similarly} \\ $$$$\frac{{CA}}{{BH}}=\mathrm{tan}\:\angle{B} \\ $$$$\frac{{AB}}{{CH}}=\mathrm{tan}\:\angle{C} \\ $$$${in}\:{current}\:{case} \\ $$$$\mathrm{tan}\:\alpha=\frac{{AB}}{{CH}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\alpha=\mathrm{30}°\:\checkmark \\ $$
Commented by fantastic last updated on 20/Sep/25
$${very}\:{good}\:{solution} \\ $$