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Question Number 196007 by sonukgindia last updated on 15/Aug/23

Answered by mr W last updated on 15/Aug/23

AC=(R/(tan (α/2)))  BC^(⌢) =(π−α)R  (π−α)R=(R/(tan (α/2)))  (π−α) tan (α/2)=1  ⇒α≈0.81047 rad ≈46.4365°

$${AC}=\frac{{R}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}} \\ $$$$\overset{\frown} {{BC}}=\left(\pi−\alpha\right){R} \\ $$$$\left(\pi−\alpha\right){R}=\frac{{R}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}} \\ $$$$\left(\pi−\alpha\right)\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow\alpha\approx\mathrm{0}.\mathrm{81047}\:{rad}\:\approx\mathrm{46}.\mathrm{4365}° \\ $$

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