Question-221958

Facebook Tweet Pin

Question Number 221958 by ajfour last updated on 13/Jun/25

Commented by mr W last updated on 14/Jun/25

$${a}\:{may}\:{not}\:{be}\:{independent}\:{from}\:{b},\:{c}. \\ $$

Commented by ajfour last updated on 14/Jun/25

Find a and R. forget a=3.

Answered by mr W last updated on 14/Jun/25

Commented by mr W last updated on 14/Jun/25

sin (β/2)=((c−b)/(c+b)) ⇒β=2 sin^(−1) ((c−b)/(c+b)) α+β=(π/2)−β ⇒α=(π/2)−2β (a/(tan (α/2)))=(c/(tan (β/2))) ⇒a=((c (1−tan β))/((1+tan β) tan (β/2))) R=(c/(sin 2β tan (β/2))) example: c=6, b=4 ⇒R≈40.7609, a≈11.8318

$$\mathrm{sin}\:\frac{\beta}{\mathrm{2}}=\frac{{c}−{b}}{{c}+{b}} \\ $$$$\Rightarrow\beta=\mathrm{2}\:\mathrm{sin}^{−\mathrm{1}} \frac{{c}−{b}}{{c}+{b}} \\ $$$$\alpha+\beta=\frac{\pi}{\mathrm{2}}−\beta \\ $$$$\Rightarrow\alpha=\frac{\pi}{\mathrm{2}}−\mathrm{2}\beta \\ $$$$\frac{{a}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}=\frac{{c}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}} \\ $$$$\Rightarrow{a}=\frac{{c}\:\left(\mathrm{1}−\mathrm{tan}\:\beta\right)}{\left(\mathrm{1}+\mathrm{tan}\:\beta\right)\:\mathrm{tan}\:\frac{\beta}{\mathrm{2}}} \\ $$$${R}=\frac{{c}}{\mathrm{sin}\:\mathrm{2}\beta\:\mathrm{tan}\:\frac{\beta}{\mathrm{2}}} \\ $$$${example}:\:{c}=\mathrm{6},\:{b}=\mathrm{4} \\ $$$$\Rightarrow{R}\approx\mathrm{40}.\mathrm{7609},\:{a}\approx\mathrm{11}.\mathrm{8318} \\ $$

Commented by mr W last updated on 14/Jun/25