Question-150683

Question Number 150683 by Jonathanwaweh last updated on 14/Aug/21

Commented by Tawa11 last updated on 14/Aug/21

$$\mathrm{Great} \\ $$

Answered by ajfour last updated on 14/Aug/21

4(AL)=R ⇒ LO=((3R)/4) (R−r)^2 =r^2 +(((3R)/4))^2 ⇒ r=((7R)/(32)) tan α=(r/((((3R)/4))))=(4/3)×(7/(32))=(7/(24)) T(((24R)/(25)), ((7R)/(25))) tan β=((24R/25)/(R−((7R)/(25))))=(4/3) x=β−(π/4) = tan^(−1) ((4/3))−(π/4) ≈ 53°−45°=8°.

$$\mathrm{4}\left({AL}\right)={R}\:\:\Rightarrow\:{LO}=\frac{\mathrm{3}{R}}{\mathrm{4}} \\ $$$$\left({R}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +\left(\frac{\mathrm{3}{R}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{r}=\frac{\mathrm{7}{R}}{\mathrm{32}} \\ $$$$\mathrm{tan}\:\alpha=\frac{{r}}{\left(\frac{\mathrm{3}{R}}{\mathrm{4}}\right)}=\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{7}}{\mathrm{32}}=\frac{\mathrm{7}}{\mathrm{24}} \\ $$$${T}\left(\frac{\mathrm{24}{R}}{\mathrm{25}},\:\frac{\mathrm{7}{R}}{\mathrm{25}}\right) \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{24}{R}/\mathrm{25}}{{R}−\frac{\mathrm{7}{R}}{\mathrm{25}}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${x}=\beta−\frac{\pi}{\mathrm{4}}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right)−\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\approx\:\mathrm{53}°−\mathrm{45}°=\mathrm{8}°. \\ $$

Commented by mr W last updated on 14/Aug/21