Question-212668

Question Number 212668 by Spillover last updated on 20/Oct/24

Answered by efronzo1 last updated on 21/Oct/24

BC^2 = 64+25−40 = 49 BC^2 = 2r^2 +r^2 ⇒r=(√((BC^2 )/3))=((7(√3))/3)

$$\:\:\mathrm{BC}^{\mathrm{2}} =\:\mathrm{64}+\mathrm{25}−\mathrm{40}\:=\:\mathrm{49} \\ $$$$\:\:\mathrm{BC}^{\mathrm{2}} =\:\mathrm{2r}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \Rightarrow\mathrm{r}=\sqrt{\frac{\mathrm{BC}^{\mathrm{2}} }{\mathrm{3}}}=\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$

Commented by Spillover last updated on 21/Oct/24

$${good} \\ $$

Answered by mr W last updated on 21/Oct/24

R=((BC)/(2 sin A)) =((√(8^2 +5^2 −2×8×5 cos 60°))/(2 sin 60°))=(7/( (√3)))≈4

$${R}=\frac{{BC}}{\mathrm{2}\:\mathrm{sin}\:{A}} \\ $$$$\:\:\:\:=\frac{\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}×\mathrm{8}×\mathrm{5}\:\mathrm{cos}\:\mathrm{60}°}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{60}°}=\frac{\mathrm{7}}{\:\sqrt{\mathrm{3}}}\approx\mathrm{4} \\ $$

Commented by Spillover last updated on 21/Oct/24