Question

260²=5F² + 4F²cos theta 180²=5F²-4F²cos theta Find the ratio of the two equation.​

Answer 1

Case-1

$\\ \sf\longmapsto 260^2=5F^2+4F^2$

$\\ \sf\longmapsto 9F^2=260^2$

$\\ \sf\longmapsto 9F=260$

$\\ \sf\longmapsto F=\dfrac{260}{9}$

$\\ \sf\longmapsto F_1=28.8$

Case-2:-

$\\ \sf\longmapsto 180^2=5F^2-4F^2$

$\\ \sf\longmapsto F^2=180^2$

$\\ \sf\longmapsto F=180$

Ratio:-

$\\ \sf\longmapsto \dfrac{F_1}{F_2}=\dfrac{28.8}{180}$

$\\ \sf\longmapsto \dfrac{F_1}{F_2}\approx\dfrac{1}{6}$

$\\ \sf\longmapsto F_1:F_2\approx=1:6$