Question

the sines of the angles of a triangle are as 4 : 5 : 6 prove that cosines of the angles are 12 : 9 :2

Answer 1

$$\begin{gathered} \mathrm{Sines}\mathrm{of}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{triangle}\mathrm{are}\mathrm{as}4:5:6 \\ \mathrm{Then}, \\ \mathrm{According}\mathrm{to}\mathrm{sine}\mathrm{law},\mathrm{ratio}\mathrm{of}\mathrm{side}\mathrm{is}\mathrm{also}4:5:6. \\ \mathrm{Now},\mathrm{consider}\mathrm{a}\mathrm{triangle}\mathrm{with}\mathrm{side}4,5\mathrm{and}6. \\ \mathrm{We}\mathrm{know}, \\ \cos \mathrm{A}=\frac{{\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2}{2\mathrm{bc}}=\frac{{\left(5\right)}^2+{\left(6\right)}^2-{\left(4\right)}^2}{2\left(5\right)\left(6\right)}=\frac{45}{60}=\frac{3}{4}=\frac{12}{16} \\ \cos \mathrm{B}=\frac{{\left(4\right)}^2+{\left(6\right)}^2-{\left(5\right)}^2}{2\left(4\right)\left(6\right)}=\frac{27}{48}=\frac{9}{16} \\ \cos \mathrm{C}=\frac{{\left(5\right)}^2+{\left(4\right)}^2-{\left(6\right)}^2}{2\left(5\right)\left(4\right)}=\frac{5}{40}=\frac{1}{8}=\frac{2}{16} \\ \mathrm{So},\cos \mathrm{A}:\cos \mathrm{B}:\cos \mathrm{C}=12:9:2 \\ \mathrm{HENCE}\mathrm{PROVED}. \end{gathered}$$