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Sagot :
WE can find the sides if the triangle by apply the Sine rule :
[tex]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}[/tex]where, a,b & c are the sides of triangle
For eg :
Consider an triangle with one side AB = 5
and angles A = 60, angle B = 45 and angle C= 75
So, substitute the value in the expression of Sine
[tex]\begin{gathered} \frac{BC}{\sin A}=\frac{AC}{\sin B}=\frac{AB}{\sin C} \\ \frac{BC}{\sin 60}=\frac{AC}{\sin 45}=\frac{5}{\sin 75} \\ \text{ Substitute the values :} \\ \frac{BC}{\sin60}=\frac{AC}{\sin45}=\frac{5}{\sin75} \\ \frac{BC}{0.866}=\frac{AC}{0.707}=\frac{5}{0.965} \\ \text{ Simplify : }\frac{AC}{0.707}=\frac{5}{0.965} \\ \frac{AC}{0.707}=\frac{5}{0.965} \\ AC=\frac{5}{0.965}\times0.707 \\ AC=3.66 \\ \text{Now, Simplify: }\frac{BC}{0.866}=\frac{5}{0.965} \\ BC=\frac{5}{0.965}\times0.866 \\ BC=4.4 \end{gathered}[/tex]The sides : AB = 5, AC = 3.66 & BC = 4.4
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