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Sagot :
Recall that the sum of all the angles in a triangle is = 180 degrees
Thus we have that
[tex]\begin{gathered} 2x+6x-5+x+5=180^0 \\ 9x=180^0 \\ x\text{ = }\frac{180^0}{9} \\ x=20^0 \end{gathered}[/tex]Hence,
[tex]\begin{gathered} Q\text{ = x + 5 } \\ =\text{ 20 + 5} \\ =25^0 \\ S\text{ = 2x } \\ =2(20^0) \\ =40^0 \\ R\text{ = 6x - 5 } \\ =6(20^0)\text{ - 5} \\ =\text{ 120 - 5} \\ =115^0 \end{gathered}[/tex]
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