+26388 solve for x
In a triangle the sum of the angles are \(180^{\circ}\)
So we have:
\(\begin{array}{|rcll|} \hline 180^{\circ} &=& 90^{\circ} + (2x-10)^{\circ}+ ( x+25)^{\circ} \quad &|\quad -90^{\circ}\\ 180^{\circ} -90^{\circ} &=& (2x-10)^{\circ}+ ( x+25)^{\circ} \\ 90^{\circ} &=& (2x-10)^{\circ}+ ( x+25)^{\circ} \\ 90 &=& (2x-10) + ( x+25) \\ 90 &=& 2x-10 + x+25 \\ 90 &=& 2x+x -10 +25 \\ 90 &=& 3x +15 \quad &|\quad :3 \\ 30 &=& x +5 \quad &|\quad -5 \\ 30-5 &=& x \\ 25 &=& x \\ \mathbf{x} & \mathbf{=} & \mathbf{25} \\\\ \text{First angle: } (2x-10)^{\circ} &=& (2\cdot 25 -10)^{\circ} \\ &=& (50-10)^{\circ} \\ &=& 40^{\circ} \\\\ \text{Second angle: } (x+25)^{\circ} &=& (25 +25)^{\circ} \\ &=& 50^{\circ} \\ \hline \end{array}\)
2x - 10 + x +25 =90
3x + 15 = 90
3x =90 - 15
3x = 75
x =75/3
x = 25 degrees, so that the 2 angles are:
2x25 - 10 =40 degrees
25 + 25 =50 degrees
+26388 solve for x
In a triangle the sum of the angles are \(180^{\circ}\)
So we have:
\(\begin{array}{|rcll|} \hline 180^{\circ} &=& 90^{\circ} + (2x-10)^{\circ}+ ( x+25)^{\circ} \quad &|\quad -90^{\circ}\\ 180^{\circ} -90^{\circ} &=& (2x-10)^{\circ}+ ( x+25)^{\circ} \\ 90^{\circ} &=& (2x-10)^{\circ}+ ( x+25)^{\circ} \\ 90 &=& (2x-10) + ( x+25) \\ 90 &=& 2x-10 + x+25 \\ 90 &=& 2x+x -10 +25 \\ 90 &=& 3x +15 \quad &|\quad :3 \\ 30 &=& x +5 \quad &|\quad -5 \\ 30-5 &=& x \\ 25 &=& x \\ \mathbf{x} & \mathbf{=} & \mathbf{25} \\\\ \text{First angle: } (2x-10)^{\circ} &=& (2\cdot 25 -10)^{\circ} \\ &=& (50-10)^{\circ} \\ &=& 40^{\circ} \\\\ \text{Second angle: } (x+25)^{\circ} &=& (25 +25)^{\circ} \\ &=& 50^{\circ} \\ \hline \end{array}\)
