Question 49: Calculate the measure of the angles of a triangle, knowing the angles are proportional to 1; 2; 3.

A. \({30^0},{60^0},{90^0}\)

B. \({40^0},{50^0},{90^0}\)

C. \({30^0},{70^0},{80^0}\)

D. \({20^0},{80^0},{70^0}\)

Let the measure of the three angles of a triangle be x, y, z (units) (Condition: x, y, z > 0)

According to the problem we have: \({x \over 1} = {y \over 2} = {z \over 3}\) and x + y + z = 180 degrees (sum of three angles of a triangle)

According to the properties of the series of equal ratios, we have: \({x \over 1} = {y \over 2} = {z \over 3} = {{180} \over 6} = 30\)

\({x \over 1} = 30 \Rightarrow x = 30;{y \over 2} = 30 \Rightarrow 2.30 = 60;{z \over 3} = 30 \Rightarrow z = 3.30 = 90\)

So the measures of the three angles of the triangle are: \({30^0},{60^0},{90^0}\)

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