In a triangle, the measure of the second angle is $20^{o}$ more than twice the first and the measure of the third angle is $55^{o}$ more than one-third of the first. Find the measure of the angles of the triangle.

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Solution

Let $x$ be the first angle of the triangle.
Therefore,
Second angle $= 20 ° + 2 x$
Third angle $= 55 ° + \frac{x}{3}$
$∴ x + (20 ° + 2 x) + (55 ° + \frac{x}{3}) = 180 ° (Sum of interior angles of triangle)$
$\Rightarrow 3 x + \frac{x}{3} = 180 ° - 75 °$
$\Rightarrow \frac{10 x}{3} = 105 °$
$\Rightarrow x = \frac{105 ° \times 3}{10} = 31.5 °$
First angle $= 31.5 °$
Therefore,
Second angle $= 20 ° + 2 \times 31.5 ° = 20 ° + 63 ° = 83 °$
Third angle $= 55 ° + \frac{31.5}{3} = 55 ° + 10.5 ° = 65.5 °$

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In a triangle, the measure of the second angle is 20o more than twice the first and the measure of the third angle is 55o more than one-third of the first. Find the measure of the angles of the triangle.

In a triangle, the measure of the second angle is $20^{o}$ more than twice the first and the measure of the third angle is $55^{o}$ more than one-third of the first. Find the measure of the angles of the triangle.