# Thread: Finding the measure of angles...

1. ## Finding the measure of angles...

For an acute angled triangle,all its angles have measure in integral degree.The smallest angle has a mesure 1/5th the measure of largest.Find all the measures of triangle>>

2. Hello anshulbshah
Originally Posted by anshulbshah
For an acute angled triangle,all its angles have measure in integral degree.The smallest angle has a mesure 1/5th the measure of largest.Find all the measures of triangle>>
You can't solve problems like this by simple algebra - you have to use a little bit of 'trial and error'.

First, let's suppose the smallest angle is $x^o$. Then the largest angle is $5x$. Since the triangle is acute-angled, $5x < 90\Rightarrow 5x \le 85$, since $x$ must be an integer, and $5x$ therefore a multiple of $5$. This gives us that $0.

Next suppose that the third (middle-sized) angle of the triangle is $y^o$. Using the angle-sum of a triangle, $y = 180 - 6x$. The smallest possible value of $y$ is when $x$ is as large as possible; i.e. $x = 17\Rightarrow y = 180 - 102 = 78$, and at this value the largest angle is $5x=85^o$.

If we make $x$ a bit smaller, say $x = 16$, then $y = 180 - 96 = 84$. But in this case $5x = 80$, which is smaller than $y$.

So the only solution is $17^o,78^o,85^o$.