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>>
Hello anshulbshahYou 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 $\displaystyle x^o$. Then the largest angle is $\displaystyle 5x$. Since the triangle is acute-angled, $\displaystyle 5x < 90\Rightarrow 5x \le 85$, since $\displaystyle x$ must be an integer, and $\displaystyle 5x$ therefore a multiple of $\displaystyle 5$. This gives us that $\displaystyle 0<x\le 17$.
Next suppose that the third (middle-sized) angle of the triangle is $\displaystyle y^o$. Using the angle-sum of a triangle, $\displaystyle y = 180 - 6x$. The smallest possible value of $\displaystyle y$ is when $\displaystyle x$ is as large as possible; i.e. $\displaystyle x = 17\Rightarrow y = 180 - 102 = 78$, and at this value the largest angle is $\displaystyle 5x=85^o$.
If we make $\displaystyle x$ a bit smaller, say $\displaystyle x = 16$, then $\displaystyle y = 180 - 96 = 84$. But in this case $\displaystyle 5x = 80$, which is smaller than $\displaystyle y$.
So the only solution is $\displaystyle 17^o,78^o,85^o$.
Grandad