Q. For any acute triangle, all its angles have measure in integral degree. The smallest angle has measure 1/5 of the measure of the greatest angle. Find the measure of the angles of the triangle?
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Q. For any acute triangle, all its angles have measure in integral degree. The smallest angle has measure 1/5 of the measure of the greatest angle. Find the measure of the angles of the triangle?
A nice problem that combines geometry and algebra.
Let x be the smallest angle; then the angles are x, 5x, and 180 - 6x. Saying that each of these angles is acute gives you two inequalities (well, three, but the third one trivially follows from one of the others). In my calculations, these inequalities have two integer solutions. However, in one of them, 5x is not the greatest angle. This leaves one solution.
You can simply "think out loud" to solve this one:Quote:
Originally Posted by akshay1995
since acute, then all angles < 90
since integers, then highest = 89
BUT since smallest is 1/5 of highest, then highest must be divisible by 5, so highest = 85
so my lowest would be 1/5(85) = 17
let's see; 180 - 85 - 17 = 78 ....YA!!