There are actually infinitely many solutions here. Given that '60' is just any one the angles, that is.
30,60,90, or 20,60,100, so on.
the sides of triangle are in arithmetic progession(AP)and of them is 60 find the angles in grade
6 answers
It said that the sides form an AP, not the angles
for angles 30, 60, 90, the sides would be 1 , √3, and 2, which do not form an AP
for angles 30, 60, 90, the sides would be 1 , √3, and 2, which do not form an AP
Oh, my apologies, I considered it to be 'angles are in an AP'
Then again, infinite solutions holds true right?
I suspect you mistyped the question, it should have been "the angles are in arithemtric..." This is an old text problem, in math texts since John Paul Jones was a midshipman.
Let the sides be 60, 60+a, 60+2a
(note that a could be negative, if 60 is not the smallest side)
Then we have
sinA/60 = sinB/(60+a) = sinC/(60+2a)
A+B+C = 180
That is only 3 equations and 4 unknowns, so there are many possible solutions. For example, let the sides be
60,60.1,60.2
Then the sides are very nearly equal, so the angles are also very nearly equal.
Or, let the sides be 60,119,178
Then the triangle is very nearly flat, so two of the angles are tiny, and one is almost 180.
And any solution in between these cases is also possible.
(note that a could be negative, if 60 is not the smallest side)
Then we have
sinA/60 = sinB/(60+a) = sinC/(60+2a)
A+B+C = 180
That is only 3 equations and 4 unknowns, so there are many possible solutions. For example, let the sides be
60,60.1,60.2
Then the sides are very nearly equal, so the angles are also very nearly equal.
Or, let the sides be 60,119,178
Then the triangle is very nearly flat, so two of the angles are tiny, and one is almost 180.
And any solution in between these cases is also possible.