Question No. 4 If N is the number of triangles of different shapes (i.e. not similar) whose angles are all integers (in degrees), then find N/300
Answers
Step-by-step explanation:
this requires us to think of various combinations of three integral angles that can be selected so as to comply with the property of triangle. From that combination we will reduce such combinations as are replicating and making the triangles similar thereby violating the condition of different shape.
Let us first try to recognise a pattern by choosing a few angles
1. suppose one angle is fixed at 1°
then rest two angles should sum up to 179°
there are various possibilities (1,178),(2,177),(3,176).......176,3),(177,2),(178,1)
we can see that there are 178 possibilities. however we can see that first and last combinations form same set of angles thereby making the triangle similar. similarly second and second last, third and third last and so son.
Actually half of the combinations are repeating. hence, actual non repeating combinations in respect of 1° is 178/2 = 89.
2. now we fix one angle at 2°
then the rest of the angles should sum up to 178°
the various combinations are (1,177),(2,176).....(89,89),,,,,,,(176,2),(177,1)
we can see that there are 177 possibilities but except one all are repeating twice. The one not repeating is (89,89)
thus non repeating combinations in this case are (177-1)/2+1
= 176/2+1
= 88+1= 89
Now, we can fix one angle anywhere from 1° to 178° (not 179°).
between 1 and 178 there are 89 odd and 89 even integers
hence the total combination will be
89+89+88+88+87+87+....1+1
= 2*(89+88+87+....+3+2+1)
= 2*89*90/2
=89*90
= 8010
Now this 8010 combinations still contain repeating combinations which we could not eliminate while calculating combination for individual degree.
for example in case of 1° , first combination is (1.178). similarly in case on 178°, one combination will be (1,1). In both the cases the combination of three angles are same i.e. 1,1 and 178.
On careful perusal we can see that there are 88 such combinations which repeat twice, one combination (60,60,60) which does not repeat and rest combinations occur thrice in 8010 combinations. For more clarity, I give one example of thrice repeating combination.
in case of 1° one combination is (2,177)
in case of 2°, one combination is (1,177)
in case of 177°, one combination is (1,2)
Now what we will do is to first remove non repeating and twice repeating combinations from 8010 and divide the balance by 3 to get non repeating combinations
1. non repeating combination = 1
2. twice repeating combinations = 88*2 = 176
balance = 8010 - 1 - 176
= 7833
hence total non repeating combinations are
7833/3 + 176/2 + 1
= 2611 + 88 + q
= 2700
hence, N = 2700
N/3 00 = 2700/300
= 9
Answer:
see above
Step-by-step explanation: