Question 2912: Find if the triangle is acute or obtuse from the sides of given length
1)5,8,9
2)24,8,(35
3)2(10,7,3
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Your typing was not clear in (2)& (3).
So far as I know, to determone if a triangle is acute or obtuse (or
even to see if it is a valid triangle) we had better use the law of
cosine. As, c^2 = a^2 + b^2 - 2 bc cos C.
So cos C = (c^2 - a^2 - b^2) / (2bc)
When c^2 < a^2 + b^2 , angle C is acute.
When c^2 > a^2 + b^2 , angle C is obtuse.
Also, the only possible obtuse angle should be the angle opposite to
the longest side.
For a valid triangle, we should have a+b > c, b+c >a and c+a > b
(1) Let (a,b,c)= (5,8,9).
At first , check 5+8 > 9 ,OK. It is a legal triangle.
(No need to check others as comparing 5+9 and 8, why?)
c= 9 is the longest. check angle C
Since 5^2 + 8^2 = 25 + 64 = 89 > 9^2 ,so angle C is acute)
Hence, this is an acute triangle.
(2) I suppose, you typed as:
(a,b,c)= (24,8,35)
Note 24+8 < 35, illegal triangle.
If I change it to (a,b,c)= (24,6,35)
c= 35 is the longest. check angle C
Since 24^2 + 6^2 = 576 + 36 = 612 < 35^2 = 1225 ,so angle C is obtuse)
Hence, this is an obtuse triangle.
Try to read my solutions carefully. And, solve similar questions
with other given triples. And always remember that the importance
is the idea not the answer.
Kenny
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