Question 1375: obtuse isosolese triangle angles 120/30/30.equal legs 7.what is formula to determine third (longest leg/base)?
Found 3 solutions by khwang, abhinav75, cleomenius:
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website!
Even we cannot have good diagram of triangle here, try to figure
out if angle BAC is 120 deg, both ABC & BCA are 30 deg, and AB=AC = 7.
A
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B D C
Let AD be the height of the base BC(note AD is perendicular to BC),
we ses that angle BAD = 120/2 =60, so BD/AB = sin 60 ,
BD = AB sin 60 = 7 * sqrt(3)/2 = 6.06
Since BD=CD,BC = 2 BD.
Hence,the longest leg BC = 2*BD = 12.12
Of course, we can use the law of cosine directly, [Note: a=BC, b= AC, c=AB]
a^2 =b^2 + c^2 - 2bc cos A,
= 7^2 + 7^2 - 2*7^2 cos(120)
= 98(1 +0.5) [Note: cos 120 = -0.5]
= 147,
So, a = sqrt(147) = 12.12 (same answer)
Answer by abhinav75(8) (Show Source):
You can put this solution on YOUR website! the median from the central angle would divide the base in two, and would be the altitude as well. (case of congruent triangles).
now, the hypotenuse is 7 for the smaller triangle. so,
cos 30 = sqrt(3)/2 = base/hypotenuse...
so, base (of the sub triangle) = hypotenuse * cos 30
so, base of the original triangle = 2 * hypotenuse * cos 30
= 2*7*sqrt(3)/2
=7*sqrt(3)
= 1*1.732
= 12.124
Answer by cleomenius(959)  (Show Source):
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