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In triangle ABC C = 120. M is the foot of perpendicular drawn from vertex A to line BC. If AC=8, BC=10, AB=?

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  • In triangle ABC C = 120. M is the foot of perpendicular drawn from vertex A to line BC. If AC=8, BC=10, AB=?


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Answer #1 | 10/02 2014 05:50
It seems to me that would be an impossible triangle if 180.
Positive: 64 %
Answer #2 | 10/02 2014 05:54
I suppose you mean: m
Positive: 58 %
Answer #3 | 10/02 2014 07:44
In triangle AMC angle M =90 degree and angle ACM =19-120=60 degree AM/ AC = sin 60 =sqrt(3)/2 AM/8 =sqrt(3)/2 AM =4*sqrt(3) ..............(i) CM/AC = cos 60 =1/2 CM/8 =1/2 CM = 4 ...............(i) Now in triangle AMB AB^2=AM^2+BM^2 AB^2 =AM^2 +(10+4)^2 =48 +196 =244 Taking sqrt AB =sqrt(244)=15.62 units ..................Ans
Positive: 38 %
Answer #4 | 10/02 2014 21:19
In △ ABC angle C = 120°. M is the foot of perpendicular drawn from vertex A to line BC. If AC = 8, BC = 10, AB = ? Since angle C is obtuse the perpendicular AM on BC is outside △ ABC. Angle CAM = 30° CM/AC = sin 30° = 1/2, CM = 1/2 AC = 4 CM/AM = tan 60° = 1.732 4/AM = 1.732 AM = 4/1.732 = 2.3094 AB^2 = AM^2 + MB^2 = 2.3094^2 + (10 + 4)^2 = 201.333 AB = 14.1892
Positive: 6 %

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