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

Answers

Answer #1 | 10/02 2014 05:50

It seems to me that would be an impossible triangle if 180.

Answer #2 | 10/02 2014 05:54

I suppose you mean: m

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

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

Answer #1| 10/02 2014 05:50