In triangle ABC, angle ACB is 50 degrees, and angle CBA is 75 degrees. Let D be the foot of the perpendicular from A to BC, O the center of the circle circumscribed about triangle ABC, and E the other end of the diameter which goes through A. Find the angle DAE, in degrees.
In triangle ABC, angle ACB is 50 degrees, and angle CBA is 75 degrees. Let D be the foot of the perpendicular from A to BC, O the center of the circle circumscribed about triangle ABC, and E the other end of the diameter which goes through A. Find the angle DAE, in degrees.
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∠ABC = 75º ∠ACB = 50º ∠BAC = 55º Let AB be 8 units
1/ By using the law of sines we find that BC = 8.555 and AC = 10.088
2/ Using Heron's formula we calculate the area of [ABC] = 30.055 u2
3/ Inradius r = A / s ==> r = 2.481 ( s -- semiperimeter)
4/ Circumradius R = (abc) / 4rs ==> R = 5.222
∠DAC = 40º ∠BAD = 15º
Angle OAC = cos-1[(AC/2) / AO] ==> ∠OAC = 15º
And finally... ∠DAE = 55 - (2*15) = 25º