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.
+1639 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
