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# geometry

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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.

Dec 4, 2020

#1
+893
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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

Dec 5, 2020