+1904 Triangle \(ABC\) has \(AC=BC\), angle \(ACB=96°\). \(D\) is a point in \(ABC\) such that angle \(DAB=18°\)and angle \(DBA=30°\). What is the measure (in degrees) of angle \(ACD\)? Please show how you got your answer.
+129907 Let AC = BC = 1
And if ACB = 96 Then CAB = CBA = 42
But ADB is a triangle and DAB = 18 and DBA = 30.......so ADB = 132
And using the Law of Cosines
AB^2 = AC^2 + BC^2 - 2 (AC)(BC)cos(96) .....so.....
AB^2 = 1^2 + 1^2 - 2(1)(1)cos(96)
AB = sqrt [2 - 2cos(96)]
And by the Law of Sines
AD / sin(30) = AB/sin(132)
AD / (1/2) = sqrt[2 - 2cos(96)] sin(132)
AD = sqrt(2 - 2cos(96)] / [2 sin(132)]
AD = [sqrt(2) (sqrt(1 - cos(96))] / [2sin(132)]
AD = [sqrt(2)/2] * [sqrt(1 - cos(96)]/ [sin(132)]
AD = [1/sqrt(2)] * [sqrt(1 - cos(96)] / [sin(132)]
AD = [ sqrt(1 - cos(96)/ sqrt(2)] / [sin(132)]
AD = sqrt [(1 - cos(96)/ 2] / sin(132)
And by a trig identity sqrt [(1 - cos(96)/ 2] = sin(48)
So
AD = sin(48)/sin(132) ...... but sin(48) = sin(132) .....therefore....
AD = 1
And angle CAB = angle DAB + angle DAC
42 = 18 + angle DAC ...therefore
Angle DAC = 42 - 18 = 24
So....in triangle CAD
AD = CA = 1
And angle DAC = 24
So......since AD = CA....the angles opposite those sides are also equal
Therefore, angle ADC = angle ACD = [180 - 24]/2 = 156/2 = 78
Therefore....Angle ACD = 78
