Triangle ABC has AC=BC, angle ACB=96 degrees. D is a point in ABC such that angle DAB=18 degrees and angleDBA=30 degrees. What is the measure in degrees of angle ACD?

I got 42 degrees!

7 answers

Hmm, the triangle seems impossible, because if you draw it, given the condition that AC = BC (thus it's an isosceles triangle), then angle ACB = 96, therefore BAC = 96 degrees also (since the sides opposite to the angles are equal). So if there are two angles which are 96, the measure of the third angle ABC is
180 - 2(96)
= 180 - 192
= -12 degrees, which is impossible.
But well I'm not sure. I guess there is a typo or a figure given?

hope this helps :3

Since the original triangle is isosceles, it is easy to calculate that angle CAB =angle CBA = 42°
then angle CAD = 24° and angle CBD = 12°
Also angle ADB = 132°

let angle ADC = x
then angle BDC = 360-x-132 = 258-x

in triangle ADC, by the sine law,
DC/sin24 = AC/sinx
DC = AC sin24/sinx

in triangle BCD,
DC/sin12 = BC/sin(22-x)
CD = BC sin12/sin(228-x)

ACsin24/sinx = BCsin12/sin(228-x)
but AC = BC, so let's divide both sides by AC

sin24/sinx = sin12/sin(228-x)
sin12(sinx) = sin24/(sin(228-x))
sin12 sinx = sin24/( (sin228cosx - cos228 sinx) )

sin12sinx = sin24sin228cosx - sin24cos228sinx
sinx(sin12 + sin24cos228) = sin24sin228cosx

sinx/cosx = sin24sin228/(sin12 + sin24cos228)
tanx = appr. 4.7046
x = 78°

then angle ACD = 180-78-24 = 78°

I have a feeling there is an easier way, but once your mind locks into a method of solution ......

I just notice two typos, but they don't affect the solution.

in the bracketed angle , it should say (228-x)

let angle ADC = x
then angle BDC = 360-x-132 = 228-x

in triangle ADC, by the sine law,
DC/sin24 = AC/sinx
DC = AC sin24/sinx

in triangle BCD,
DC/sin12 = BC/sin(228-x)

it's 78

nju≈fgygkchfvfdc

96°

78°