ABC is an isosceles triangle with AB=AC, ∠BAC=96∘. D is a point such that ∠DCA=48∘, AD=BC and angle DAC is obtuse. What is the measure (in degrees) of ∠DAC?
I placed D outside the triangle top-right of A, joined CD and AD
In triangle ABC , it is easy to see that angle ACB = 42°
and if angle DCA= 48° , then angle BCD = 90°
(this helps to make a reasonable sketch)
Suppose we let AC = 5 (anything will do)
then by the sine law:
BC/sin96 = 5/sin42
BC = 7.43145 (I stored that in my calculator)
now switch to triangle ACD
sinD/AC = sin48/AD , but AD = BC = 7.431...
and AC = 5
SinD/5 = sin48/7.431..
sinD = .5
D = 30° , ahhh!!!
then angle DAC = 180+48+30 = 102°
To find the measure of angle ∠DAC, we can use the given information about the isosceles triangle ABC and the point D.
We are given that ABC is an isosceles triangle with AB = AC. Since AB = AC, angles ∠ABC and ∠ACB are also equal. Let's denote this common angle as x, so ∠ABC = ∠ACB = x.
Since ∠BAC = 96°, we know that ∠ABC + ∠ACB + ∠BAC = 180° (angles in a triangle sum to 180 degrees). Plugging in the values we have, we get:
x + x + 96° = 180°
2x + 96° = 180°
2x = 180° - 96°
2x = 84°
x = 42°
So, the measure of angles ∠ABC and ∠ACB is 42° each.
Now, let's focus on point D. We are given that ∠DCA = 48° and AD = BC. Since ABC is an isosceles triangle, BC = AC. Let's denote the measure of ∠DAC as y.
Since ∠DCA = 48° and ∠ACB = 42°, we can subtract these angles from ∠DAC to get:
∠DAC = ∠DCA - ∠ACB
∠DAC = 48° - 42°
∠DAC = 6°
So, the measure of ∠DAC is 6 degrees.