Not Enough Angles

Consider adding point E inside ABC, where triangle EBC is an equilateral triangle. Note $EC=BC=AD$ , $\angle ECA=\angle DAC = 20^\circ$ . Thus, we have $\triangle ECA \equiv \triangle DAC$

Due to symmetry, E lies on the angle bisector of $\angle A$ . Thus $\angle EAC=10^\circ$ .

Calculating, $\angle ADC=\angle CEA=180^\circ-20^\circ-10^\circ=150^\circ$ .

Construct a point $E$ such that it is on the opposite side of $AC$ as $B$ and that triangle $BCE$ is congruent to triangle $ADC$ . Therefore, we were trying to find angle $BCE$ . After connecting $EA$ , we see that angle $ABE = 80-20 = 60$ degrees. Since $AB = AE = BE = AC$ , triangle $ABE$ is an equilateral triangle. Thus angle $CAE = 60-20 = 40$ degrees. Since triangle $CAE$ is an isosceles triangle, angle $ACE = 70$ degrees. So angle $BCE = 80+70 = 150$ degrees.

$\angle A + \angle B + \angle C = \pi \Rightarrow \angle C = 80$

Triangle ABC is isosceles, so AB = AC = a

Say AD = x $\Rightarrow$ DB = a - x

Let $\Theta$ be the angle $\angle$ ADC

Take the triangle BDC: $\angle$ BDC = $\pi - \Theta$

Law of Sines on triangle BDC

$\frac{x}{\sin\theta} = \frac{a-x}{\sin(\theta - 80)} = \frac{a}{\sin\theta + sin(\theta - 80)} \Rightarrow \frac{a}{x} = \frac{\sin\theta + sin(\theta - 80)}{sin\theta}$

Using the addition formulas for trigonometry we have:

$\frac{a}{x} = \frac{sin\theta + sin\theta cos80 -sin80cos\theta}{sin\theta} = 1 + cos80 -sin80cotg\theta$

Law of Sines on triangle ADC

$\frac{a}{sin80} = \frac{x}{sin20} \Rightarrow \frac{a}{x} = \frac{sin80}{sin20}$

$\Rightarrow \frac{sin80}{sin20} = 1 + cos80 -sin80cotg\theta$

Isolating $cotg\theta$ , we have:

$cotg\theta = \frac{cos80sin20 + sin20 - sin80}{sin80sin20}$

Using the formula $sinp -sinq = 2cos(\frac{p+q}{2}) sin(\frac{p-q}{2})$

$cotg\theta = \frac{-2cos50 sin30 + cos80sin20}{sin80sin20} = \frac{cos80 sin20 - sin40}{sin80sin20}$

Using the fact that $sin2\alpha = 2sin\alpha cos\alpha$

$cotg\theta = \frac{cos80 -2cos20}{sin80} = \frac{(cos80 - cos20) - cos20}{cos10}$

Now we use the formula: $cosp - cosq = -2sin(\frac{p+q}{2}) cos(\frac{p-q}{2})$

$cotg\theta = \frac{-2sin30sin50 - cos20}{cos10} = \frac{-(cos40 + cos 20)}{cos10}$

And finally we use the formula: $cosp + cosq = 2cos(\frac{p+q}{2})cos(\frac{p-q}{2})$

$cotg\theta = -\frac{2cos30cos10}{cos10} = -\sqrt{3}$

$\Rightarrow \theta = 150°$ , as $0° \leqslant \theta \leqslant 180°$

Excuse me for eventual english mistakes.

It is clear that ABC is isoceles, with AB = AC.

Construct triangle DAE on AD such that triangle DAE is congruent to triangle BAC. Then AE = AC, and angle DAE = 80 degees. It follows that angle EAC = 80 - 20 = 60 degrees, and that angle ACE = angle AEC = (180 - 60)/2 = 60 degrees.

Thus, triangle EAC is equilateral and ED = EC.

Angle DEC = angle AEC - angle AED = 60 - 20 = 40 degrees. Since ED = EC, angle EDC = angle ECD = (180 - 40)/2 = 70 degrees.

Therefore, angle ADC = angle EDC + angle ADE = 80 + 70 = 150 degrees.

WLOG we assume $BC=1$ . Applying sine law on $\triangle ABC$ we get $AB=\frac{\sin 80^{\circ}}{\sin 20^{\circ}}$ . Construct point $D'$ on $AB$ such that $\angle AD'C=150^{\circ}$ . So $\angle BD'C=30^{\circ},\angle D'CB=\angle 70^{\circ}$ . Applying sine law on $\triangle BCD'$ we get $BD'=2\sin 70^{\circ}$ . Now $\sin 40^{\circ}+\sin 20^{\circ}=\sin (30+10)^{\circ}+\sin (30-10)^{\circ}$ $=2\sin 30^{\circ}\cos 10^{\circ}=\sin 80^{\circ}$ $\Rightarrow \frac{\sin 80^{\circ}-\sin 40^{\circ}}{\sin 20^{\circ}}=1$ $\Rightarrow \frac{\sin 80^{\circ}-2\sin 20^{\circ}\cos 20^{\circ}}{\sin 20}=1$ $\Rightarrow \frac{\sin 80^{\circ}}{\sin 20^{\circ}}-2\sin 70^{\circ}=1$ $\Rightarrow AB-BD'=BC$ which implies $D=D'$ . So $\angle ADC=150^{\circ}$ .

ANGLE C =80 Let angle ADC be x then, In triangle BDC, sin(180-x)/BC = sin80/CD 1/CD = sinx / {BC * sin80} .....i In triangle ADC, sin20/CD = sin(160-x)/AD sin20 *sinx / { BC *sin80} =sin(160-x)/AD SINCE BC=AD, sin20/sin80 = sin(160-x)/sinx sin20/sin80 = {sin160 * cosx -cos160 *sinx} /sinx sin20/sin80 = sin160 * cotx - cos160 sin20/sin80 +cos160 = sin 160 *cotx cot x= -1.73205 We also know that 80<x<180 So, x=150

Clearly, the triangle is isosceles. Let $l$ be the perpendicular bisector of $AC$ . E is the point that is symmetric to $C$ , over the line $l$ . $\triangle ADE$ is equilateral because it has an angle of $60^\circ$ and 2 equal sides around it. Because both $\triangle ADE$ and $\triangle ACE$ are at least isosceles, and share the same base, their altitudes are collinear. That means that $D$ lies on the altitude from $C$ , and $\angle ADC = 180^\circ - \angle ACD - \angle CAD = 180^\circ - 10^\circ - 20^\circ = 150^\circ$

in triangle ABC the third angle C is also equal to 80 degrees hence the triangle is isosceles

and thus AB=AC=x

hence $BC^2$ = 2 $x^2$ - 2 $x^2$ (cos20) (by cosine rule)

$BC^2$ = 2 $x^2$ (1-cos20)

and BC is the square root of that quantity on the right side

since AD=BC

AD= $x \times(\sqrt{2[1-cos20]}$ )

$DC^2$ = $AD^2$ + $x^2$ -2 $x\times AD$ (cos20)

after plugging in the values for AD,pulling $x^2$ common from the expression and then taking the square root. we find that

DC=0.684 x

using sine rule in triangle ADC

let angle ADC be equal to K

$\frac{DC}{sin20}$ = $\frac{x}{sin K}$

$\frac{0.684 x}{sin20}$ = $\frac{x}{sin K}$

after simplification we arrive at

K=arcsin (0.5)

now there are 2 cases

Case 1

K=30

if this were true

in triangle ADC the sum of all angles must be 180

20+30 + angleACD = 180

angleACD = 130

but we know that angle C which is greater than angle ACD is 80

hence angle ACD cannot be 130 and thus case 1 is not true

case2

K = 180-30 [sin K = sin (180-K)]

since case 1 is not true case 2 must be true and can easily be verified to be true.

and therefore K=150

We take a point E such that AE = EC. So, $\angle CAE = \angle ACE = 20 ^ \circ$ .

Now, we draw a parallel to AC by D. This line cuts to EC in F. So, $\angle CAE = \angle FDE = 20 ^ \circ$ .

DF // AC, then AD = CF, and CF = BC... So, BCF is an isosceles triangle. But, $\angle BCF = \angle ACB - \angle ACE = 80 ^ \circ - 20 ^ \circ = 60 ^ \circ$ . Then, BF = CF = BC = AD.

$\angle ABF = \angle ABC - \angle CBF = 80 ^ \circ - 60 ^ \circ = 20 ^ \circ$ . Then BFD is an isosceles triangle because $\angle ABF = \angle FDB = 20 ^ \circ$ and BF = DF = CF = BC = AD.

ACE is an isosceles triangle and AC // DF, then DEF is an isosceles triangle. $\angle DFE = \angle EDF = 20 ^ \circ$ . But $\angle DFE$ is the exterior angle of the triangle DFC, (is an isosceles triangle) then $\angle FDC = \angle FCD = 10 ^ \circ$ .

$\angle ACD = \angle ACE - \angle DCF = 20 ^ \circ - 10 ^ \circ = 10 ^ \circ$

$\angle ADC = 180 ^\circ - \angle ACD - \angle CAD = 180 ^ \circ - 10 ^ \circ - 20 ^ \circ = 150 ^ \circ$

AC=CE=AE=ED

Given that ∠A=20, ∠B=80, using that ∠A+∠B+∠C=180, when substituted in ∠C=80 as well. Now that ∠B=∠C. I can assume that Triangle ABC is an isosceles triangle.

Given point D on line segment AB, where AD=BC. I will show that ∠ACD=10.

By the Law of Sines in Triangle ACD, AD/sin(10)=CD/sin(20).

By the Double Angle Formula, sin(20)=2 sin(10) cos(10), so that CD=2 AD cos(10).

In triangle BCD, BC/sin(30)=CD/sin(80)=CD/cos(10), so that CD=2 BC cos(10).

Implying AD=BC and D=D. Thus, ∠ACD=10 and ∠ADC=180-20-10=150.

Initially, there doesn't seem to be much that we can work with. Construct the point $E$ (on the same side of $BC$ as $A$ ) such that triangles $ADC$ and $BCE$ are congruent. Since $\angle ABC = \angle ACB = 80^\circ$ , $ABC$ is an isosceles triangle and $AB = AC =BE$ . Since $AB=BE$ and $\angle ABE = \angle ABC - \angle EBC = 80^\circ - 20^\circ = 60^\circ$ , it follows that $ABE$ is an equilateral triangle. As such, $AE = AB = AC$ , so $\angle ACE = 70^\circ$ . Thus $\angle ADC = \angle BCE = \angle BCA + \angle ACE = 80^\circ + 70^\circ = 150 ^\circ$ .

Note: This solution follows by considering the 18-gon. If $B$ and $C$ are consecutive vertices, then $E$ is one removed from $C$ .

$WLOG ~let ~AB=AC=1. ~~Since~\Delta \text{ ABC is isosceles. } ~~\\ Given~AD=BC=2*Sin\frac 1 2 *20=2*sin10. \\ Consider ~ \Delta ~ ADC:-\\ Applying~ Cos ~ Rule ~ ~ DC^2=AD^2+AC^2 - 2*AD*AC*Cos20.\\ Applying~ Sin ~ Rule ~ \angle ADC=Sin^{-1}(AC*\dfrac{Sin20}{DC})\\ =Sin^{-1}\left ( 1* \dfrac{Sin20}{\sqrt{(2*Sin10)^2+1^2 - 2*(2*Sin10)*1*Cos20})} \right )=30.014.\\ Range~ of ~Sin^{-1}\alpha~ is~ ~ -90<\alpha<90, ~answer~ is~ 30~ or ~ 150.~ We~ know~ the~ angle ~ is ~ obtuse.\\ \therefore ~ \angle ADC=150^o.\\~~\\$
In fact I have used TI-83 Plus and split my calculations as follows and got the answer as 30.
$X=2Sin10, .........Y=X^2+1^2 - 2*X*1*Cos20, ....\angle ~ ADC=Sin^{-1}\dfrac {Sin20}{\sqrt Y }=30$ .

In this triangle the three angles are 20,80,80 degrees.Thus it is an isosceles triangle.Now in this triangle a line is drawn parallel to BC which intersects AC at E. Now triangles ADE and ABC are similar to each other by AAA Similarity rule.-(1) The required angle(ADC) is divided by this parallel line into two angles namely angle(ADE) and angle(EDC). From (1), angle(ADE)=80 degrees. Similarly,we can also find that angle(EDC)=70 degrees. Therefore,angle(ADC)=angle(ADE)+angle(EDC). =80+70=150 degrees.