10-20-30-40 What Comes Next

Note that $\triangle{ABC}$ is isosceles, since $\angle{CAB}$ and $\angle{BCA}$ are both $40^\circ$ . Drop altitude $CE$ from $C$ to $AB$ . Reflect point $D$ over altitude $CE$ (to points $F$ ). Since $\triangle{ACE}$ is a right triangle, it follows that $\angle{DCE}=30$ , which means that $\triangle{DCF}$ is equilateral since the triangle is also isosceles. Note that $\angle{CFB}=180^\circ-20^\circ-10^\circ=150^\circ$ , which implies that $\angle{DFB}=360^\circ-150^\circ-60^\circ=150^\circ$ , from which it follows that $\triangle{DFB} \cong \triangle{CFB}$ by SAS. Thus, $\angle{BDC}=\angle{BDF}+\angle{CDF}=60^\circ+20^\circ=\boxed{80^\circ}$ .

Notice that angle DCB is 80 degrees, since the sum of the angles in ABC must be 180. Thus, by the sine Ceva's, sin 30 * sin 20 * sin(100-x) = sin 10 * sin 80 * sin (x - 60), where x is the desired angle. Since sin 30 = 1/2, and sin 20 = 2 sin10 sin80, one notes that sin (100 - x) = sin (x - 60), which implies 100-x = x - 60 (since they sum to 160 not 180), from which one obtains x = 80.

First, notice that $\angle DAC + \angle DAB = \angle BAC = \angle ABC$ . Hence, ABC is an isosceles triangle in C, and $\angle ACB = 100 ^ \circ$ . Furthermore, from the hypothesis we can calculate that $\angle BCD = \angle ACB - \angle ACD = 80 ^ \circ$ .

Now, we construct a point E on AD such that $\angle EBA = \angle DAB = 30 ^ \circ$ . This implies that the triangle EBA is also isosceles in E.

From the above assumption, it follows that:

_ $CE \perp BC \rightarrow \angle DCE = \frac{1}{2} \angle BCA - \angle DCA = 30 ^ \circ$ . On the other hand, $\angle CDE = \angle angle DCA + \angle DAC = 30 ^ \circ = \angle DCE$ , which proves that CDE is an isosceles triangles in E

_ $\angle CBE = \angle CBA - \angle EBA = 10^ \circ \rightarrow \angle CBE + \angle BCD = 90 ^ \circ$ . Hence, $BE \perp CD$ . In addition, the triangle CDE is isosceles in E, so BE is the perpendicular bisector of CD.

From the above two results, it follows that CDB is an isosceles triangle in B, which gives the answer $\angle BDC = \angle BCD = 80 ^ \circ$ .

First, note that CÂB=DÂC+DÂB=40º=(angle) (CBA), where AC=AB. Second, (angle) (ACB)=180º-40º-40º=100º, where (angle) (DCB)=100º-20º=80º and still (angle) (ADC)=180º-20º-10º=150º.

Assume, without lost generality, that AC=AB=1.

For the sine law in the triangle ACD, we have

AC/sin(150º)=CD/sin (10º), where CD=2*sin(10º) because AC=1, sin(150º)=1/2.

Let E be the foot perpendicular to CD by B.

We have, in the right triangle CEB, (angle) (EBC)=90º-80º=10º and
EC=1*sin (10º)=sin (10º), where ED=CD-EC=sin (10º).

Thus, BE is average and height about the side CD of the triangle ACD.

So we conclude that the triangle CBD is isosceles with base CD, i. e., (angle)(CDB)=(angle)( DCB)=80º.

Observation. Here, (angle)(XAY)=XÂY.

Still, we use many times the sum of the interior angles of a triangle is 180º .

Notice that angle $DCB=80^\circ$ . Let $AC=BC=1$ and angle $BDC=x$ . Therefore, angle $DBC=100^\circ-x$ .

By the law of sines on triangle $ADC$ , $\sin 150^\circ /1=\sin 10^\circ /DC$ . It follows that $DC=2\sin 10^\circ$ . By the law of sines on triangle $DBC$ , $\sin x/1=\sin (100^\circ -x) / (2\sin 10^\circ )$ .

After cross multiplying and the sine subtraction formula, $2\sin10^\circ \sin x=\sin 100^\circ \cos x- \cos 100^\circ \sin x$ . By the co-function identities, the right hand side becomes $\cos 10 \cos x+\sin 10 \sin x$ , which simplifies to $\tan x=\cot 10^\circ = \tan 80^\circ$ , so our final answer is 80.

[Latex edits]

Extend $CD$ to intersect $AB$ at $E$ , and draw the altitude $CO$ from $C$ to $AB$ . Note that $\triangle ABC$ is isosceles, because $\angle CAB = \angle DAC + \angle DAB = 10^\circ + 30^\circ = 40^\circ = \angle ABC$ ; hence $\angle DCB = 180^\circ - 2(40^\circ) - 20^\circ = 80^\circ$ , and $\angle OCE = \frac{1}{2} \angle DCB - \angle DCA = 50^\circ - 20^\circ = 30^\circ$ . Furthermore, $\triangle AED$ is isosceles, since $\angle ADE = \angle DAC + \angle DCA = 30^\circ = \angle DAE$ . Now suppose $AC = BC = 1$ . Then $OC = \sin 40^\circ$ , and since $\triangle EOC$ is a 30-60-90 right triangle, we have $CE = \frac{2}{\sqrt{3}} \sin 40^\circ.$ Also, by the Law of Sines, $\frac{AE}{\sin 20^\circ} = \frac{CE}{\sin 40^\circ} = \frac{2}{\sqrt{3}}$ , so $DE = AE = \frac{2}{\sqrt{3}} \sin 20^\circ$ . Consequently, $\begin{aligned} CD &= CE - DE = \frac{2}{\sqrt{3}}(\sin 40^\circ - \sin 20^\circ) \\ &= \frac{2}{\sqrt{3}} \cdot 2 \cos 30^\circ \sin 10^\circ \\ &= 2 \sin 10^\circ. \end{aligned}$ Now let $\angle BDC = \theta,$ so that $\angle DBC = 100^\circ - \theta$ . Then again by the Law of Sines, $\frac{\sin \theta}{BC} = \frac{\sin(100^\circ - \theta)}{CD}$ , or $\begin{aligned} 2 \sin 10^\circ \sin \theta &= \sin(100^\circ - \theta) = \cos(10^\circ - \theta) \\ &= \cos 10^\circ \cos \theta + \sin 10^\circ \sin \theta. \end{aligned}$ It follows that $\sin 10^\circ \sin \theta = \cos 10^\circ \cos \theta$ , or equivalently, $\cos(10^\circ + \theta) = 0$ , or $\theta = 80^\circ.$

We observe triangle ABC is isosceles with its base angles equaling to 40 each. Clearly, we can see angle DCB=80 (by angle sum property of a triangle) Now, we draw a line LM || AB through D.
We can obviously say that D can’t be the mid-point of LM , since then angle ACD will become equal to angle BCD (which is not so actually). Therefore, D lies somewhere else on the line-segment LM other than its mid-point. Now, we know that angle LAD = 10 < angle DAB = 30, we observe that as we place point D farther from the mid- pt. of LM, from L , i.e. when we consider LD > MD, then angle LAC becomes larger than angle EAB contradicting the given values. Therefore, we need to take, LD < MD . Now, we mark another point E on line ML such that ME=LD, and so angle ECM=20 and angle EBM = 10 as they then become symmetrical. We thus get, angle EBA = 30.Also, angle ADM=150(since angles on the same side of the transversal equals 180, so 180 – 30 = 150) .Also, ADC = 150(from triangle ACD). Hence, angle CDB = 360 – ( 150+150 ) =60. Similarly, angle CED = 60. Therefore, angle DCE = 60 ( angle ACD=angle BCE = 20 ). Hence, triangle DEC is equilateral. Now, in triangles CEB and DEB, we have,

EB = EB ( Common )

DF = FC ( triangle DEC is equilateral )

Angle DFB = Angle BFC = 150 Therefore, the two triangles are congruent. Hence, angle BDE = 20 ( corresponding parts of congruent triangles are equal) Hence, angle BDC = angle CDE + angle BDE = 60 + 20 = 80

A little thinking (and drawing) shows that ABC is a isosceles triangle whose angles are 40degree,40 degree,and 100 degree,.... . So draw it and put in D. Now, i am considering another point E symmetrically placed on the other side from D. The point E must be within the triangle BCD... So ∠ DCE become 60°. Since D and E are symmetrically positioned, CDE becomes equilateral.[like the triangle ACD ] Note that ∠CEB = 180 – 20 – 10 = 150°. So ∠BED will be 360 – 150 – 60 = 150°. Now join BD. It is easy to show that BED and BEC are congruent triangles. So BDC is isosceles and so ∠BDC = ∠BCD = 80°.

Let x= angle BDC. We notice that angle BAC =40 degrees, so the triangle ABC is isosceles with AC =AB =a. We also get that angle ACD=100 degrees and the angle CDA is 150 degrees. From triangle ADB and the law of sines we get: BD/sin30 =AB/sin(210-x). From triangle BDC and the law of sines we also get: a/sin x =BD/sin 80, so BD= a* sin80 / sin x (1) . Let K be the middle of AB. The altitude from C to AB passes from K. So, cos 40 = AK/AB = AK/a = sin50. Therefore BD/sin 30 = AB/sin(210-x) = 2 a sin50/sin(210-x) and since sin 30 =1/2 we get that BD=a* sin50/sin(210-x) (2). From equalities (1) and (2) we get: sin 50 * sin x = sin 80 * sin(210-x) or using trigonometric identities we can see that: sin x * cos 40 = sin(x-30) * cos 10. But sin x * cos 40 = ( sin(x+40) + sin(x-40))/2 and sin(x-30) * cos 10 = (sin(x-20) + sin(x-40))/2. In other words: sin(x + 40)=sin(x-20). Solving this equation for 0< x<180 we get x=80.

Let angle BDC be a. By sine rule,

BC/sin(a) = BD/sin(80) => BC/BD = sin(a)/sin(80).

AB/sin(210 - a) = BD/sin(30) => AB/BD = sin(210 - a)/sin(30)

BC/sin(40) = AB/sin(100) => BC/AB = sin(40)/sin(100)

BC/AB = sin(40)/sin(100) = [sin(a)sin(30)]/[sin(80)sin(210 - a)]

sin(40)/sin(100) = sin(40)/sin(80)

[sin(a)sin(30)]/[sin(80)sin(210 - a)] = [0.5sin(a)]/[sin(80)sin(a - 30)]

So, 2sin(40) = sin(a)/sin(a - 30)

Multiplying both sides by cos(a),

sin(80)cos(120 - a) = sin(a)cos(40).

Since there is only one solution, and a = 80 is clearly a solution. So angle BDC = 80.

We find that <DCB=80. Let X be <ABD. Then by trig Ceva, we have sin80/sin(40-x) sinx/sin30 sin10/sin20=1.

Notice that sin20=2sin10cos10 by the double angle formula. Then we have sin80/sin(40-x)*sinx/cos10=1. But sin80=cos10, so sin(40-x)=sin(x). Here it follows that x=20, and <BDC=80.

Firstly we construct vectors, corresponding to the edges of the triangle. As the angles do not depend on the linear sizes and orientation, we can scale and rotate the construct arbitrarily, say $\overline{AC} = (1,0)$ .

$\angle BAC = \angle DAB + \angle DAC = 40^\circ$ , and thus $\overline{AB} = k_1 (\cos{40^\circ}, \sin{40^\circ})$ . At the same time $\angle BCx = \angle CAB + \angle ABC = 80^\circ$ as an external angle of $\triangle ABC$ and $\overline{CB} = k_2 (\cos{80^\circ}, \sin{80^\circ})$ . $k_1$ and $k_2$ can be obtained from the fact that $\overline{AB}-\overline{CB}=\overline{AC}=(1, 0)$ , which can be written as simultaneous equations $\begin{cases} k_1 \cos{40^\circ} - k_2 \cos{80^\circ} = 1 \\ k_1 \sin{40^\circ} - k_2 \sin{80^\circ} = 0 \end{cases},$ and so $k_1=\frac{\sin{80^\circ}}{\sin{40^\circ}} = 2\cos{40^\circ}$ and $k_2=\frac{\sin{40^\circ}}{\sin{40^\circ}} = 1$ .

The same approach is applied to segments $AD$ and $CD$ , obtaining vectors $\overline{AD} = k_3 (\cos{10^\circ}, \sin{10^\circ})$ and $\overline{CD} = k_4 (\cos{160^\circ}, \sin{160^\circ})$ such that $\overline{AD}-\overline{CD}=\overline{AC}=(1, 0)$ . Thus scaling factors can again be found from simultaneous equations $\begin{cases} k_3 \cos{10^\circ} - k_4 \cos{160^\circ} = 1 \\ k_3 \sin{10^\circ} - k_4 \sin{160^\circ} = 0 \end{cases},$ giving $k_3=\frac{\sin{160^\circ}}{\sin{150^\circ}}$ and $k_4=\frac{\sin{10^\circ}}{\sin{150^\circ}}$ .

We now find $\overline{BD}=\overline{CD}-\overline{CB}=\frac{\sin{10^\circ}}{\sin{150^\circ}}(\cos{160^\circ}, \sin{160^\circ}) - (\cos{80^\circ}, \sin{80^\circ})$ . Its norm is $\begin{aligned} \left(\frac{\sin{10^\circ}}{\sin{150^\circ}}\cos{160^\circ}-\cos{80^\circ}\right)^2 + \left(\frac{\sin{10^\circ}}{\sin{150^\circ}}\cos{160^\circ}-\cos{80^\circ}\right)^2 = \\ \left(\frac{\sin{10^\circ}}{\sin{150^\circ}}\right)^2 + 1 - 2\frac{\sin{10^\circ}}{\sin{150^\circ}}(\cos{160^\circ}\cos{80^\circ}-\sin{160^\circ}\sin{80^\circ}) = \\ 4\sin^2{10^\circ}+1-4\sin^2{10^\circ}= & 1. \end{aligned}$

This means that $\triangle BDC$ is equilateral, and so $\angle{BDC}=\angle{BCD}=80^\circ$ .

Of course, more straightforward solution involves direct construction of the triangle $ABC$ and the point $D$ . The angle $\angle{BDC}$ can then be measured from the sketch.

The following angles are obtained easily: $\angle DCB = 80^\circ , \angle CDA = 150^\circ.$

Also, note that triangle $ABC$ is isosceles with $AC = BC$ . If we let $x = \angle BDC$ , we can use the sine law on triangles $ADC$ and $DBC$ as follows:

$\frac{AC}{\sin 150^\circ} = \frac{DC}{\sin 10^\circ}$ $\frac{DC}{\sin(100^\circ-x)} = \frac{BC}{\sin x} = \frac{AC}{\sin x}$

from which we get the following equation in $x$ :

$\frac{\sin 10^\circ}{\sin 150^\circ} = \frac{\sin(100^\circ-x)}{\sin x}$

Using the fact that $\sin (100^\circ -x) = \sin (80^\circ + x)$ , the sine of sum identity and the cofunction identity for sine and cosine, we obtain:

$2\sin x \cos 80^\circ = \sin 80^\circ \cos x + \sin x\cos 80^\circ$

or

$\tan x = \tan 80^\circ$ .

Since $x <180^\circ$ , it follows that $x = 80^\circ$ .

I used construction....