Angle Chasing, huh?

Yes, you can totally angle chase this problem. First, recognize that the large triangle is an isosceles triangle, and then draw in the lines as shown in graphic. The rest is easy angle chasing, especially after identifying a couple more isosceles triangles.

I used the sine rule a couple of times to determine that the right-most of the inner triangles was isosceles. Given that the outer triangle was also isosceles it was plain sailing with the sum of angles in a triangle

I will use the law of sines on three triangles to get to the solution.

Identity I: $\frac{sin(20°)}{\overline{CM}} = \frac{sin(100° - \theta)}{\overline{BM}}$ on triangle BCM, which can be written as:

$\overline{BM} = \overline{CM}*\frac{sin(100° - \theta)}{sin(20°)}$

Identity II: $\frac{sin(10°)}{\overline{CM}} = \frac{sin(\theta)}{\overline{AM}}$ on triangle ACM, which can be written as:

$\frac{1}{\overline{AM}} = \frac{sin(\theta)}{sin(10°)*\overline{CM}}$

Identity III: $\frac{sin(30°)}{\overline{BM}} = \frac{sin(20°)}{\overline{AM}}$ on triangle ABM, which can be written as:

$\frac{\overline{BM}}{\overline{AM}} = \frac{sin(30°)}{sin(20°)}$

Now, if we multiply the first two identities, we get: $\frac{\overline{BM}}{\overline{AM}} = \frac{sin(10°)*sin(100° - \theta)}{sin(20°)*sin(\theta)}$

From here, we derive that: $\frac{sin(30°)}{sin(20°)} = \frac{sin(10°)*sin(100° - \theta)}{sin(20°)*sin(\theta)}$ , which can be simplified to: $sin(30°)*sin(\theta) = sin(10°)*sin(100° - \theta)$ .

Now, $sin(100° - \theta) = sin(90° + 10° - \theta)$ , which also can be written as $cos(10° - \theta)$ . Plugging it back into the previous equality:

$sin(30°)*sin(\theta) = sin(10°)*cos(10° - \theta)$ .

The left hand side can be broken into the sum of two sines divided by two:

$sin(10°)*cos(10° - \theta) = \frac{1}{2}*(sin(20° - \theta) + sin(\theta))$

And since we also know that $sin(30°) = \frac{1}{2}$ , we can rewrite the equality again yielding:

$\frac{sin(\theta)}{2} = \frac{sin(20° - \theta) + sin(\theta)}{2}$

Therefore, $\frac{sin(20° - \theta)}{2} = 0 \rightarrow sin(20° - \theta) = 0$

This would potentially yield an infinitude of solutions, yet since we know that theta is an interior angle of a triangle (therefore $0 < \theta < 180$ ) as well as it is an angle no greater than $100°$ , we see that the only possible solution is $\theta = 20°$ .

In the above question let the required angle be $\theta$ , then using Trigonometric form of Ceva's theorem, we have

( $sin \theta$ )( $sin 30^\circ$ )( $sin 20^\circ$ )=( $sin100^\circ-\theta$ )( $sin10^\circ$ )( $sin20^\circ$ )

This amounts to ( $sin \theta$ )=2( $sin 100^\circ-\theta$ )( $sin 10^\circ$ )

which upon simplifying using 2( $sinA$ )( $sinB$ )=( $cos A-B$ ) - ( $cosA+B$ ) gives

$cos(110^\circ-\theta$ )=0

$\implies \theta=20^\circ$

And yes, it is equivalent to using the sine rule many times.

The whole triangle has angles 40, 40, 100 so is isosceles. Let the two equal sides be x

Use the cosine rule to get the third side to be x multiplied by root(2-2cos100) = 1.532x

Use the sine rule to get the length from the point inside the main triangle to the right corner to be 1.532x times sin(30)/ sin(130) = x

Now this makes the internal triangle on the right isosceles with apex angle 20, so base angles both 80

Since the main triangle has an angle of 100 at the top, the red part must be 100-80 = 20

Call the centre O,so OA/OB=sin(20°)/sin(30°) sine theorem for triangle ABO,

Also from sine theorem for triangles AOC and BCO we have

put x=<OCA,then <BCO=180°-(20°+20°+30°+10°+x)=100°-x

so AO/OC=sin(x)/sin(10°) , OB/OC=sin(100°-x)/sin(20°)

so AO/OB=(sin(x)/sin(10°))/(sin(100°-x)/sin(20°))=sin(20°)/(0.5)

so 0.5 sin(x)sin(20°)=sin(10°)sin(20°) sin(100°-x)

so 0.5sin(x)=sin(10°)*sin(100°-x)

sin(x)=2sin(10°) (cos(10°) cos(x)+sin(10°)sin(x))

sin(x) cos(20°)=sin(20°) cos(x)

tg(x)=tg(20°)

so x=20° or x=200° but because x<100°

we choose x=20°

angle {bca}={cab}=40° so side cb=CA=p. put side AB=u ,,cos(40)=p^2+u^2-p^2÷(2pu.cos40) so u=2cos40.p. I used a graphical solution by assuming side p & draw the triangle 3 times with 3 different values of side p and every time by using angular scaler the unknown angle equal to 20 degree .