Finding an unknown angle

Question

Geometry: Auxiliary Lines

As shown in the figure:

Answer 1

This is not the desired method, but it does show that $50^\circ$ is the result.

Call the fourth point D, and suppose A is at $(0,0)$ and C is at $(1,0)$. Then B is at $(x_B,y_B)=\left(\frac{\tan(70^\circ)}{\tan(58^\circ)+\tan(70^\circ)},\frac{\tan(58^\circ)\tan(70^\circ)}{\tan(58^\circ)+\tan(70^\circ)}\right) \approx (0.631921861,1.011286374)$ while D is at $(x_D,y_D)=\left(\frac{\tan(62^\circ)}{\tan(12^\circ)+\tan(62^\circ)},\frac{\tan(12^\circ)\tan(62^\circ)}{\tan(12^\circ)+\tan(62^\circ)}\right) \approx (0.898457801,0.190973101)$.

Then the length of AD is $a=\sqrt{x_D^2+y_D^2} \approx 0.918529883$ and of BD is $b = \sqrt{(x_D-x_B)^2+(y_D-y_C)^2} \approx 0.862528419$. Using the sine rule, the angle ABD is $$\sin^{-1}\left(\frac{a}{b}\sin(46^\circ)\right) = 50^\circ$$ at least to the precision of my calculations.

For what it is worth, an accurate diagram looks like

Answer 2

Let central point is $O$.

$\angle OBC = 180^\circ - (46^\circ+12^\circ) - (62^\circ+8^\circ) - X = 52^\circ-X$.

Using the sine formula, we have: $$\frac{AO}{\sin 62^\circ} = \frac{CO}{\sin 12^\circ},$$ $$\frac{AO}{\sin X} = \frac{BO}{\sin 46^\circ},$$ $$\frac{CO}{\sin (52^\circ-X)} = \frac{BO}{\sin 8^\circ},$$ so, $$\sin 12^\circ \cdot \sin 8^\circ \cdot \sin X = \sin 62^\circ \cdot \sin 46^\circ \cdot \sin (52^\circ-X).$$

(Here we can find that $X=50^\circ$. Uniqueness of solution is discussed in comments).

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In fact we need to prove that $$\sin 8^\circ \cdot \sin 12^\circ \cdot \sin 50^\circ =^{???} \sin 2^\circ \cdot \sin 46^\circ \cdot \sin 62^\circ.$$ Using formulas $$2\cdot \sin\alpha \cdot \sin\beta = \cos(\alpha-\beta) - \cos(\alpha+\beta),$$ $$2\cdot \sin\alpha \cdot \cos\beta = \sin(\alpha-\beta) + \sin(\alpha+\beta),$$ $$\sin(180^\circ-\varphi) = \sin(\varphi), \quad \cos(180^\circ-\varphi) = -\cos(\varphi),$$

we have $$\sin 8^\circ \cdot (\cos 38^\circ - \cos 62^\circ) =^{???} \sin 2^\circ \cdot (\cos 16^\circ + \cos 72^\circ),$$ $$-\sin 30^\circ + \sin 46^\circ + \sin 54^\circ - \sin 70^\circ =^{???} -\sin 14^\circ + \sin 18^\circ - \sin 70^\circ + \sin 74^\circ,$$ $$-\sin 30^\circ + \sin 46^\circ + \sin 54^\circ + \sin 14^\circ - \sin 18^\circ - \sin 74^\circ =^{???} 0,$$ $$(\sin 14^\circ + \sin 46^\circ - \sin 74^\circ) + ( - \sin 18^\circ + \sin 54^\circ -\frac{1}{2} )=^{???} 0,$$ $$(\sin 14^\circ + \sin 134^\circ + \sin 254^\circ) + \frac{1}{2}( - \sin 18^\circ + \sin 54^\circ +\sin 126^\circ + \sin 198^\circ + \sin270^\circ )=^{???} 0,$$ it is equality, because left side is equal to $$\mathrm{Im} \left( p(\omega_3^0 + \omega_3^1 + \omega_3^2) + \frac{q}{2}( \omega_5^0 + \omega_5^1 +\omega_5^2 + \omega_5^3 + \omega_5^4) \right)=$$ $$\mathrm{Im} \left( p \cdot \frac{\omega_3^3-1}{\omega_3-1} + \frac{q}{2}\cdot \frac{\omega_5^5-1}{\omega_5-1} \right)= \mathrm{Im} \left( p \cdot \frac{1-1}{\omega_3-1} + \frac{q}{2}\cdot \frac{1-1}{\omega_5-1} \right)= \mathrm{Im} \left( p \cdot 0 + \frac{q}{2} \cdot 0 \right)= 0,$$ where $p = \exp(i\pi 7/90)$, $q = \exp(-i\pi /10)$, $w_{...}$ $-$ roots of unity:

$w_3 = \exp(i 2 \pi / 3) = \cos 2\pi/3 + i \sin 2\pi/3 = \cos 120^\circ + i \sin 120^\circ$,

$w_5 = \exp(i 2 \pi / 5) = \cos 2\pi/5 + i \sin 2\pi/5 = \cos 72^\circ + i \sin 72^\circ$.

Answer 3

Edit: Typically, we'll use the fact that the sum of the angles in any triangle is $180^\circ$, and the sum of the angles about the middle vertex is $360^\circ$.

In this case, though, that isn't enough to get the answer on its own.

Second Edit: Henry's comment (specifically, the phrase "given $A$ and $B$") gave me the idea that should get us (at least most of the way) there. Let's call the unlabeled vertex $M$. By $AB$, I denote the length of the segment from $A$ to $B$, and similar for $BC$, $AC$, $AM$, $BM$, $CM$.

Note that $\angle ABC$ has measure $52^\circ$ by the $180^\circ$ rule. Thus, by Law of Sines applied to triangle $ABC$, $$\frac{AB}{\sin 70^\circ}=\frac{AC}{\sin 52^\circ},$$ so $$AC=\frac{\sin 52^\circ}{\sin 70^\circ}AB.$$ Applying Law of Sines to triangle $ACM$ gives us $$\frac{AM}{\sin 62^\circ}=\frac{AC}{\sin 106^\circ},$$ so $$AM=\frac{\sin 62^\circ}{\sin 106^\circ}AC=\frac{\sin 62\sin 52}{\sin 106^\circ\sin 70}AB.$$

Now, by the $360^\circ$ rule, $\angle AMB$ has measure $134^\circ-X$, so applying the Law of Sines (and the previous work) to triangle $ABM$ gives us $$\frac{AB}{\sin(134^\circ-X)}=\frac{AM}{\sin X}=\frac{\sin 62^\circ\sin 52^\circ}{\sin X\sin 106^\circ\sin 70^\circ}AB,$$ and so $$\sin X\sin 106^\circ\sin 70^\circ=\sin(134^\circ-X)\sin 62^\circ\sin 52^\circ.$$ Now, observe that $\sin\theta=\sin(180^\circ-\theta)$ for any angle $\theta$. This allows us to rewrite the above as $$\sin X\sin 74^\circ\sin 70^\circ=\sin(X-46^\circ)\sin 62^\circ\sin 52^\circ.$$ Having just looked at the clock, I'm going to get some sleep. I'll think on it more later today, and see if I can figure out how to get the rest of the way from here.