Yet another adventitious triangle

Question

The new training field of the Gryffindor Quidditch team is in the shape of a triangle, with small towers in its corners $A,B,C$. The angles at $A$, $B$, $C$ are respectively $50^{\circ}$, $50^{\circ}$, $80^{\circ}$. George Weasley is standing in a point $G$ somewhere in the middle of the field, and notes that $\angle GAB=30^{\circ}$ and that $\angle ABG=10^{\circ}$.

Question: How large is the angle $\angle BGC$?

(The answer to this question will be an integer. A good solution will clearly explain the reason why an integer number shows up here.)

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This puzzle continues and is closely related to the following puzzles: $*$ A triangular Quidditch field $*$ The adventitious 18-gon $*$ https://puzzling.stackexchange.com/questions/30174/another-adventitious-triangle-for-ivo-beckers

Answer 1

The angle is

$70^{\circ}$

For this, we will need the law of sines:

(Law of Sines) In any triangle $ABC$ it holds that

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=d_{ABC}$$

where $d_{ABC}$ is a constant, the diameter of $ABC$'s circumcenter.

I provide an auxiliary image, drawn to scale, to help us navigate the answer. We will use $a$ to denote the side of $ABC$ opposite $A$, and similarly for $b$ and $c$.

We have $\angle BAG = 30^{\circ}$ so $d_{ABG}=\frac{r}{\sin 30^{\circ}} = 2r$. Using the law of sines on $ABG$ once again, we find that $c = 2r \sin 140^{\circ} = 2r \sin 40^{\circ}$ ($40^{\circ}$ and $140^{\circ}$ are supplementary).

Now, the angle at $C$ is $80^{\circ}$, so $d_{ABC} = \frac{c}{\sin 80^{\circ}} = \frac{2r \sin 40^{\circ}}{\sin 80^{\circ}}$. By the double angle formula $\sin 2\theta = 2 \sin\theta\cos\theta$, it follows that $d_{ABC} = \frac{r}{\cos 40^{\circ}}$.

We use the law of sines once again on $ABC$. Since the angle at $A$ is $50^{\circ}$, we find that$$\frac{a}{\sin 50^{\circ}} = \frac{r}{\cos 40^{\circ}}$$

Since $40^{\circ}$ and $50^{\circ}$ are complementary, this implies $a = r$, so that $BCG$ is isoceles. Finally, $\angle ABG = 10^{\circ}$ and the angle at $B$ is $50^{\circ}$, so $\angle CBG = 40^{\circ}$ and thus $\angle BGC = 70^{\circ}$.

EDIT: In the spirit of the previous problems, perhaps I should mention that this too is a problem inscribed in an $18$-gon (or perhaps more appropriately, a $36$-gon, as f'' observed).

Answer 2

Brute-forcing with the law of sines works, but here's a different approach:

Let's make a new point $D$ such that $BCD$ is an equilateral triangle. Then draw the line $DA$, as in the picture:

$\angle ACD = 140^{\circ}$, so as $\triangle ACD$ is isosceles $\angle CAD = 20^{\circ} = \angle CAG$. Therefore $A$, $G$ and $D$ all lie in the same line.

$\angle GBD = 40^{\circ} + 60^{\circ} = 100^{\circ}$, and $\angle GDB = 60^{\circ} - 20^{\circ} = 40^{\circ}$, so $\angle DGB = 180^{\circ} - 100^{\circ} - 40^{\circ} = 40^{\circ}$. Therefore $\triangle BGD$ is an isosceles triangle, and $BG = BD = BC$, which implies $\triangle BCG$ is also an isosceles triangle.

From the fact that $\angle GBC = 40^{\circ}$, the base angles of the isosceles triangle $\triangle BCG$ each equal $70^{\circ}$.