Given a triangle $\triangle ABC$, with an internal point $K$, find $\angle AKC$.

Question

As title suggests, the question is to find the measure of $\angle AKC$ in the triangle $\triangle ABC$, with an internal point $K$ and some given angles.

This is a pretty challenging problem, I'm going to post my solution below as an answer, but I'm not sure if my answer is correct and if my method is correct. Please share your own answers and approaches. I want to see if my answer is correct and if there are any other ways to arrive at an answer.

Answer 1

We have $\angle ABC = 66^\circ$ and $\angle AKC = 150^\circ$. Assume $BC = 1$, then $AC = \frac{\sin 48^\circ}{\sin 66^\circ} = \frac{\sin 48^\circ}{\cos 24^\circ} = 2 \sin 24^\circ$ and $KC = \frac{\sin 12^\circ}{\sin 150^\circ} = 2\sin 12^\circ$.

Now $$AK^2 = AC^2 + KC^2 - 2AC \cdot KC \cdot \cos \angle ACK = 4\sin^2 24^\circ + 4\sin^2 12^\circ - 8 \sin 12^\circ \sin 24^\circ \cos 48^\circ.$$

Note that $\sin^2 12^\circ = \frac{(1 - \cos 24^\circ)}{2}$ and $\sin 12^\circ \cos 48^\circ = \frac{(\sin 60^\circ - \sin 36^\circ)}{2}$. Now everything is in $24^\circ$ and $36^\circ$ and the values can be calculated. The result is $\frac{(3 - \sqrt{5})}{2}$. So $AK = \frac{(\sqrt{5} - 1)}{2}$.

$$\cos X = \frac{AK^2 + KC^2 - AC^2}{2AK \cdot KC}$$

Calculate again and get $\cos X = -\frac{\sqrt{7 - \sqrt{5} - \sqrt{30 - 6\sqrt{5}}}}{4}$. Compare this with a cos table and get $X = 102^\circ$.

Answer 2

This is my approach:

1.) Using $BC$ as the base, construct an equilateral triangle $\triangle BDC$ such that point $D$ lies outside of $\triangle ABC$, therefore $AB=BC=BD=CD$. Notice that $\angle BKC=150^\circ$. This means that, according to the inscribed angle theorem, $DK=BD=CD=BC$ (can be proven easily by extending the equilateral triangle below and then using the properties of cyclic quadrilaterals).

2.) Above implies that, also according to inscribed angle theorem, that $\angle KDC=24^\circ$ and that $\angle KDB=36^\circ$. Notice that $\angle ABD=108^\circ$ as well as $\angle KDB=36^\circ$, and $AB=BD$, this directly implies that $AKD$ is a straight line (this is intuitive enough but it can also be proven easily via contradiction). Therefore, $\triangle ABD$ is an isosceles triangle, so $\angle BAD=36^\circ$ and $\angle KAC=30^\circ$, therefore $X=\angle AKC=102^\circ$

Answer 3

Your answer is correct.

Non-synthetic, trigonometric way. By three times law of sines in the figure we can have $$\sin18^{\circ}\sin36^{\circ}\sin(132^{\circ}-X)=\sin12^{\circ}\sin48^{\circ}\sin(X-66^\circ).$$ By means of the observation $\sin12^{\circ}\sin48^{\circ}=\sin18^{\circ}\sin30^{\circ}$, we can see that the solution is $X=102^{\circ}$.

Answer 4

Here’s the Trig Ceva method, as you requested. Let $\theta=\angle CAK$. By a simple angle chase, $\angle BAC = 66$, so that $\angle BAK=66-\theta$. Then by Trig Ceva:

$$\frac{\sin(66-\theta)}{\sin\theta}\frac{\sin 48}{\sin18}\frac{\sin 12}{\sin36}=1.$$ $$\frac{\sin(66-\theta)}{\sin\theta}=\frac{\sin18\sin36}{\sin48\sin12}$$ As Bob Dobbs notes, this reduces to $$\frac{\sin(66-\theta)}{\sin\theta}=\frac{\sin 36}{\sin 30}$$ From which it is extremely obvious that $\theta=30$, so that $\angle AKC = 102$, as wanted.