euclidean-geometrygeometrytrianglestrigonometry
The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead anywhere. My actual approach, which I will post as an answer below, uses the law of Cosines. Please share your own approaches especially if they use a different method!
In the figure, $\Delta AED$ together point $C$ constitute the given figure.
The constructions are:
(1) Equilateral triangle $BAC$,
(2) A circle with center $B$ and radius $BA$,
(3) Note that the circle passes through $E$ because $\angle AEC=\frac{\angle ABC}{2}$
(4) Join $DB$
Note that we do not assume that $D, E, B$ are collinear.
We are going to $\color{red}{prove}$ that $\color{red}{D, E, B}$ are $\color{red}{collinear}$.
$\color{red}{Proof:}$
Note that $\angle EAB=45^o$ and $BA=BE \implies \angle ABE=90^o$
i.e. $\color{red}{BE \bot BA}$
On the other hand, $AC=CB=CD \implies \angle ABD=90^o$
i.e. $\color{red}{BD \bot BA}$
The $2$ statements implies that $B, E, D$ are collinear.
Since $CB=CD, \angle BCA=60^o $ and $BCD$ is a trianlge, $\angle CDB=30^0$
A very simple approach, based on your picture, using the law of sines:
$$\angle CAO=40-x,\angle COA=140^{\circ} \implies \frac{\sin (40-x)^{\circ}}{\sin 140^{\circ}}=\frac {\sin (20+x)^{\circ}}{\sin 40^{\circ}}=\frac{\sin (20+x)^{\circ}}{\sin 140^{\circ}}$$
$$\implies \sin (40-x)^{\circ}=\sin (20+x)^{\circ}\implies 40-x=20+x\implies x=10^{\circ}.$$
Best Answer
Here's my approach:
1.) First, we draw a line $AD$ from $A$ such that $\angle ABD=\angle DAB=\beta$, this means that $\angle DAC=\alpha-\beta$ ($\angle BAC=\alpha$). This implies that $AD=BD=x$.
Now, we know that:
$$\cos(A-B)=\frac{7}{8}$$.
We can apply the Law of Cosines in $\triangle DAC$:
$$\cos(A-B)=\frac{x^2+4^2-(5-x)^2}{8x}$$
$$\frac{7}{8}=\frac{x^2+16-25+10x-x^2}{8x}$$
$$7x=10x-9$$
Therefore, $x=3$.
Now, we can apply the Law of Cosines again in $\triangle DAC$:
$$\cos(C)=\frac{4+16-9}{16}$$
$$\cos(C)=\frac{11}{16}$$