Find the area of given triangle

Question

In a triangle $ABC$, where $a=8$, $c=1$ and $\cos (A-C) ={16\over 65}$.

Where $(a,b,c)$ are sides and $(A,B,C)$ are angles.

How can we find the area of the triangle $ABC$?

A hint will be appreciated.

Answer 1

Hint: The Law of Sines says $$ \sin(A)=8\sin(C)\tag{1} $$ Then $$ \begin{align} \frac{16}{65} &=\cos(A)\cos(C)+\sin(A)\sin(C)\tag{2}\\ \left(\frac{16}{65}-8\sin^2(C)\right)^2 &=\left(1-64\sin^2(C)\right)\left(1-\sin^2(C)\right)\tag{3}\\ \frac{256}{4225}-\frac{256}{65}\sin^2(C)+64\sin^4(C) &=1-65\sin^2(C)+64\sin^4(C)\tag{4}\\ \sin^2(C) &=\frac1{65}\tag{5}\\ \end{align} $$ This means that $$ \sin^2(A)=\frac{64}{65}\quad\cos^2(A)=\frac1{65}\quad\sin^2(C)=\frac1{65}\quad\cos^2(C)=\frac{64}{65}\tag{6} $$ and therefore, $$ \begin{align} \cos(A+C) &=\cos(A)\cos(C)-\sin(A)\sin(C)\\ &=\frac8{65}-\frac8{65}\\[2pt] &=0\tag{7} \end{align} $$ What does this say about $B$, the angle between $a$ and $c$?

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Note: Since $A,C\le\pi$, we have $\sin(A),\sin(C)\ge0$. However, only given $(6)$, $\cos(A),\cos(C)$ may not both be positive. Therefore, we know that $\sin(A)\sin(C)=\frac8{65}$, but we only know that $|\cos(A)\cos(C)|=\frac8{65}$. However, $(2)$ tells us that $\cos(A)\cos(C)=\frac8{65}$.

Answer 2

Given

$$ c,a,\quad \delta = C-A $$

$$ C-A = \gamma- \alpha = \delta $$

Construct double angle $ 2 \alpha $ at vertex A.

Law of Sines on $ \Delta ABC $

$$ \frac{c}{a}= \frac{\sin (\delta+\alpha)}{\sin \alpha} \tag1$$

Trig expand and algebraically simplify

$$ \tan \alpha= \frac {\sin \delta}{( c/a -\cos \delta )} \rightarrow \alpha_{1,2} \tag2 $$

$$ \alpha_2 =\alpha_1 + \pi \tag3$$

$$ \beta = (\pi- \delta- 2 \alpha _{1,2}) \tag4 $$

$$ \gamma = (\pi- \alpha _{1,2} -\beta) \tag5 $$

One value suffices.

Area

$$ \frac12 \,a \,c \sin \beta = .. $$

Also checked all the results numerically

EDIT1:

If we use supplementary angle for angles assuming it as an ambiguous case, all conditions given could be correctly realized, but this should be seen in a construction.