Solve for $a$ given $\frac{\sin a}{\sin (66+a) }=\frac{\sin 48}{\sin 66}$ $$\dfrac{(\sin a)}{(\sin 66+a) }=\dfrac{(\sin 48)}{(\sin 66) }$$
So basically i came across this equation while solving a geometry problem and am having a hard time solving it.
I tried using the cyclic functions of trigonometric functions which but still couldn't solve it any help would be appreciated (especially simple to understand solutions if any)
 A: Hint:
$$\frac{\sin(a)}{\sin(a+66)}=\frac{\sin(a)}{\sin(a)\cos(66)+\cos(a)\sin(66)}=\frac1{\cos(66)+\cot(a)\sin(66)}.$$
Make $\cot(a)$ the subject.
A: Rearrange the equation and then factorize
\begin{align}
& \ \sin a \sin 66 - \sin (66+a) \sin48\\
=& \ \frac12[\cos(66-a)-\cos(66+a)-\cos(18+a)+\cos(114+a)]\\
 =& -\frac12[\cos(66+a)+\cos(18+a)]
 =-\cos24\cos(42+a)=0\\
\end{align}
Thus, $\cos(42+a)=0$, yielding $ a= 48+180n$
A: Hint
Find an equivalent equation of the type
$$A \sin a + B \cos a = 0$$
that you can solve. For that use the formula $\sin(x+y) = \sin x \cos y + \cos x \sin y$.
Note: this answer only works if the response to Gae. S comment is positive.
A: Solution 1
$$\dfrac{(\sin a)}{\sin (66^{\circ}+a) }=\dfrac{(\sin 48^{\circ})}{(\sin 66^{\circ}) }=2\sin24^{\circ}$$
equivalent to
$$\frac{1}{2} \sin a=\sin24^{\circ}\sin(66^{\circ}+a)=\sin24^{\circ}\sin66^{\circ}\cos a+\sin24^{\circ}\cos66^{\circ}\sin a\\
=\frac{1}{2}\sin48^{\circ}\cos a+ (\sin24^{\circ})^2\sin a$$
equivalent to
$$\frac{1-2(\sin24^{\circ})^2}{2}\sin a= \frac{\sin48^{\circ}\cos a}{2}$$
equivalent to
$$\cos48^{\circ} \sin a= \sin48^{\circ}\cos a$$
equivalent to
$$0=\cos48^{\circ} \sin a- \sin48^{\circ}\cos a=\sin(a-48^{\circ})$$
hence $a=48^{\circ}+ n\pi $ for $n\in \mathbb{Z}$
A: Solution 2
By @Yves Daoust 's hint,
$$\dfrac{(\sin a)}{(\sin 66^{\circ}+a) }=\frac{1}{\cos66^{\circ}+\sin66^{\circ}\cot a}$$
It is strictly increasing in $[-\pi/2,\pi/2]$ and so it has to have unique solution. Applying $a=48^{\circ}$ gives us the solution. Because $\dfrac{(\sin a)}{(\sin 66^{\circ}+a) }$ is also periodic, all solutions are $a= 48^ {\circ}+ n\pi $ for $n\in \mathbb {Z}$
