不使用计算器，求整数 $a$、$b$ 和 $c$ 的值，使得方程 $5x - \sqrt{3} = 4 + \sqrt{3}\,x$ 的解为 $\frac{a + b\sqrt{3}}{c}.$ 已知 $\sin A = \tfrac{3}{5}$，其中 A 为锐角，$\tan B = \tfrac{5}{12}$，且 A 和 B 在不同的象限。不使用计算器，计算以下各项的值： $\sin\left(\dfrac{\pi}{2} - B\right)$ $\tan 2A$

1. Group terms involving $x$ on one side of the equation

$5x - \sqrt{3}\,x = 4 + \sqrt{3}$

2. Factorise $x$ and isolate it

$x(5 - \sqrt{3}) = 4 + \sqrt{3}$
$x = \frac{4 + \sqrt{3}}{5 - \sqrt{3}}$

3. Rationalise the denominator

$x = \frac{(4 + \sqrt{3})(5 + \sqrt{3})}{(5 - \sqrt{3})(5 + \sqrt{3})}$

4. Expand and simplify to find $a$, $b$, and $c$

$x = \frac{20 + 4\sqrt{3} + 5\sqrt{3} + 3}{5^2 - (\sqrt{3})^2}$
$x = \frac{23 + 9\sqrt{3}}{25 - 3}$
$x = \frac{23 + 9\sqrt{3}}{22}$
Comparing with $\frac{a + b\sqrt{3}}{c}$, we have $a = 23$, $b = 9$, $c = 22$.

1. Determine the quadrant of angle B

Since $A$ is acute, it lies in the 1st quadrant. We are given that $\tan B = \frac{5}{12} > 0$ and that $A$ and $B$ are in different quadrants. Therefore, $B$ must lie in the 3rd quadrant.

2. Find the value of cos B

For $\tan B = \frac{5}{12}$, the opposite side is $5$ and the adjacent side is $12$. The hypotenuse is $\sqrt{5^2 + 12^2} = 13$.
Since $B$ is in the 3rd quadrant, $\cos B$ is negative.
$\cos B = -\frac{12}{13}$

3. Evaluate the required expression using complementary angles

Using the complementary angle identity (or expansion formula):
$\sin\left(\frac{\pi}{2} - B\right) = \cos B$
$\sin\left(\frac{\pi}{2} - B\right) = -\frac{12}{13}$

1. Find the value of $\tan A$

Since $\sin A = \frac{3}{5}$ and $A$ is acute, we can draw a right-angled triangle with opposite side $3$ and hypotenuse $5$. The adjacent side is $\sqrt{5^2 - 3^2} = 4$.
Therefore, $\tan A = \frac{3}{4}$.

2. Apply the double angle formula to evaluate $\tan 2A$

$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$
$= \frac{2\left(\frac{3}{4}\right)}{1 - \left(\frac{3}{4}\right)^2}$
$= \frac{\frac{3}{2}}{1 - \frac{9}{16}}$
$= \frac{\frac{3}{2}}{\frac{7}{16}}$
$= \frac{3}{2} \times \frac{16}{7} = \frac{24}{7}$