如图,在△ABC中,∠A,∠B,∠C的对边分别用a,b,c表示.

【基础设问】
(1)若∠C=90°,则sin A = ,cos B = ,tan B = .(用含a,b,c的式子表示)
(2)若∠B是锐角,且cos(∠B + 15°) = $\frac{1}{2}$,则∠B的度数为 .
(3)若∠C=90°,b=3,tan A = $\frac{8}{3}$,则a的值为 ,c的值为 .
【能力设问】
(4)若∠C=90°,∠A=2∠B,则
①∠A = °,∠B = °.
②求$\cos^2 B + \sin A · \tan B$的值.
(5)若∠C=90°,sin A = $\frac{2}{3}$,求tan B和cos B的值.
(6)若BC=12,tan A = $\frac{3}{4}$,∠B=30°,求AC和AB的长.
【拓展设问】
(7)过点C作CD⊥AB于点D,设∠ACD = α,∠BCD = β(0° ＜ α + β ＜ 90°).从沪科版九上教材P132例2中,我们知道$S_{△ ABC} = \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C$,请你利用该结论证明:$\sin(α + β) = \sin α · \cos β + \cos α · \sin β$.
(8)利用(7)中结论解决下面的问题:如图所示,一艘渔船从位于C海岛南偏东15°方向,距C海岛$55(\sqrt{6} - \sqrt{2})$海里的B处出发,欲前往正西方向的灯塔A处.已知在C海岛周围50海里水域内有暗礁,则这艘渔船在航行过程中有触礁的危险吗?请说明理由.

(1) $\frac{a}{c}$ $\frac{a}{c}$ $\frac{b}{a}$
(2) $45°$ 由题意,得$∠ B + 15° = 60°, \therefore ∠ B = 45°$.
(3) $8$ $\sqrt{73}$ $\because ∠ C = 90°, \tan A = \frac{a}{b}, \therefore a = b\tan A = 3 × \frac{8}{3} = 8$,由勾股定理,得$c = \sqrt{b^2 + a^2} = \sqrt{73}$.
(4) ① $60$ $30$
②解:$\cos^2 B + \sin A · \tan B$
$= \cos^2 30° + \sin 60° · \tan 30°$
$= \frac{3}{4} + \frac{\sqrt{3}}{2} × \frac{\sqrt{3}}{3}$
$= \frac{5}{4}$.
(5)解:$\because ∠ C = 90°, \sin A = \frac{a}{c} = \frac{2}{3}, \therefore$ 设 $a = 2k(k ＞ 0)$,则 $c = 3k$,
由勾股定理,得 $b = \sqrt{c^2 - a^2} = \sqrt{5}k$,
$\therefore \tan B = \frac{b}{a} = \frac{\sqrt{5}k}{2k} = \frac{\sqrt{5}}{2}$.
$\cos B = \cos(90° - ∠ A) = \sin A = \frac{2}{3}$.
(6)解:解题思路:
过点 $C$ 作 $CH ⊥ AB$ 于点 $H$(图略).
在 $\mathrm{Rt}△ BCH$ 中,$BC = 12, ∠ B = 30°$,
$\therefore CH = \frac{1}{2}BC = 6, \therefore BH = \sqrt{BC^2 - CH^2} = 6\sqrt{3}$.
在 $\mathrm{Rt}△ ACH$ 中,$\tan A = \frac{3}{4} = \frac{CH}{AH}, \therefore AH = 8$,
$\therefore AC = \sqrt{AH^2 + CH^2} = 10, AB = AH + BH = 8 + 6\sqrt{3}$.
(7)证明:由题意可知,$S_{△ ACD} = \frac{1}{2}AC · CD · \sin α, S_{△ BCD} = \frac{1}{2}BC · CD · \sin β, S_{△ ABC} = \frac{1}{2}AC · BC · \sin ∠ ACB = \frac{1}{2}AC · BC · \sin(α + β)$.
$\because S_{△ ABC} = S_{△ ACD} + S_{△ BCD}$,
$\therefore \frac{1}{2}AC · BC · \sin(α + β) = \frac{1}{2}AC · CD · \sin α + \frac{1}{2}BC · CD · \sin β$,
等式两边同时除以 $\frac{1}{2}AC · BC$,
得 $\sin(α + β) = \frac{CD}{BC} · \sin α + \frac{CD}{AC} · \sin β$,
在 $\mathrm{Rt}△ ACD$ 中,$\cos α = \frac{CD}{AC}$,
在 $\mathrm{Rt}△ BCD$ 中,$\cos β = \frac{CD}{BC}$,
$\therefore \sin(α + β) = \sin α · \cos β + \sin β · \cos α$.
(8)解:这艘渔船在航行过程中没有触礁的危险.理由如下:
过点 $C$ 作 $CD ⊥ AB$ 于点 $D$(图略),由题意可知 $∠ CBD = 75°$.
由(7) 可知 $\sin ∠ CBD = \sin 75° = \sin(30° + 45°) = \sin 30° · \cos 45° + \cos 30° · \sin 45° = \frac{1}{2} × \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} × \frac{\sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$.
在 $\mathrm{Rt}△ BCD$ 中,$BC = 55(\sqrt{6} - \sqrt{2})$ 海里,$\sin ∠ CBD = \frac{CD}{BC}$,
$\therefore CD = BC · \sin 75° = 55(\sqrt{6} - \sqrt{2}) × \frac{\sqrt{6} + \sqrt{2}}{4} = 55$(海里),
$\because 55 ＞ 50, \therefore$ 这艘渔船在航行过程中没有触礁的危险.