图中显示了一个三角形 ABC，延长 AB 至 D。$AB = 18\,\mathrm{cm}$，$BC = 20\,\mathrm{cm}$，$AC = 22\,\mathrm{cm}$，$DC = 30\,\mathrm{cm}$ 且 $\angle ADC = 39^\circ$。 计算 计算 $\angle CAB$。 计算 $\angle CBD$。 计算三角形 ACD 的面积。 图中显示了一个边长为 18 cm 的菱形 ABCD。 CBD 是以 C 为圆心的扇形。 ABD 是以 A 为圆心的扇形。 角 BCD = 30°。 将 30° 转换为弧度，答案用 π 表示。 计算阴影部分的面积，将你的答案表示为 a + bπ 的形式，其中 a 和 b 为整数。

1. Identify the triangle and choose the cosine rule for the required angle

In $\triangle ABC$, the required angle is $\angle CAB$, between sides $AB=18$ and $AC=22$. The side opposite $\angle CAB$ is $BC=20$.

Use
$a^2=b^2+c^2-2bc\cos A.$

2. Substitute the three side lengths into the cosine rule

$20^2=18^2+22^2-2(18)(22)\cos \angle CAB$

$400=324+484-792\cos \angle CAB$

3. Solve for the angle

$792\cos \angle CAB=324+484-400$

$\cos \angle CAB=\frac{408}{792}=0.515151\ldots$

$\angle CAB=\cos^{-1}(0.515151\ldots)=58.98\ldots^\circ$

1. Identify the known angle in triangle BCD and use the sine rule to find angle BCD

Since A, B and D are collinear, $\angle BDC=\angle ADC=39^\circ$.

In $\triangle BCD$,
$\frac{\sin \angle BCD}{BD}=\frac{\sin 39^\circ}{BC}$
$\frac{\sin \angle BCD}{BD}=\frac{\sin 39^\circ}{20}$
Also, by the sine rule,
$\frac{BD}{\sin \angle BCD}=\frac{30}{\sin \angle CBD}$
This is not directly usable yet because $BD$ and $\angle CBD$ are unknown.

2. Find angle BCD using the sine rule with known opposite pair

In $\triangle BCD$, $CD=30$ is opposite $\angle CBD$, while $BC=20$ is opposite $\angle BDC=39^\circ$.

Let $\angle CBD=x$.
$\frac{\sin x}{30}=\frac{\sin 39^\circ}{20}$
$\sin x=\frac{30\sin 39^\circ}{20}$
$x=\sin^{-1}\left(\frac{30\sin 39^\circ}{20}\right)$

3. Select the geometrically appropriate obtuse angle from the sine rule ambiguity

$x=\sin^{-1}\left(\frac{30\sin 39^\circ}{20}\right)=70.7^\circ \text{ or } 180^\circ-70.7^\circ$
$x=70.7^\circ \text{ or } 109.3^\circ$

From the diagram, $\angle CBD$ is obtuse, so
$x=109.3^\circ$

1. Find the included angle between $AC$ and $CD$ in triangle $ACD$

Since $A$, $B$ and $D$ are collinear,
$\angle CAD=\angle CAB=60.1^\circ$.

$\angle ACD=180^\circ-60.1^\circ-39^\circ=80.9^\circ$

2. Use two sides and the included angle to calculate the area of triangle $ACD$

$\text{Area of }\triangle ACD=\frac12(22)(30)\sin80.9^\circ$

$=325.8\ldots$

$\approx 326\text{ cm}^2$

1. Convert degrees to radians using $180^\circ=\pi$ radians

$30^\circ=30\times \frac{\pi}{180}=\frac{\pi}{6}$

1. Find the area of the rhombus using two adjacent sides and the included angle

In rhombus $ABCD$, $AB=AD=18\text{ cm}$ and $\angle BAD=30^\circ$.

$\text{Area of rhombus}=18\times18\times\sin30^\circ$

$=324\times\frac12=162\text{ cm}^2$

2. Find the total area of the two sectors

Each sector has radius $18\text{ cm}$ and angle $\dfrac{\pi}{6}$ radians.

$\text{Area of one sector}=\frac12 r^2\theta$

$=\frac12(18)^2\left(\frac{\pi}{6}\right)=27\pi\text{ cm}^2$

There are two equal sectors, so

$\text{total sector area}=2(27\pi)=54\pi\text{ cm}^2$

3. Subtract the sector areas from the rhombus area and write in the required form

$\text{Shaded area}=162-54\pi$

$=162-54\pi\text{ cm}^2$

This is in the form $a+b\pi$, where $a=162$ and $b=-54$.