圆心为 O 的圆在 D 处有一条切线 XY。∠ODC = 70°，∠BCD = 120°，且 ∠ODA = 15°。 求 ∠CDY。 求 ∠DAB。 求 ∠OBA。 该图显示了一个圆心为 O 的圆的扇形 OABC 和一个等腰三角形 OCD。OC = OD = 10 cm 且 ∠COD = 1.2 弧度。 阴影弓形 ABC 的面积， 长度 CD。

1. Find the angle between chord $CD$ and the tangent $DY$.

Since $XY$ is a tangent at $D$, $OD \perp XY$.

So $\angle ODY = 90^\circ$.

Given $\angle ODC = 70^\circ$,
$\angle CDY = 90^\circ - 70^\circ = 20^\circ.$

1. Use opposite angles of cyclic quadrilateral $ABCD$

Since $A$, $B$, $C$ and $D$ lie on the circumference, $ABCD$ is a cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary, so
$\angle DAB + \angle BCD = 180^\circ.$
Given $\angle BCD = 120^\circ$,
$\angle DAB = 180^\circ - 120^\circ = 60^\circ.$

1. Find the central angle $\angle DOA$ using triangle $ODA$.

Since $OA=OD$ (radii), triangle $ODA$ is isosceles.

$\angle OAD=\angle ODA=15^\circ$

$\angle DOA=180^\circ-15^\circ-15^\circ=150^\circ$

2. Find the central angle subtended by chord/arc $DB$ using the earlier result $\angle DAB=60^\circ$.

From part (ii),

$\angle DAB=60^\circ$

The angle at the centre standing on the same arc $DB$ is twice the angle at the circumference.

$\angle DOB=2\times 60^\circ=120^\circ$

3. Find $\angle AOB$, then use isosceles triangle $AOB$ to find $\angle OBA$.

Angles around point $O$ add to $360^\circ$.

$\angle AOB=360^\circ-150^\circ-120^\circ=90^\circ$

Since $OA=OB$ (radii), triangle $AOB$ is isosceles.

$\angle OBA=\frac{180^\circ-90^\circ}{2}=45^\circ$

1. Find the angle of sector $AOC$ in radians.

Since $A, O, D$ lie on a straight line,
$\angle AOC+\angle COD=\pi.$
Hence
$\angle AOC=\pi-1.2.$

2. Find the area of sector $AOC$.

Radius $=10$ cm.
$\text{Area of sector }AOC=\frac12 r^2\theta$
$=\frac12(10)^2(\pi-1.2)$
$=50(\pi-1.2).$

3. Subtract the area of triangle $AOC$ to obtain the shaded segment area.

$\text{Area of }\triangle AOC=\frac12 r^2\sin\theta$
$=\frac12(10)^2\sin(\pi-1.2)$
$=50\sin 1.2.$

Therefore,
$\text{Area of shaded segment}$
$=50(\pi-1.2)-50\sin1.2$
$=26.0718\ldots$
$\approx 26.1\text{ cm}^2.$

1. Apply the cosine rule to triangle $OCD$.

In $\triangle OCD$,

$CD^2=OC^2+OD^2-2(OC)(OD)\cos\angle COD$

$CD^2=10^2+10^2-2(10)(10)\cos(1.2)$

2. Evaluate and round the length of $CD$.

$CD^2=200-200\cos(1.2)$

$CD=\sqrt{200-200\cos(1.2)}$

$CD=11.288\ldots$

So, to 3 significant figures,

$CD=11.3\text{ cm}$