A tangent and a radius meet at a right angle, so triangles OAB and OCB are congruent, right-angled triangles.

The line OB bisects the angle of the sector.

$\begin{array}{rcl} Angle A O B & = & \frac{120}{2} \\ = & 60 ° \end{array}$

[1]

Work out the length AB using SOHCAHTOA.
You know the adjacent and you want to find out the opposite so use $\tan θ = \frac{O p p}{A d j}$ .

4-6-2

$\begin{array}{rcl} \tan 60 & = & \frac{A B}{10} \\ 10 \tan 60 & = & A B \end{array}$

[1]

$\tan (60) = \sqrt{3}$ .

$\begin{array}{rcl} A B & = & 10 \sqrt{3} m \end{array}$

Calculate the area of the quadrilateral ABCO by calculating the area of the triangle ABO and doubling it.

$\begin{array}{rcl} A_{t r i a n g l e} & = & \frac{1}{2} \times 10 \times 10 \sqrt{3} \\ = & 50 \sqrt{3} cm \end{array}$

[1]

$\begin{array}{rcl} A_{q u a d r i l a t e r a l} & = & 2 \times 50 \sqrt{3} \\ = & 100 \sqrt{3} m^{2} \end{array}$

Calculate that area of the sector OAC, using the formula: $A = \frac{θ °}{360 °} \times π r^{2}$ , where $r$ is the radius and $θ$ is the angle of the sector.

$\begin{array}{rcl} A_{s e c t o r} & = & \frac{120}{360} \times π \times 10^{2} \end{array}$

[1]

$\begin{array}{rcl} A_{s e c t o r} & = & \frac{100}{3} π m^{2} \end{array}$

Find the area of the shaded section by subtracting the area of the sector OAC from the area of the quadrilateral ABCO.

$\begin{array}{rcl} (100 \sqrt{3} - \frac{100}{3} π) \end{array} {cm}^{2}$ [1]

Circles, Arcs & Sectors (CIE IGCSE Maths: Extended)

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