In the diagram, AB is a tangent to the circle with centre O, ∠CAB = 58° and CDE is a straight line. Find, stating your reasons clearly,

1. Identify the angle in the same segment of the left circle

In the left circle, $A,B,C,D$ lie on the circumference. Angles $\angle CAB$ and $\angle CDB$ stand on the same chord $CB$.

Hence
$\angle CDB = \angle CAB = 58^\circ$
(angles in the same segment).

2. Use the straight line $CDE$ to obtain the required angle

Since $C,D,E$ is a straight line, $DC$ and $DE$ are the same straight line in opposite directions. The required angle $\angle BDE$ is the angle between $DB$ and $DE$, which corresponds to the angle found from the diagram:
$\angle BDE = 58^\circ$

1. Relate the tangent angle to the angle at the circumference standing on chord $BE$

Since $AB$ is a tangent to the right-hand circle at $B$, and $DE$ is the same straight line as $CE$,

$\angle BED = \angle CAB = 58^\circ$

by the tangent-chord theorem / alternate segment theorem.

2. Find the obtuse central angle subtended by chord $BE$

The angle at the centre standing on the same arc $BE$ is twice the angle at the circumference:

$\angle BOE = 2 \times \angle BED$

$= 2 \times 58^\circ = 116^\circ$.

1. Use equal radii and the angle sum of triangle $BOE$ to find the base angle

Since $OB = OE$ (radii of the same circle), $\triangle BOE$ is isosceles.

From part (ii), $\angle BOE = 116^\circ$.

Hence
$\angle EBO = \frac{180^\circ - 116^\circ}{2} = 32^\circ.$

1. Identify the right angle between the tangent and the radius

Since $AB$ is a tangent to the circle at $B$ and $OB$ is a radius,

$\angle ABO = 90^\circ$.

2. Subtract the known angle to find $\angle ABE$

From part (iii), $\angle EBO = 32^\circ$.

Hence

$\angle ABE = \angle ABO - \angle EBO$

$= 90^\circ - 32^\circ$

$= 58^\circ$.

1. Write the shaded area as outer sector area minus inner sector area

Using sector area $=\dfrac{1}{2}r^2\theta$, where $\theta$ is in radians,

$\text{Area of shaded region}=\frac{1}{2}y^2(0.5)-\frac{1}{2}x^2(0.5).$

2. Simplify the expression in terms of $x$ and $y$

$\text{Area}=\frac{1}{4}y^2-\frac{1}{4}x^2=\frac{1}{4}(y^2-x^2).$

1. Write the lengths of the two arcs using arc length formula

Since $\theta=0.5$ radians and arc length $=r\theta$,

inner arc $AB=0.5x$,

outer arc $CD=0.5y$.

2. Form the perimeter equation for the shaded region

The two straight sides each have length $y-x$.

Given perimeter $=120$,
$0.5x+0.5y+(y-x)+(y-x)=120.$

3. Simplify and rearrange to show the required result

$0.5x+0.5y+2y-2x=120$
$2.5y-1.5x=120$
Multiply by $2$:
$5y-3x=240$
$5y=240+3x$
$y=\frac{1}{5}(240+3x).$