# Circle Geometry Homework Help!! (1 Viewer)

#### alussovsky

##### Member
Hi! So these questions are from the year 12 Cambridge X1 book--

I've been going around in circles (figuratively and literally) for the past almost hour on this whole circle geo unit...

#### Jonomyster

##### New Member
$\text{(a)}$
$OP = OT \text{ (equal radii)}$
$\angle OTA = 90^{\circ} \text{ tangent perpendicular to radius}$
$\cos(\alpha) = OT / (OP + PA) = 1/2 \rightarrow \alpha = 60^{\circ}$
$\text{Then, } \gamma = 30^{\circ}$
$\text{Now, } \angle OPT = \angle OTP = 90^{\circ} \text{ (isoceles triangle and angle sum of triangle)}$
$\text{Then, using angle in a semicircle is a right angle, } \beta = \alpha/2 = 30^{\circ}$

$\text{(b)}$
$\text{Reflex } \angle QOT = 220^{\circ} \text{ (angle at centre is twice angle at circumference)}$
$\alpha = 360^{\circ} - 220^{\circ} = 140^{\circ}$
$\text{Now, } \angle OTA = 90^{\circ} \text{ (tangent perpendicular to radius)}$
$\angle OTP = 30^{\circ}$
$\beta = 360^{\circ} - 110^{\circ} - 30^{\circ} - 140^{\circ} = 80^{\circ} \text{ (angle sum of quad)}$

$\text{(c)}$
$\angle OSA = 45^{\circ} \text{ (tangent perpendicular to radius)}$
$\alpha = 360^{\circ} - 90^{\circ} - 90^{\circ} - 50^{\circ} = 130^{\circ} \text{ (angle sum of quad)}$
$\text{Now since we have } \alpha \text{, } \beta = (\text{Reflex } \angle SOT)/2 = 115^{\circ} \text{ (angle at the centre is twice angle at circumference)}$
$\text{Use angle sum of quadrilateral OSAT to get } \gamma$

$\text{(d)}$
$\text{Let O be the centre of the circle and let D be the unnamed point with the } 140^{\circ} \text{ angle}$
$\text{Using tangent perpendicular to radius, } \angle OCD = 80^{\circ} \text{, } \angle OBD = 90^{\circ} - \beta$
$\text{Reflex } \angle COB = 280^{\circ} \text{ (angle at centre is twice angle at circumference)}$
$\angle COB = 360^{\circ} - 280^{\circ} = 80^{\circ}$
$\text{Use angle sum of quad to get } \beta \text{ and } \alpha$