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Question

In the following figure, AD is the diameter of the circle with centre O. Chords AB,BC and CD are equal. If DEF = 110o, calculate :

(i) AEF, (ii) FAB.


Solution

Given AB = BC = CD

(i) AOD  is the diameter

\angle AED = 90^o  (angle in semicircle)

\angle AEF + \angle AED = \angle FED c4.jpg

\angle AEF = 110-90=20^o

(ii) Since AB = BC = CD  (given)

\angle AOB = \angle BOC = \angle COD  (equal arcs subtend equal angles at center)

We know \angle AOD = 180^o

\Rightarrow \angle AOB = \angle BOC = \angle COD = 60^o

In \triangle AOB

OA = OB  (radius of the same circle)

Therefore \angle OAB = \angle OBA

\angle OAB + \angle OBA = 180-\angle AOB = 180-60 = 120^o

\Rightarrow \angle OAB = \angle OBA = 60^o

ADEF  is a cyclic quadrilateral

\angle DEF + \angle DAF = 180^o  (opposite angles are supplementary)

\angle DAF = 180-110=70^o

Now \angle FAB = \angle DAF + \angle OAB = 70+60=130^o

\\


Mathematics
Concise Mathematics
Standard X

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