Understanding Engineering Mathematics

Solution to review question 5.1.9

(i) Referring to Figure 5.8,a= 360 °− AOB,and AOB= 2 ACB=

2 × 40 °= 80 °.Soa= 280 °.

(ii) Equal angles subtended by equal arcs gives ACB=c=ADB=

30 °. Angles in triangleABDthen givesa= 180 °− 30 °− 70 °= 80 °.

Equal angles subtended by equal arcs now gives DAC= 20 °=

DBC=b.Soa= 80 °,b= 20 °,c= 30 °.

(iii)OA=OB, so triangleABOis isosceles and thereforea= 35 °and

b= 180 °− 35 °− 90 °= 55 °, since the chord bisectorOPis perpen-

dicular to the chordAB.

(iv) The angleDABis subtended by a diameter and is therefore 90°.

So from triangleABD, ABD= 180 °− 30 °− 90 °= 60 °.Butthe

angles subtended by the chordADatBandCare the same and so

a= 60 °also.

5.2.10 Cyclic quadrilaterals

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The results so far essentially rely on fitting triangles into circles – we say thecircle
circumscribesthe triangle. In general, when a circle circumscribes a polygon the vertices
of the polygon lie on the circle. (A circle iscircumscribedby a figure, orinscribedin
a figure when all sides of the figure touch the circle.) The next step up is to consider
circumscribed quadrilaterals – i.e. quadrilaterals whose vertices lie on a circle, and sides
inside the circle. This is called acyclic quadrilateral. See Figure 5.25 showing the cyclic
quadrilateralABCD.

A

D

C

B

O

2 a

2 c

a

c

Figure 5.25Cyclic quadrilateral.

The main result on cyclic quadrilaterals is that:

• The opposite angles of a cyclic quadrilateral are supplementary –
  i.e. add up to 180°.

The proof is interesting and not difficult. In Figure 5.25, with obvious notation we have

2 a+ 2 c= 360 °= 2 (a+c)

= 2 ( DAB+ DCB)

Hence
DAB+ DCB= 180 °