(a) In the diagram, AB // CD and BC // FE. \(\stackrel\frown{CDE} = 75°\) and \(\stackrel\frown{DEF} = 26°\). Find the angles marked x and y. (b) The diagra...

Answer Details

(a) Finding x and y. From the diagram \(AB\parallel CD\), \(BC\parallel FE\), \(\widehat{CDE}=75^\circ\) and \(\widehat{DEF}=26^\circ\).

Draw a line through \(D\) parallel to \(BC\) and \(FE\). Because \(DE\) is a transversal of the parallel lines \(FE\) and this new line, the angle \(DE\) makes with the parallel family at \(D\) equals the angle it makes at \(E\):

\[\text{angle between } DE \text{ and the } BC\text{-family}=\widehat{DEF}=26^\circ.\]

Since \(\widehat{CDE}=75^\circ\) is the angle between \(DC\) and \(DE\), the angle between \(DC\) and the \(BC\)-family is

\[75^\circ-26^\circ=49^\circ.\]

Now \(x=\widehat{ABC}\) is the angle at \(B\) between \(BA\) (in the \(CD\)-family, since \(AB\parallel CD\)) and \(BC\). This is exactly that same angle between the two families:

\[\boxed{x=49^\circ.}\]

By alternate angles (\(AB\parallel CD\), transversal \(BC\)), \(\widehat{BCD}=\widehat{ABC}=49^\circ\). The marked angle \(y\) is the reflex angle at \(C\) (the loop over the peak):

\[y=360^\circ-49^\circ=\boxed{311^\circ.}\]

(b) Circle \(ABCD\), centre \(O\), radius \(7\text{ cm}\). Reflex \(\widehat{AOC}=190^\circ\), so the ordinary \(\widehat{AOC}=360^\circ-190^\circ=170^\circ\).

(i) \(\widehat{ABC}\) stands on the arc \(ADC\) (the arc corresponding to the reflex \(190^\circ\)). Angle at circumference \(=\tfrac12\) angle at centre:

\[\widehat{ABC}=\tfrac12(190^\circ)=95^\circ.\]

(ii) \(\widehat{ADC}\) stands on the arc \(ABC\) (central angle \(170^\circ\)):

\[\widehat{ADC}=\tfrac12(170^\circ)=85^\circ.\]

(Check: \(ABCD\) is cyclic, \(95^\circ+85^\circ=180^\circ\).)

(c)(i) Arc \(ABC\). Central angle \(=170^\circ\), \(r=7\), \(\pi=3.142\):

\[\text{arc}=\frac{170}{360}\times 2\times 3.142\times 7=\frac{170}{360}\times 43.988=20.77\approx 20.8\text{ cm}.\]

(c)(ii) Chord \(AD\). In \(\triangle OAD\), \(OA=OD=7\) so it is isosceles with base angles \(\widehat{DAO}=\widehat{ODA}=35^\circ\). Hence \(\widehat{AOD}=180^\circ-2(35^\circ)=110^\circ\).

\[AD=2r\sin\!\left(\tfrac{110^\circ}{2}\right)=2(7)\sin 55^\circ=14\times 0.8192=11.47\approx 11.5\text{ cm}.\]