(a) Without using calculator or mathematical tables, evaluate \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6})\) (b) In the diagram, O is the ...

Answer Details

(a) Evaluate \(\dfrac{3}{\sqrt3}\left(\dfrac{2}{\sqrt3}-\dfrac{\sqrt{12}}{6}\right)\) without tables.

First simplify each surd. \(\dfrac{3}{\sqrt3}=\dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{3\sqrt3}{3}=\sqrt3.\)

Inside the bracket: \(\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}\) and \(\dfrac{\sqrt{12}}{6}=\dfrac{2\sqrt3}{6}=\dfrac{\sqrt3}{3}.\)

\[\frac{2\sqrt3}{3}-\frac{\sqrt3}{3}=\frac{\sqrt3}{3}.\]

Therefore

\[\sqrt3\times\frac{\sqrt3}{3}=\frac{3}{3}=\boxed{1.}\]

(b) Circle, centre \(O\); \(AB\) produced to \(E\), \(\widehat{ACB}=49^\circ\), \(\widehat{CBE}=68^\circ\).

Since \(A,B,E\) are collinear, \(\widehat{ABC}\) and \(\widehat{CBE}\) are angles on a straight line:

\[\widehat{ABC}=180^\circ-68^\circ=112^\circ.\]

In \(\triangle ABC\):

\[\widehat{BAC}=180^\circ-\widehat{ABC}-\widehat{ACB}=180^\circ-112^\circ-49^\circ=19^\circ.\]

(i) Interior angle \(AOC\). \(\widehat{ABC}=112^\circ\) is the angle at the circumference standing on chord \(AC\); it subtends the major arc \(AC\), whose central angle (reflex \(AOC\)) is \(2\times112^\circ=224^\circ.\) Hence the interior (non-reflex) angle is

\[\widehat{AOC}=360^\circ-224^\circ=\boxed{136^\circ.}\]

(ii) Angle \(BOC\). \(\widehat{BAC}=19^\circ\) at the circumference subtends arc \(BC\); the angle at the centre on the same arc is twice as large:

\[\widehat{BOC}=2\times19^\circ=\boxed{38^\circ.}\]

(Check: \(\widehat{ACB}=49^\circ\Rightarrow\widehat{AOB}=98^\circ,\) and \(98^\circ+38^\circ+224^\circ=360^\circ.\))