(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 centre of the circle. The side AB is produced to E, < ACB = 49^\circ and < CBE = 68^\circ. Calculate,
(i) the interior angle AOC ; (ii) < BOC.
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a) \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6})\)
= \(\frac{6}{\sqrt{9}} - \frac{3\sqrt{12}}{6\sqrt{3}}\)
= \(2 - \frac{2\sqrt{3}}{2\sqrt{3}}\)
= \(2 - 1 = 1\)
(b)(i) \(< ABC = 180^\circ - 68^\circ = 112^\circ\)
\(< AOC = 2 \times 112^\circ = 224^\circ\)
\(\hat{O} = < AOC (obtuse)\)
= \(360^\circ - 224^\circ = 136^\circ\)
(ii) In \(\Delta AOC\), OA = OC (radii)
\(\therefore \Delta AOC = isosceles\)
\(\hat{A} = < OAC ; \hat{C} = < ACO ; \hat{A} = \hat{C} = x\)
\(\hat{O} + \hat{A} + \hat{C} = 180^\circ\)
\(136^\circ + x + x = 180^\circ \implies 2x = 44^\circ\)
\(x = 22^\circ\)
In \(\Delta BOC\), OB = OC (radii)
\(\therefore \Delta BOC = isosceles\)
\(< OCB = 49^\circ + 22^\circ= 71^\circ\)
\(< OBC = < OCB = 71^\circ\)
\(< OBC + < OCB + < BOC = 180^\circ\)
\(71^\circ + 71^\circ + < BOC = 180^\circ\)
\(< BOC = 180^\circ - 142^\circ = 38^\circ\)
Source: Myschool Past Questions Archive (Record #53319)