(a) Solve for x and y in the following equations :
\(2x - y = \frac{9}{2}\)
\(x + 4y = 0\)
(b)
In the diagram, TA is a tangent to the circle at A. If \(\stackrel\frown{BCA} = 40^\circ\) and \(\stackrel\frown{DAT} = 52^\circ\), find \(\stackrel\frown{BAD}\).
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a) \(2x - y = \frac{9}{2} .... (1)\)
\(x + 4y = 0 ....... (2)\)
From (2), x = - 4y. Put into (1), we have
\(2(- 4y) - y = \frac{9}{2}\)
\(-8y - y = \frac{9}{2} \implies - 9y = \frac{9}{2}\)
\(y = \frac{\frac{9}{2}}{-9} = -\frac{1}{2}\)
\(x = - 4y = -4(-\frac{1}{2})\)
\(x = 2\)
\((x, y) = (2, -\frac{1}{2})\)
(b)
In \(\Delta ACD, \stackrel\frown{ACD} = 52^\circ \) \(\stackrel\frown{ACD} = \stackrel\frown{DAT}\)
\(\therefore \stackrel\frown{BCD} = 40^\circ + 52^\circ = 92^\circ\)
Note \(\stackrel\frown{BCD}\) and \(\stackrel\frown{BAD}\) are supplementary = 180^\circ
\(\therefore \stackrel\frown{BAD} = 180^\circ - 92^\circ = 88^\circ\)
Source: Myschool Past Questions Archive (Record #53016)