(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 \(\stackrel\frown{BCA} = 40°\) and \(\stackrel\frown{DAT} = 52°\), find \(\stackrel\frown{BAD}\).
\(2x - y = \frac{9}{2}\)
\(x + 4y = 0\)
(b)
In the diagram, TA is a tangent to the circle at A. If \(\stackrel\frown{BCA} = 40°\) and \(\stackrel\frown{DAT} = 52°\), find \(\stackrel\frown{BAD}\).
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Correct Answer: Option
Explanation:
(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 \(\Delta ACD, \stackrel\frown{ACD} = 52° \) \(\stackrel\frown{ACD} = \stackrel\frown{DAT}\)
\(\therefore \stackrel\frown{BCD} = 40° + 52° = 92°\)
Note \(\stackrel\frown{BCD}\) and \(\stackrel\frown{BAD}\) are supplementary = 180°
\(\therefore \stackrel\frown{BAD} = 180° - 92° = 88°\)
(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 \(\Delta ACD, \stackrel\frown{ACD} = 52° \) \(\stackrel\frown{ACD} = \stackrel\frown{DAT}\)
\(\therefore \stackrel\frown{BCD} = 40° + 52° = 92°\)
Note \(\stackrel\frown{BCD}\) and \(\stackrel\frown{BAD}\) are supplementary = 180°
\(\therefore \stackrel\frown{BAD} = 180° - 92° = 88°\)