(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 \(\sta...

Answer Details

(a) Simultaneous equations.

\[2x-y=\frac{9}{2}\quad(1),\qquad x+4y=0\quad(2).\]

From (2), \(x=-4y\). Substitute into (1):

\[2(-4y)-y=\frac{9}{2}\]

\[-8y-y=\frac{9}{2}\]

\[-9y=\frac{9}{2}\ \Rightarrow\ y=-\frac{1}{2}.\]

Then \(x=-4y=-4\left(-\dfrac{1}{2}\right)=2\).

\[\boxed{x=2,\quad y=-\tfrac{1}{2}}.\]

(b) Tangent TA to the circle at A.

From the diagram, \(B, C, D, A\) lie on the circle, \(TA\) is the tangent at \(A\), \(\angle BCA=40^\circ\) (at \(C\)) and \(\angle DAT=52^\circ\) (between the tangent \(AT\) and chord \(AD\)).

Chord AB. By the alternate segment theorem, the tangent-chord angle between the tangent (on the side towards \(X\), opposite \(T\)) and chord \(AB\) equals the angle in the alternate segment \(\angle BCA\):

\[\angle BAX=\angle BCA=40^\circ.\]

Straight tangent line. The points \(X, A, T\) lie on the straight tangent line, so the three angles at \(A\) on one side sum to \(180^\circ\):

\[\angle BAX+\angle BAD+\angle DAT=180^\circ.\]

\[40^\circ+\angle BAD+52^\circ=180^\circ\]

\[\angle BAD=180^\circ-40^\circ-52^\circ=\boxed{88^\circ}.\]