Step 1: Set up the figure.
Let $D$ be the foot of the tower, $C$ the top of the tower (bottom of the flagstaff), and $A$ the top of the flagstaff. Let $P$ be the point on the ground with $PD = 30\text{ m}$. The wire to the tower top makes $\angle CPD = 30^{\circ}$, and the wire to the flagstaff top makes $\angle APD = 60^{\circ}$.
Step 2: Find the height of the tower using the tangent ratio.
In right triangle $PDC$:
\[ \tan 30^{\circ} = \frac{CD}{PD} \implies CD = 30\tan 30^{\circ} = \frac{30}{\sqrt{3}} = 10\sqrt{3}\text{ m} \]
Using $\sqrt{3}=1.73$:
\[ CD = 17.3\text{ m} \]
Step 3: Find the wire lengths using the Pythagoras theorem instead of the cosine ratio.
For the wire to the bottom of the flagstaff, $PC$ is the hypotenuse of right triangle $PDC$:
\[ PC^{2} = PD^{2} + CD^{2} = 30^{2} + (10\sqrt{3})^{2} = 900 + 300 = 1200 \]
\[ PC = \sqrt{1200} = 20\sqrt{3} = 20(1.73) = 34.6\text{ m} \]
Step 4: Find the height up to the flagstaff top, then the second wire length.
In right triangle $PDA$:
\[ \tan 60^{\circ} = \frac{AD}{PD} \implies AD = 30\tan 60^{\circ} = 30\sqrt{3} = 30(1.73) = 51.9\text{ m} \]
By Pythagoras in triangle $PDA$:
\[ PA^{2} = PD^{2} + AD^{2} = 30^{2} + (30\sqrt{3})^{2} = 900 + 2700 = 3600 \]
\[ PA = \sqrt{3600} = 60\text{ m} \]
Final Answer:
The height of the tower is $17.3$ m, and the wires attached to the bottom and top of the flagstaff measure $34.6$ m and $60$ m respectively.
\[ \boxed{\text{Tower height} = 17.3\text{ m},\ PC = 34.6\text{ m},\ PA = 60\text{ m}} \]