The interior angle of a regular polygon is 20^ exto more than three times the exterior angle. Find the number of sides of the polygon. 5. Simplify the expression (x2 - 5xy + 6y2)/(18y2 - 2x2).

Let $n$ be the number of sides of the regular polygon.
Let $I$ be the interior angle and $E$ be the exterior angle.
The relationship between the interior and exterior angles is $I + E = 180^\circ$.
The given condition is $I = 3E + 20^\circ$.

Substitute $I = 180^\circ - E$ into the given condition:

The formula for the exterior angle of a regular polygon is $E = \frac{360^\circ}{n}$.
Substitute the value of $E$:

The number of sides of the polygon is $\boxed{9}$.

5. Simplify the expression.

The numerator is a quadratic expression:

The denominator is $18y^2 - 2x^2$. Factor out the common term 2:

Recognize the difference of squares pattern, $a^2 - b^2 = (a-b)(a+b)$, where $a=3y$ and $b=x$:

To match a term in the numerator, we can rewrite $(3y - x)$ as $-(x - 3y)$:

Cancel the common factor $(x - 3y)$:

This can also be written as:

The simplified expression is $\boxed{\frac{2y - x}{2(x + 3y)}}$.

6. Solve for x given.

Notice that $9^x = (3^2)^x = 3^{2x} = (3^x)^2$.
Let $y = 3^x$. Substitute $y$ into the equation:

Factor the quadratic expression. We need two numbers that multiply to 81 and add to -82. These numbers are -1 and -81.

This gives two possible values for $y$:

Case 1: $y = 1$

Since $3^0 = 1$, then $x = 0$.

Case 2: $y = 81$

Since $81 = 3^4$, then $x = 4$.
The solutions for $x$ are $\boxed{0 \text{ and } 4}$.

7. A man on top of a tower 300m sees two cars P and Q on a straight level road. The angle of depression of P was $48^\circ$ and that of Q was $28^\circ$. Calculate the distance between the two cars. (Give your answer to 2.d.p)

Let the height of the tower be $H = 300$ m.
Let $B$ be the base of the tower and $T$ be the top.
The angles of depression are measured from the horizontal line at $T$.
The angle of depression to P is $48^\circ$, so the angle of elevation from P to T is $\angle TPB = 48^\circ$.
The angle of depression to Q is $28^\circ$, so the angle of elevation from Q to T is $\angle TQB = 28^\circ$.
We have two right-angled triangles: $\triangle TBP$ and $\triangle TBQ$.

In $\triangle TBP$:

In $\triangle TBQ$:

Since P and Q are on a straight level road and P is closer to the tower than Q (because $48^\circ > 28^\circ$), the distance between the cars is the difference between BQ and BP.

The distance between the two cars is $\boxed{294.11 \text{ m}}$.