O 是等边三角形的中心。 三角形 ABC 是一个棱锥的底面。
顶点 X 垂直位于 O 的正上方，且 AX = 18 cm。

1. Find the height of the equilateral triangle from vertex $B$ to side $AC$.

Let $M$ be the midpoint of $AC$.

Since $ABC$ is equilateral, $AB=AC=8\text{ cm}$ and $AM=4\text{ cm}$.

In right-angled triangle $ABM$,
$BM=\sqrt{8^2-4^2}$
$BM=\sqrt{64-16}=\sqrt{48}$

2. Use the centre property of an equilateral triangle to find $OA$.

The centre $O$ of an equilateral triangle lies $\dfrac{2}{3}$ of the way along the median from a vertex.

So
$OA=\frac{2}{3}BM$
$OA=\frac{2}{3}\sqrt{48}$
$OA=4.6188\ldots$
$OA=4.619\text{ cm} \quad \text{to 3 d.p.}$

1. Find the area of the equilateral triangular base $ABC$.

For equilateral triangle $ABC$, each side is $8\text{ cm}$.

Height of the base triangle:
$h=\sqrt{8^2-4^2}=\sqrt{64-16}=\sqrt{48}$

Area of base:
$\frac12\times 8\times \sqrt{48}=27.7128\ldots\text{ cm}^2$

2. Find the area of one triangular side face of the pyramid.

Each side face, for example triangle $XAC$, is isosceles with
$XA=XC=18\text{ cm},\quad AC=8\text{ cm}.$

The perpendicular height of triangle $XAC$ bisects $AC$, so half the base is $4\text{ cm}$.

Height of one side face:
$\sqrt{18^2-4^2}=\sqrt{324-16}=\sqrt{308}$

Area of one side face:
$\frac12\times 8\times \sqrt{308}=70.180\ldots\text{ cm}^2$

3. Add the base area and the areas of the three congruent side faces.

Total surface area
$=27.7128\ldots+3(70.180\ldots)$
$=238.253\ldots$

$\approx 238\text{ cm}^2$

1. Apply the cosine rule to angle $PQR$.

In $\triangle PQR$, angle $PQR$ is opposite side $PR$.

Using the cosine rule,
$PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos \angle PQR$
$3.5^2 = 5^2 + 4^2 - 2(5)(4)\cos \angle PQR$

2. Solve for the angle $PQR$.

$\cos \angle PQR = \frac{5^2+4^2-3.5^2}{2(5)(4)}$
$\cos \angle PQR = \frac{25+16-12.25}{40} = 0.71875$
$\angle PQR = \cos^{-1}(0.71875) = 44.047\ldots^\circ$

So, to 1 decimal place,
$\angle PQR = 44.0^\circ$

1. Find $AQ$ and hence $OQ$ in the base triangle.

Since $AC=8$ cm and $AQ:QC=3:1$,
$AQ=\frac{3}{4}\times 8=6\text{ cm}.$
From part (a), $OA=4.619$ cm and $O$ lies on the line of symmetry of the equilateral triangle. Hence $O$ is horizontally $4$ cm from $A$ along $AC$ and vertically above the midpoint of $AC$.

Since $Q$ is on $AC$ and $AQ=6$ cm, while the midpoint of $AC$ is $4$ cm from $A$,
$OQ=6-4=2\text{ cm}.$

2. Find the vertical height $XO$ of the pyramid.

Since $X$ is vertically above $O$, $XO\perp OA$.

In right-angled triangle $XOA$,
$XO^2=AX^2-OA^2$
$XO=\sqrt{18^2-4.619^2}$
$XO=17.397\text{ cm}\;\text{(approximately)}.$

3. Use the tangent ratio to find the angle of elevation from $Q$ to $X$.

Let the angle of elevation of $X$ from $Q$ be $\theta$.

In right-angled triangle $XOQ$,
$\tan \theta=\frac{XO}{OQ}$
$\tan \theta=\frac{17.397}{2}.$

4. Evaluate the angle and round appropriately.

$\theta=\tan^{-1}\left(\frac{17.397}{2}\right)$
$\theta=83.44^\circ\;\text{(approximately)}.$