(a) In the diagram, \(\Delta\) ABD is right-angled at B. |AB| = 3 cm, |AD| = 5 cm, \(\stackrel\frown{ACB}\) = 61° and \(\stackrel\frown{DAC}\) = x°. Calcula...

(a) Value of x

In the diagram \(\triangle ABD\) is right-angled at \(B\), with \(|AB| = 3\ \text{cm}\) and hypotenuse \(|AD| = 5\ \text{cm}\). Point \(C\) lies on \(BD\), \(\angle ACB = 61^\circ\) and \(\angle DAC = x^\circ\).

Step 1 - the whole angle at A. In right triangle \(ABD\), \(AB\) is adjacent to \(\angle BAD\) and \(AD\) is the hypotenuse, so

\[\cos(\angle BAD)=\frac{|AB|}{|AD|}=\frac{3}{5}=0.6,\qquad \angle BAD=\cos^{-1}(0.6)=53.13^\circ.\]

Step 2 - the lower part of that angle. Triangle \(ACB\) is also right-angled at \(B\), so its three angles give

\[\angle BAC=180^\circ-90^\circ-61^\circ=29^\circ.\]

Step 3 - subtract. Since \(C\) lies on \(BD\), the angle \(\angle DAC\) is the difference between the whole angle at \(A\) and \(\angle BAC\):

\[x=\angle BAD-\angle BAC=53.13^\circ-29^\circ=24.13^\circ.\]

Correct to one decimal place, \(x = \mathbf{24.1}\).

(b) The pyramid

\(OABCD\) is a pyramid on a square base of side \(2\ \text{cm}\) with a slant height of \(4\ \text{cm}\). In a pyramid the slant height is the distance from the apex \(O\) down the middle of a triangular face to the midpoint of a base edge (call that midpoint \(M\)). It is not the sloping corner edge \(OA\); mixing the two up is the usual error here. The foot \(N\) of the vertical height is the centre of the square base.

(i) Vertical height

\(N\) is the centre of the base and \(M\) is the midpoint of a base edge, so \(NM\) is half the side:

\[NM=\tfrac{1}{2}\times 2=1\ \text{cm}.\]

The vertical height \(h=ON\), the slant height \(l=OM=4\ \text{cm}\) and \(NM=1\ \text{cm}\) form a right triangle at \(N\). By Pythagoras' theorem,

\[l^{2}=h^{2}+NM^{2}\ \Rightarrow\ h^{2}=4^{2}-1^{2}=16-1=15,\]

\[h=\sqrt{15}=3.873\ \text{cm}.\]

Correct to three significant figures, \(h = \mathbf{3.87\ \text{cm}}\).

(ii) Volume

\[V=\frac{1}{3}\times(\text{base area})\times h=\frac{1}{3}\times(2\times 2)\times\sqrt{15}=\frac{4\sqrt{15}}{3}.\]

\[V=\frac{4\times 3.873}{3}=\frac{15.49}{3}=5.164\ \text{cm}^{3}.\]

Correct to three significant figures, \(V = \mathbf{5.16\ \text{cm}^{3}}\).