\(AC\) is a diagonal of the rectangular base \(ABCD\). Triangle \(ABC\) is right-angled at \(B\), with \(AB = 7\) cm and \(BC = AD = 14\) cm.

\(AC^2 = 7^2 + 14^2 = 49 + 196 = 245\)

\(AC = \sqrt{ 245 } = 15.7\) cm

\(AG\) runs from one corner of the cuboid to the opposite one. Triangle \(ACG\) is right-angled at \(C\), because \(CG\) is vertical and \(AC\) lies flat in the base.

Use \(AC^2 = 245\) rather than the rounded \(15.7\) — rounding early is the commonest way this mark is lost.

\(AG^2 = AC^2 + CG^2 = 245 + 484 = 729\)

\(AG = \sqrt{ 729 } = 27\) cm

The longest straight line inside a cuboid runs from one corner to the opposite corner, and its square is the sum of the squares of all three edges.

\(d^2 = 25^2 + 13^2 + 18^2 = 625 + 169 + 324 = 1118\)

\(d = \sqrt{ 1118 } = 33.4\) cm

The knitting needle is \(37\) cm and the longest that fits is \(33.4\) cm, so the answer is No.

For any cuboid the square of the corner-to-corner diagonal is the sum of the squares of the three edges:

\(d^2 = 6^2 + 14^2 + h^2\)

\(31^2 = 6^2 + 14^2 + h^2\)

\(961 = 36 + 196 + h^2\)

\(h^2 = 961 - 36 - 196 = 729\)

\(h = \sqrt{ 729 } = 27\)

\(h = 27\) cm

Take the difference in each direction. Those three differences are the edges of a cuboid with \(AB\) as its corner-to-corner diagonal.

\(x\): \(13 - 10 = 3\)    \(y\): \(17 - 21 = -4\)    \(z\): \(10 - 22 = -12\)

For any cuboid the square of that diagonal is the sum of the squares of the three edges. A difference that comes out negative does no harm, because it is squared:

\(AB^2 = (3)^2 + (-4)^2 + (-12)^2 = 9 + 16 + 144 = 169\)

\(AB = \sqrt{ 169 } = 13\)

The angle wanted is \(\angle GAC\): drop \(G\) straight down onto the base and it lands on \(C\), so \(AC\) is the shadow of \(AG\) on the base.

First find \(AC\), the diagonal of the base. Triangle \(ABC\) is right-angled at \(B\), with \(BC = AD = 6\) cm.

\(AC^2 = 196 + 36 = 232\), so \(AC = \sqrt{ 232 } = 15.23\) cm

Triangle \(ACG\) is right-angled at \(C\), with \(CG = 15\) cm opposite the angle and \(AC\) adjacent to it, so use tangent — and keep \(\sqrt{ 232 }\) in the calculator rather than the rounded value.

\(\tan \angle GAC = \frac{ 15 }{ \sqrt{ 232 } }\)

\(\angle GAC = 44.6^\circ\)

Every sloping edge runs from a corner of the base to \(V\), and \(V\) is above the centre of the square. So the right-angled triangle to use has the sloping edge as its hypotenuse, the height as one short side and half a diagonal of the base as the other.

The diagonal of the square base is \(AC\), and triangle \(ABC\) is right-angled at \(B\):

\(AC^2 = 10^2 + 10^2 = 200\)

Half of that diagonal is what the triangle needs, and halving a length squares to a quarter:

\(\left(\frac{ AC }{2}\right)^2 = \frac{ 200 }{4} = 50\)

\(x^2 = 256 + 50 = 306\)

\(x = \sqrt{ 306 } = 17.5\) cm

\(V\) drops onto the centre of the base, and the centre lies on the diagonal through that corner, so the angle wanted is the angle between the sloping edge and half of that diagonal.

The height of \(16\) cm is opposite the angle and half the diagonal, \(7.07\) cm, is adjacent to it, so use tangent — and work from \(\sqrt{ 50 }\) rather than from the rounded answer to part (a).

\(\tan(\text{angle}) = \frac{ 16 }{ \sqrt{ 50 } }\)

The angle is \(66^\circ\), to the nearest degree.

The height drops from \(V\) to the centre of the base. Join that centre out to \(A\), and three lengths close up into a right-angled triangle: the height, half the diagonal of the base, and the sloping edge as the hypotenuse.

What makes this the awkward way round is that the hypotenuse is the length being given and a short side is the one being asked for.

Start with the whole diagonal of the square base. Triangle \(ABC\) is right-angled at \(B\):

\(AC^2 = 20^2 + 20^2 = 800\)

Only half of that diagonal is in the triangle, and halving a length quarters its square:

\(\left(\frac{ AC }{2}\right)^2 = \frac{ 800 }{4} = 200\)

Now write Pythagoras with the sloping edge as the hypotenuse and rearrange it, rather than adding two squares as usual:

\(44^2 = h^2 + 200\)

\(h^2 = 1936 - 200 = 1736\)

\(h = \sqrt{ 1736 } = 41.7\) cm

Call the centre of the base \(O\), so \(VO\) is the perpendicular height.

The angle between two flat surfaces is measured where they meet, along a line drawn at right angles to that meeting edge in each surface. The face \(VAB\) meets the base along \(AB\), and \(VM\) and \(OM\) are both perpendicular to \(AB\) — so the angle wanted is \(\angle VMO\).

Triangle \(VOM\) is right-angled at \(O\). \(OM\) is half a base edge, not half a diagonal, because \(M\) is the middle of a side:

\(OM = 24 \div 2 = 12\) cm

\(VO = 46\) cm is opposite the angle and \(OM\) is adjacent to it, so use tangent:

\(\tan \angle VMO = \frac{ 46 }{ 12 }\)

\(\angle VMO = 75.4^\circ\)

Half the diagonal would give the angle a sloping edge makes with the base, which is a smaller angle and a different question.

\(AF\) is not an edge of the prism — it crosses the sloping face, from a bottom corner at one end to the top corner at the other. Two Pythagoras steps reach it, and the first is the one that is easy to miss.

First the sloping side of one end, \(AC\). The perpendicular height of an isosceles triangle meets the base at its midpoint, so the triangle to use has a short side of \(14 \div 2 = 7\) cm:

\(AC^2 = 49 + 225 = 274\), so \(AC = 16.55\) cm, to 2 d.p.

\(CF\) is the length of the prism, and it is at right angles to the whole of the end face — so it is at right angles to \(AC\), and triangle \(ACF\) has its right angle at \(C\):

\(AF^2 = AC^2 + CF^2 = 274 + 676 = 950\)

Carry \(AC^2 = 274\) forward whole. Squaring \(16.55\) and working from that loses accuracy before the second step even starts.

\(AF = \sqrt{ 950 } = 30.8\) cm

The two ropes point in different directions, so nothing links them across the ground. What they share is the radio mast itself, and that is the way in: work out its height from the first rope and carry it into the second.

Triangle \(QPA\) is right-angled at \(P\), with \(QA\) as the hypotenuse and the height \(PQ\) opposite the angle of \(36^\circ\):

\(PQ = 25 \times \sin 36^\circ = 14.69\) m, to 2 d.p.

Triangle \(QPB\) is right-angled at \(P\) as well, because the radio mast is vertical and so is at right angles to every line on the ground. The same height \(PQ\) is opposite the angle wanted, and \(QB\) is the hypotenuse:

\(\sin(\text{angle}) = \frac{ PQ }{ 27 }\)

Keep the unrounded height in the calculator rather than typing \(14.69\) back in.

The angle is \(33^\circ\), to the nearest degree.