2024 Paper 1, Q18
The diagram shows some dimensions of a large storage container that is a rectangular prism. The angle $ABC$ is $60^\circ$.
A person requires a container that is at least 4 metres in height.
Make a justified decision about whether this storage container meets the person’s requirements.
Two base edges are given and the height is not. The only other fact is an angle, and it is an angle you cannot see on any single face.
The angle tells you the height
Keep the base at 4 m by 3 m and stretch the container upwards. Angle $ABC$ opens up as the box gets taller, so being told the angle is $60^\circ$ is the same as being told the height.
the height runs 2 m to 7 m and back, pausing where the angle reaches $60^\circ$
Three sides on three different faces, with one unknown
$AB$ lies across the top, over the 4 m and 3 m edges, so $AB=\sqrt{4^2+3^2}=5$. There is no $h$ in it at all.
$BC$ is the diagonal of the 4 m by $h$ face, so $BC=\sqrt{16+h^2}$.
$AC$ is the diagonal of the 3 m by $h$ face, so $AC=\sqrt{9+h^2}$.
You normally use it to find an angle. Here the angle is the given and the side is the unknown. And because $h^2$ appears on both sides, it cancels.
One height makes the angle exactly 60°
Drag the height. The angle is $\cos^{-1}\!\left(\frac{3.2}{\sqrt{16+h^2}}\right)$, so it gets bigger as $h$ increases. Only one value of $h$ gives $60^\circ$, and it is just under 5 m.
Why can you not just use right-angle trigonometry?
QCAA marking guide · 5 marks
Let $h$ be the height of the rectangular prism.
$$BC^2=4^2+h^2\ \Rightarrow\ BC=\sqrt{16+h^2}$$
Marker: correctly expresses $BC$ in terms of the height. Any variable is accepted, and QCAA accepts the expressions written straight onto the diagram.
$$AC=\sqrt{3^2+h^2}=\sqrt{9+h^2},\qquad AB=\sqrt{4^2+3^2}=5$$
Marker: correctly expresses $AC$ in terms of the height. Note each of the three sides lives on a different face. That is the part of this question that is genuinely three-dimensional.
$$9+h^2=\left(16+h^2\right)+25-2\times 5\sqrt{16+h^2}\times\tfrac12$$
$$5\sqrt{16+h^2}=32\ \Rightarrow\ \sqrt{16+h^2}=\frac{32}{5}\ \Rightarrow\ h=\sqrt{\left(\frac{32}{5}\right)^2-16}$$
Marker: determines an expression for the height of the container. The $h^2$ terms cancel, so what looked like a quadratic is a one-step rearrangement. FT marks allowed.
$$h=\sqrt{6.4^2-16}>\sqrt{6^2-16}=\sqrt{20}>\sqrt{16}=4$$
The height is more than 4 m, so the container does meet the requirement
Marker: decides if the container meets the requirements. QCAA’s own response never evaluates $6.4^2$. It replaces 6.4 by 6 to get a lower bound of $\sqrt{20}$, which already beats 4. The exact height is $\frac{4\sqrt{39}}{5}\approx 5.00$ m.
“Make a justified decision” is the instruction. Your decision sentence has to be there, and it has to follow from working that supports it.
Marker: shows logical organisation communicating key steps, this mark can only be awarded if previous evidence supports the decision. Label the lengths, define $h$, substitute into the named formula, and write the conclusion as a sentence.
Proving the container is tall enough without a calculator
$\frac{4\sqrt{39}}{5}\approx 5.00$ m
$BC=\frac{32}{5}=6.4$ m exactly, so $h^2=\frac{624}{25}$. Comfortably over the 4 m the person needs.
On a technology-free paper you do not need $6.4^2=40.96$. Any lower bound above 4 settles the decision, and $\sqrt{20}$ takes one line.
QCAA also accepts a vector method for the height. And accepts the expressions written directly onto the diagram rather than restated in prose.
The question never mentions a triangle, a diagonal or the cosine rule. You have to find triangle $ABC$ inside the solid, notice that its three sides sit on three different faces, and write all three in terms of the one thing you were not given. Then you use the cosine rule backwards, because the angle is given and a side is the unknown. The last hidden step is the one students skip when they are short of time. The question says “make a justified decision”, and a height with no sentence after it does not get the final mark. I would always finish with a sentence that answers the question, even if you are rushing.
Question wording and marking-guide steps are from the 2024 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.