Question 16
In the diagram DC represents a 60 metre vertical tower.
A and B are two points in the same horizontal plane as the foot C of the tower.
The angle above the horizontal from A to D is 28° and the angle above the horizontal from B to D is 35°.
The bearing of C from A is 050°T and the bearing of C from B is 300°T.
Determine the distance between A and B to the nearest metre.
[4 marks]Two upright triangles, then one flat one
The tower gives you the ground distances. Then you look straight down, and the bearings give you the angle between them.
The upright triangles only give you lengths. The angle you need lives on the ground, and it comes from the bearings.
Build the ground triangle from above
This is the bird’s-eye view, with north up. First choose which two lengths are sides of triangle ABC. Then move the bearings and watch how the angle at C is built from the angles at A and B.
Start with the sides. Triangle ABC lies flat on the ground, so both of its sides at C have to be on the ground too.
AD and BD run from the ground up to the top of the tower. They are not on the ground, so they are not sides of triangle ABC. Using them gives AB ≈ m, which is too long. These come from $\frac{60}{\sin 28^\circ}$ and $\frac{60}{\sin 35^\circ}$. You want tan, because 60 m is opposite the angle and the ground distance is adjacent.
These are the question’s bearings. The angle at C is $50^\circ+60^\circ=110^\circ$, so the cosine rule gives AB ≈ 163.37 m, which is 163 m to the nearest metre.
With these bearings the angle at C is °, so AB ≈ m. The two side lengths never change. Only the angle between them does, and that changes AB.
North lines are parallel. The blue angle at A reappears at C between CA and due south, and the orange angle at B reappears between CB and due south. They are alternate angles, so the angle at C is their sum.
The bearing of C from B is measured clockwise from north, so the orange angle at B is 360° minus the bearing.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The QCAA’s sample response labels the ground distances $b=AC$ and $a=BC$, and calls the answer $c=AB$.
Start on the ground. Draw a north line at A and at B, as the sample response does. The bearing of C from A is 50° clockwise from north. The bearing of C from B is 300°T, which is 60° anticlockwise from north. Draw a north line through C as well. North lines are parallel, so both angles reappear at C on either side of due south.
$$\begin{aligned}360^\circ-300^\circ&=60^\circ\\\angle ACB&=50^\circ+60^\circ\\&=110^\circ\end{aligned}$$
Back bearings give the same angle. The bearing of A from C is 230°T and the bearing of B from C is 120°T, and $230^\circ-120^\circ=110^\circ$. This mark may be implied by your later working, for example by a correct $\cos110^\circ$ in the cosine rule.
Marker: correctly establishes the angles using given bearings
Now the two upright triangles, ACD and BCD. Each is right-angled at C. The 60 m tower is opposite the angle of elevation and the ground distance is adjacent, so use tan.
$$\begin{aligned}\tan28^\circ&=\frac{60}{b}\\b&=\frac{60}{\tan28^\circ}\\&\approx112.84\ \text{m}\end{aligned}$$
$$\begin{aligned}\tan35^\circ&=\frac{60}{a}\\a&=\frac{60}{\tan35^\circ}\\&\approx85.69\ \text{m}\end{aligned}$$
Leaving the distances in trigonometric form is accepted, for example $AC=\frac{60}{\tan28^\circ}$. Keep the full calculator values for the next step.
Marker: correctly determines the distance $a$ and $b$
In triangle ABC you now have two sides and the angle between them. That is the cosine rule.
$$\begin{aligned}c^2&=a^2+b^2-2ab\cos C\\&=85.69^2+112.84^2\\&\quad-2\times85.69\times112.84\\&\qquad\times\cos110^\circ\\&\approx26\,690\\c&\approx163.37\ \text{m}\end{aligned}$$
Equivalent decimal values are accepted. If your angle or distances had a slip, you can still earn this mark by follow-through.
Marker: determines distance between A and B
The last mark is for how you set it out. Label your diagram with $a$, $b$ and $c$ and the angles, use equals signs properly, and connect the steps with short statements such as “Distance from A to the tower”. Then finish with a sentence that answers the question, rounded as asked.
Marker: shows logical organisation communicating key steps
Each phrase of the question gave you something
No single step here is hard. What makes it complex is that three triangles, each in its own plane, have to be joined up. The two upright triangles give you lengths but no useful angle. The ground triangle needs those lengths plus an angle that only the bearings can give you. You have to see that chain before you start. It is still familiar, because the diagram is drawn for you and the ground triangle ends up as the standard cosine rule setup, with two sides and the angle between them.
If I had to guess where you might lose marks, it would be the angle at C. Using 50° or 60° on its own, or 250° from subtracting the two given bearings, puts the wrong angle in the cosine rule. A sketch from above with three north lines settles it. The other risk is the mark for communication. Name $a$, $b$ and $c$, say which triangle each line comes from, and finish with a sentence that gives 163 m.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2021 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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