KBAT: 3D Triangle Problems

Q: How do I know which angle to put inside the cosine rule?

Use the included angle, the one between the two sides you already know. Here FAFAFA and FBFBFB are known and the 120∘120^{\circ}120∘ sits between them at FFF, so it is the correct angle for AB2=FA2+FB2−2 FA⋅FBcos⁡120∘AB^{2}=FA^{2}+FB^{2}-2\,FA\cdot FB\cos 120^{\circ}AB2=FA2+FB2−2FA⋅FBcos120∘. Choosing an angle not between the two known sides is the most common slip on these questions.

Q: Do I lose marks if I round AB too early?

You risk it. Keep AB2=604AB^{2}=604AB2=604 exact and only round at the very end, or carry several decimals through part (c). Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, so method marks reward a correct chain of working, but a value rounded too soon can push the final angle or area outside the accepted range.

What makes this a KBAT question

A routine solution-of-triangles question hands you a single triangle and one rule to apply. A three-dimensional KBAT question hides several triangles inside one figure, a vertical mast, two lines drawn on the ground, two slanting wires, and never tells you which triangle to solve first or which rule fits it.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: each individual step is Form 4 work, Pythagoras, the cosine rule, the area formula, but seeing the flat diagram as a real object, spotting which triangles are right-angled and which are oblique, and matching the right tool to each is left to you. Students who sketch the situation and mark a right angle where the pole meets level ground turn a crowded 3D picture into a short chain of ordinary 2D triangles they already know how to finish.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Q1[10 marks]

A vertical mast $VF$ stands on horizontal ground, with top $V$ and foot $F$ ; its height is $VF = 24$ m. Two guy wires run from the top $V$ down to anchor points $A$ and $B$ pegged on the level ground, where $FA = 18$ m and $FB = 10$ m.

On the ground, the angle between the two anchors, $\angle AFB$ , is $120^{\circ}$ . (a) Find the length of each guy wire, $VA$ and $VB$ .

(b) Find the distance $AB$ between the two anchor points. (c) Find the angle $\angle AVB$ between the two wires at the top of the mast, and hence the area of triangle $VAB$ .

Show worked solution

Understand. The mast is vertical, so it meets the horizontal ground at right angles: $\angle VFA = \angle VFB = 90^{\circ}$ .

That makes triangles $VFA$ and $VFB$ right-angled, ideal for Pythagoras. Triangle $AFB$ lies flat on the ground and is oblique, since its angle is $120^{\circ}$ , so it needs the cosine rule.

The top triangle $VAB$ is a slanting, oblique triangle standing in space.

Plan. Use Pythagoras in the two vertical right triangles to find the wires $VA$ and $VB$ .

Use the cosine rule in the ground triangle $AFB$ to find $AB$ . Then, with all three sides of triangle $VAB$ known, use the cosine rule again for $\angle AVB$ , and the area formula $\frac{1}{2}\,VA\cdot VB\,\sin(\angle AVB)$ .

Execute and check. (a) In the vertical right triangles, apply Pythagoras:

VA=\sqrt{VF^{2}+FA^{2}}=\sqrt{24^{2}+18^{2}}=\sqrt{900}=30

VB=\sqrt{VF^{2}+FB^{2}}=\sqrt{24^{2}+10^{2}}=\sqrt{676}=26

So the guy wires are $VA = 30$ m and $VB = 26$ m.

(b) In the ground triangle $AFB$ , the $120^{\circ}$ is the included angle between $FA$ and $FB$ , so use the cosine rule with $\cos 120^{\circ}=-\tfrac{1}{2}$ :

AB^{2}=FA^{2}+FB^{2}-2\,FA\cdot FB\,\cos 120^{\circ}=18^{2}+10^{2}-2(18)(10)\left(-\tfrac{1}{2}\right)=424+180=604

AB=\sqrt{604}\approx 24.58\ \text{m}

(c) Now triangle $VAB$ has all three sides: $VA=30$ , $VB=26$ , $AB^{2}=604$ . Rearrange the cosine rule for the angle at $V$ :

\cos(\angle AVB)=\frac{VA^{2}+VB^{2}-AB^{2}}{2\,VA\cdot VB}=\frac{900+676-604}{2(30)(26)}=\frac{972}{1560}=0.6231

\angle AVB\approx 51.46^{\circ}

The area of the slant triangle then follows from the area formula:

\text{Area}=\tfrac{1}{2}\,VA\cdot VB\,\sin(\angle AVB)=\tfrac{1}{2}(30)(26)\sin 51.46^{\circ}\approx 390(0.7822)\approx 305\ \text{m}^{2}

Check. Each wire is longer than the mast itself ( $30>24$ and $26>24$ ), as any slanting wire must be, a quick sanity check that the right triangles were set up correctly.

The angle $\angle AVB\approx 51.5^{\circ}$ is smaller than the $120^{\circ}$ on the ground, which fits the picture: seen from high above, the two anchors appear drawn closer together. The cosine value $0.6231$ lies between $-1$ and $1$ , so the angle is valid.

Finding a sensible first step

With a three-dimensional figure the reliable first move is to redraw it as separate flat triangles, each on its own, before reaching for any rule. Start where the certainty is: a vertical mast meets level ground at a right angle, so mark $90^{\circ}$ at the foot $F$ in both vertical planes.

That single observation tells you triangles $VFA$ and $VFB$ are right-angled, so Pythagoras, not the cosine rule, is the quick tool for the wires. Only the flat ground triangle carries the given $120^{\circ}$ , so it is the one that needs the cosine rule.

Redrawing each triangle by itself, with its known sides and angle labelled, stops you from mixing a slant length with a ground length, and turns one crowded picture into three clean, familiar problems.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a three-dimensional triangle question a marker looks for:

A clear labelled diagram, or separate sketches, showing the right angles where the mast meets the ground.
Pythagoras applied correctly in each vertical triangle: $VA=\sqrt{24^{2}+18^{2}}=30$ and $VB=\sqrt{24^{2}+10^{2}}=26$ .
The cosine rule set up with the correct included angle, using $\cos 120^{\circ}=-\tfrac{1}{2}$ .
$AB^{2}=604$ kept exact until the final line, then rounded to $AB\approx 24.58$ m.
The cosine rule rearranged correctly for the angle, giving $\cos(\angle AVB)=\tfrac{972}{1560}$ .
The area from $\tfrac{1}{2}\,VA\cdot VB\,\sin(\angle AVB)$ , with the angle itself, not its cosine, placed inside the sine.
Units (metres, square metres) and sensible rounding shown at the end.

How a teacher helps

Three-dimensional questions reward students who slow down to redraw, and that habit grows fastest with a teacher watching over your shoulder. In a one-to-one lesson our teachers ask you to pull each triangle out of the figure and name its right angle before choosing a rule, so you never fire the cosine rule at a triangle Pythagoras would settle in one line.

We rehearse the order, vertical triangles first, ground triangle next, slant triangle last, until it feels automatic. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

Why is Pythagoras enough for the wires but not for

AB

Because a vertical mast meets level ground at a right angle, triangles $VFA$ and $VFB$ are right-angled, and Pythagoras handles right triangles directly. The distance $AB$ sits in the ground triangle $AFB$ , whose angle is $120^{\circ}$ , an oblique triangle, so it needs the cosine rule instead.

Matching the right tool to each triangle is the heart of the question.

How do I know which angle to put inside the cosine rule?

Use the included angle, the one between the two sides you already know. Here $FA$ and $FB$ are known and the $120^{\circ}$ sits between them at $F$ , so it is the correct angle for $AB^{2}=FA^{2}+FB^{2}-2\,FA\cdot FB\cos 120^{\circ}$ .

Choosing an angle not between the two known sides is the most common slip on these questions.

Do I lose marks if I round

AB

too early?

You risk it. Keep $AB^{2}=604$ exact and only round at the very end, or carry several decimals through part (c).

Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, so method marks reward a correct chain of working, but a value rounded too soon can push the final angle or area outside the accepted range.

Does a diagram really earn marks?

A labelled sketch is not usually a mark on its own, but it is the fastest way to earn the marks that follow. Marking the right angles at the foot of the mast shows the examiner why you reached for Pythagoras, and separating the ground triangle makes the correct cosine-rule setup obvious.

A clear figure turns a confusing 3D problem into steps you can score.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

KBAT: Three-Dimensional Triangle Problems