Heights and Distances, and the Properties of Triangles
These two topics are applied trigonometry, and they need no identity the shared trigonometry chapter has not already given you. What they need is a drawing. Almost every mark lost here is lost before any trigonometry begins, in translating a sentence into a figure: which angle is at the observer, which length is the unknown, and whether the two observation points are on the same side of the object or opposite sides. So the method is fixed. Draw first, label second, write the equations third, and only then solve. For heights and distances that means marking the vertical object, putting the observer on the ground and writing tan(angle) = height / distance for each observation — two observations give two equations and two unknowns, and the algebra is always short. For triangle properties it means labelling sides a, b, c opposite to angles A, B, C and then choosing between the sine rule and the cosine rule on one question: do I know an angle and the side opposite it? If yes, sine rule. If I know three sides, or two sides and the angle between them, cosine rule. That single decision is most of the topic.
Properties of Triangles — Which Rule, and When
| Rule | Statement | Use it when |
|---|---|---|
| Sine rule | a/sin A = b/sin B = c/sin C = 2R | You know an angle and the side opposite it, plus one more part. Also the source of the circumradius, since R = a/(2 sin A). |
| Cosine rule | c² = a² + b² − 2ab·cos C | You know three sides, or two sides and the included angle. Rearranged: cos C = (a² + b² − c²)/(2ab). |
| Area | ½·ab·sin C, or Heron’s √(s(s−a)(s−b)(s−c)) | Two sides and the included angle → the sine form. Three sides → Heron. |
| Projection | a = b·cos C + c·cos B | Asked verbatim as a "which of these is true" option more often than it is used to compute anything. |
🎯 Why the choice of rule is the whole question
Given three sides and asked for an angle, the sine rule cannot start — it needs an angle to begin with, and you have none. Given an angle and its opposite side, the cosine rule will work but takes three times as long. So the first line you write decides how long the question takes, and reading the given parts before choosing is worth the ten seconds. The test is one question: is there an angle paired with the side opposite it?Key takeaways
- Draw first, label second, write the tangent equations third, solve last. Most marks lost in heights and distances are lost in the translation from sentence to figure.
- The angle of elevation from A to B equals the angle of depression from B to A, so a problem set at the top of a tower can be redrawn from the ground.
- tan 45° = 1, so at 45° the horizontal distance equals the height. tan 30° = 1/√3 and tan 60° = √3 are the only other values the paper uses.
- The two-observation problem is one subtraction: write both distances as h/tan θ, set their difference to the given gap, and solve for h.
- Choose the rule by one test: an angle paired with the side opposite it means sine rule; three sides, or two sides and the included angle, means cosine rule.
- Area is ½·ab·sin C when you have the included angle and Heron’s formula when you have three sides. The circumradius comes free from the sine rule: R = a/(2 sin A).
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The angle of elevation of the top of a tower from a point 30 m away on level ground is 45°. The height of the tower is:
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Answer: B — 30 m
tan 45° = height / distance, and tan 45° = 1, so the height equals the distance: 30 m. This is the reason examiners use 45° — the arithmetic disappears entirely and the question becomes a test of whether you drew the figure the right way up. Options C and D are what you get from using tan 60° or tan 30° instead, which is the error of reading the angle off the wrong vertex.A ladder 15 m long leans against a wall, making an angle of 60° with the ground. The height up the wall that it reaches is:
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Answer: C — 15√3/2 m
The ladder is the hypotenuse, not a leg, so the height is 15·sin 60° = 15 × (√3/2) = 15√3/2 ≈ 12.99 m. Option B, 15√3, comes from using tan 60° — which would be right if 15 were the distance along the ground rather than the ladder itself, and confusing the two is the single error this question exists to catch. Option A is 15·cos 60° = 7.5, the distance of the foot of the ladder from the wall, which is a real quantity but not the one asked for.The angle of elevation of the top of a tower is 60° from a point on the ground, and 30° from a point 40 m further away in the same straight line. The height of the tower is:
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Answer: B — 20√3 m
Write both ground distances in terms of the height h. From the near point, tan 60° = h/d₁ so d₁ = h/√3. From the far point, tan 30° = h/d₂ so d₂ = h√3. The far point is 40 m beyond the near one, so h√3 − h/√3 = 40. Combine over √3: h(3 − 1)/√3 = 2h/√3 = 40, giving h = 20√3 ≈ 34.6 m. Check it: d₁ = 20√3/√3 = 20 and d₂ = 20√3·√3 = 60, and 60 − 20 = 40 as stated. Option C is what you get from adding the two expressions instead of subtracting.From the top of a lighthouse, the angle of depression of a boat is 30°. The angle of elevation of the top of the lighthouse from the boat is:
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Answer: A — 30°
The horizontal at the top of the lighthouse and the horizontal at the boat are parallel, and the line of sight is a transversal, so the angle of depression from the top and the angle of elevation from the boat are alternate angles and therefore equal — 30°. No height is needed, which is why option D is wrong. Option B, 60°, comes from treating the two angles as complementary, which they are not: they are equal, and the 90° − θ relationship belongs to the other acute angle of the right triangle, at the top of the tower.In a triangle, a = 7, b = 8 and the included angle C = 60°. The area of the triangle is:
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Answer: B — 14√3 square units
Two sides and the angle between them is exactly the case for the sine form: area = ½·ab·sin C = ½ × 7 × 8 × sin 60° = 28 × (√3/2) = 14√3 ≈ 24.25 square units. Option D, 28√3, forgets the ½. Option C, 28, uses sin 90° = 1 in place of sin 60°, which is what happens if you read "included angle" as "right angle". Heron’s formula would also work here but only after finding the third side by the cosine rule first — three times the work for the same answer.In a triangle with sides 5, 7 and 8, the angle opposite the side of length 7 is:
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Answer: C — 60°
Three sides and no angle means the cosine rule, in its rearranged form. With the 7 opposite the angle we want, the other two sides are 5 and 8: cos θ = (5² + 8² − 7²)/(2 × 5 × 8) = (25 + 64 − 49)/80 = 40/80 = 1/2, so θ = 60°. The sine rule cannot even be started here — it needs an angle to begin with, and none is given, which is the point of the question. Option D is worth ruling out on sight: 5² + 7² = 74 ≠ 64, so the triangle is not right-angled anywhere.In a triangle, a = 5 and A = 30°. If B = 90°, then b equals:
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Answer: C — 10
An angle with its opposite side is given, so the sine rule applies directly: a/sin A = b/sin B, giving b = a·sin B/sin A = 5 × sin 90°/sin 30° = 5 × 1/(1/2) = 10. The result is worth recognising without the formula: in a right triangle the side opposite the right angle is the hypotenuse, and the side opposite a 30° angle is half the hypotenuse — so a = 5 forces b = 10. Option B is the third side, c = 5√3, which follows from C = 60°.In a triangle, a = 10 and A = 30°. The radius of its circumcircle is:
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Answer: B — 10
The sine rule carries the circumradius in its last term: a/sin A = b/sin B = c/sin C = 2R. So 2R = 10/sin 30° = 10/(1/2) = 20, and R = 10. Option C is 2R itself, which is what you get by stopping one step early — and it is the commonest wrong answer here, because the extended sine rule is usually memorised as "= 2R" and the 2 is then forgotten. Note that one side and its opposite angle are enough: the circumradius does not depend on the other two sides at all.