Every triangle hides a neat set of relationships between its three sides and three angles. In the NDA exam, Properties of Triangles is a dependable scoring chapter because the questions plug straight into standard formulas like the sine rule and cosine rule. Learn six or seven results well and you can secure 2–4 marks almost every year with very little working.
Why This Chapter Rewards You Quickly
Properties of Triangles sits at the heart of NDA trigonometry, and the questions are usually direct and formula-based. Once you recognise which two pieces of information you are given — sides, angles, or a mix — the right formula almost picks itself, and the answer follows in two or three lines.
The topic also links neatly with Height and Distance, the sine and cosine of standard angles, and even coordinate geometry. So time spent mastering the triangle relations quietly pays off across several chapters of the paper. Cavalier teachers treat this as a foundation block: get the core six formulas solid and a whole family of questions becomes routine.
The entire chapter rests on a standard notation and roughly seven key results. In any triangle ABC, the angles are A, B, C and the sides opposite them are a, b, c. Keeping this opposite-side convention straight is the single most important habit for this chapter.
Always label first: side a is opposite angle A, side b opposite B, side c opposite C. Half the mistakes here come from mixing up which side faces which angle.
Standard Notation and Basic Facts
Before any formula, fix the language. In triangle ABC:
- The three angles satisfy A + B + C = 180° (or π radians).
- Side a = BC, side b = CA, side c = AB.
- The semi-perimeter is written s = (a + b + c)/2.
- The area of the triangle is denoted by Δ (capital delta).
The semi-perimeter s appears in a surprising number of formulas, so it is worth computing right at the start of any problem that gives you all three sides. Because the angles add to 180°, you can always find a third angle once two are known, and useful identities like sin(A + B) = sin C drop out of this fact.
Since A + B + C = 180°, we get sin(A + B) = sin C and cos(A + B) = −cos C. These two substitutions rescue many proof-style questions.
The Sine Rule (Law of Sines)
The sine rule links each side to the sine of its opposite angle. It is the formula to reach for whenever you know an angle and its opposite side.
In any triangle ABC: a/sin A = b/sin B = c/sin C = 2R
Here R is the radius of the circumscribed circle (the circumradius).
The constant ratio equalling 2R is a powerful bonus: the sine rule does not just relate the sides and angles, it also hands you the circumradius directly. So a question that gives one side and its opposite angle quietly gives you R as well.
Use the sine rule when the data is of the type angle-side-angle or angle-angle-side. Because every term has a clear opposite, you simply pair the known side with its known angle and cross-multiply to find the unknown.
If a problem mentions or asks for the circumradius R, the sine rule is almost always the fastest route, since R = a/(2 sin A).
The Cosine Rule (Law of Cosines)
The cosine rule connects the three sides to one angle. It is the tool for situations where the sine rule cannot start — that is, when you do not have a matching angle-and-opposite-side pair.
cos A = (b2 + c2 − a2) / (2bc)
cos B = (c2 + a2 − b2) / (2ca)
cos C = (a2 + b2 − c2) / (2ab)
Equivalently, a2 = b2 + c2 − 2bc·cos A. Notice that if angle A is 90°, then cos A = 0 and the formula collapses to a2 = b2 + c2 — the familiar Pythagoras theorem. So the cosine rule is really Pythagoras generalised to any angle.
Reach for the cosine rule when you are given two sides and the included angle (SAS) or all three sides (SSS). With three sides known, you can find every angle one by one.
Watch the sign: the term is minus 2bc·cos A. If the angle is obtuse its cosine is negative, which makes that side longer — do not drop the negative sign by accident.
The Projection Formula
The projection formula expresses one side as the sum of the projections of the other two sides onto it. It is short and easy to forget, yet examiners enjoy testing it.
a = b·cos C + c·cos B
b = c·cos A + a·cos C
c = a·cos B + b·cos A
The picture behind it is simple. Drop a perpendicular from the top vertex to the base; the base then splits into two pieces, and each piece is one of the other sides multiplied by the cosine of the angle next to it. Adding the pieces gives back the whole base.
In each projection line, the side on the left does not appear on the right. The right-hand side mixes only the other two sides and angles. That pattern is the quickest way to recall the formula.
Area of a Triangle — Four Ways
The area Δ can be found from whatever data you have. Keep all four forms ready.
Two sides and included angle: Δ = ½ ab·sin C = ½ bc·sin A = ½ ca·sin B
Heron's formula (three sides): Δ = √[s(s − a)(s − b)(s − c)], where s = (a + b + c)/2
Using circumradius: Δ = abc / (4R)
Using inradius: Δ = r·s
The first form, half the product of two sides times the sine of the angle between them, is the workhorse for SAS data. Heron's formula is the go-to when only the three sides are known and no angle is given.
The last two forms quietly connect area to the two special radii of a triangle. They let you slide between Δ, R and r without re-solving the whole triangle, which is exactly the kind of shortcut the NDA paper rewards.
If a question gives all three sides, compute s first, then plug into Heron's formula. Trying to find an angle before the area usually wastes time.
Circumradius R and Inradius r
Two circles sit naturally with every triangle. The circumcircle passes through all three vertices, with radius R. The incircle touches all three sides from inside, with radius r.
Circumradius: R = abc / (4Δ), and also R = a/(2 sin A)
Inradius: r = Δ / s
A neat half-angle form: r = (s − a) tan(A/2)
These two radii appear constantly because they bundle together the sides, the area and the angles. If a problem hands you the three sides, you can compute s, then Δ by Heron, then both R and r in a couple of lines.
A clean special case worth memorising: in an equilateral triangle of side a, R = a/√3 and r = a/(2√3), so R = 2r. That ratio R = 2r is a popular one-line objective question.
Do not confuse the two radii. R uses abc in the numerator (R = abc/4Δ), while r uses the area over the semi-perimeter (r = Δ/s). Mixing them up is a classic exam slip.
Half-Angle Formulas
When a question involves the half of an angle, these formulas, all built from the sides and semi-perimeter s, are exactly what you need.
sin(A/2) = √[(s − b)(s − c) / (bc)]
cos(A/2) = √[s(s − a) / (bc)]
tan(A/2) = √[(s − b)(s − c) / (s(s − a))]
Because they use only the sides through s, these formulas let you find an angle's half without first finding the angle itself. They are especially handy in proofs and in questions that ask for tan(A/2) directly.
The corresponding formulas for B/2 and C/2 follow the same pattern — just rotate the letters. Once you see the structure of the A/2 version, you never need to memorise the others separately.
Worked Example: Solving a Triangle
Let us solve a typical SSS problem the way you should in the exam hall.
In triangle ABC, a = 7, b = 8 and c = 9. Find cos A and then the area Δ.
Notice how the cosine rule handled the angle from three sides, while Heron's formula gave the area without needing any angle at all. Choosing the right tool for the given data is the whole skill here.
Common Mistakes to Avoid
Most marks lost in this chapter come from small slips, not hard ideas.
- Pairing the wrong side with an angle — always keep a opposite A, and so on.
- Dropping the minus sign in the cosine rule's −2bc·cos A term.
- Forgetting to compute the semi-perimeter s before using Heron's formula or the half-angle formulas.
- Swapping the R and r formulas (R = abc/4Δ versus r = Δ/s).
When you use the sine rule to find an unknown angle, remember an angle can have two possible values (acute or obtuse) with the same sine. Check against the triangle's other facts before committing to one answer.
Previous-Year Style Practice
Here is a question modelled on the NDA exam pattern. Try it before reading the solution.
Q. In a triangle ABC, if a = 2, b = 2√3 and angle A = 30°, then what is the value of angle B?
Answer: By the sine rule, a/sin A = b/sin B, so sin B = (b·sin A)/a = (2√3 × sin 30°)/2 = (2√3 × ½)/2 = √3/2. Therefore B = 60° or 120°. Both keep A + B < 180°, so B = 60° or 120° — the classic ambiguous (two-triangle) case.
Notice how the sine rule connected the matching pair (a, A) with (b, B) instantly, and how the question hid a subtle two-answer trap. Spotting that ambiguity is exactly what examiners test.
Quick Revision Before the Exam
Glance over these the night before your paper and the morning of it.
- Sine rule: a/sin A = b/sin B = c/sin C = 2R.
- Cosine rule: cos A = (b2 + c2 − a2)/(2bc); use for SAS or SSS.
- Projection: a = b·cos C + c·cos B (left side absent on the right).
- Area: ½ ab·sin C; Heron's √[s(s−a)(s−b)(s−c)]; abc/4R; r·s.
- R = abc/4Δ, r = Δ/s; equilateral gives R = 2r.
Drill 10–15 mixed triangle problems daily for a week before the exam. Quick, accurate use of the sine and cosine rules frees time for the heavier calculus questions.
Frequently asked questions
When should I use the sine rule versus the cosine rule?
Use the sine rule when you have an angle paired with its opposite side, such as angle-side-angle data. Use the cosine rule when you have two sides with the included angle (SAS) or all three sides (SSS), because there is no matching angle-side pair to start with.
How many questions come from Properties of Triangles in the NDA exam?
Typically 2 to 4 questions appear across the NDA Maths paper, often blended with height and distance. They are mostly direct formula-based problems, making this a reliable scoring area once the standard formulas are memorised.
What is the difference between the circumradius R and the inradius r?
The circumradius R is the radius of the circle passing through all three vertices, found from R = abc/(4Δ). The inradius r is the radius of the circle touching all three sides from inside, found from r = Δ/s, where s is the semi-perimeter.
Why does the sine rule sometimes give two possible angles?
Because two different angles, one acute and one obtuse, can share the same sine value (for example, sin 60° equals sin 120°). This is the ambiguous case; you must check the result against the triangle's other facts to decide which angle is valid.
Which area formula is best when only the three sides are known?
Heron's formula is best: first compute the semi-perimeter s = (a + b + c)/2, then use Δ = √[s(s - a)(s - b)(s - c)]. It needs no angle at all, so it is faster than finding an angle and then the area.
Related NDA Maths topics
Want a teacher to walk you through NDA Maths?
Cavalier's NDA batches break every topic into classroom sessions with daily practice, tests and doubt-clearing.