Properties of triangles for NDA

Properties of triangles is where trigonometry stops being about angles on a unit circle and starts measuring real figures. The NDA questions here are mostly one-formula questions: a missing side, an area, a circumradius, the type of a triangle. The skill is choosing the right formula from what is given. With two sides and the included angle, reach for the cosine rule or ½bc sin A. With all three sides, use Heron. With angles and one side, use the sine rule.

Typical question types are:

• finding a side or an angle with the sine or cosine rule;
• the area from three sides, or from two sides and an angle;
• R and r for a given triangle, especially right-angled and equilateral ones;
• deciding whether a triangle is acute, right or obtuse from its sides;
• identifying the triangle from a condition such as a cos A = b cos B;
• sides in the ratio of the sines of given angles.

Notation

In triangle ABC, a, b and c are the sides opposite the angles A, B and C. s = (a + b + c)/2 is the semi-perimeter, Δ is the area, R the circumradius and r the inradius. Always A + B + C = 180°.

The sine rule

a/sin A = b/sin B = c/sin C = 2R.

Why 2R: draw the circumcircle and the diameter BD through B. Angle BDC equals A, because both stand on the chord BC (angles in the same segment). Angle BCD is 90°, because it is the angle in a semicircle. In the right triangle BCD, sin A = BC/BD = a/(2R).

Use the sine rule when you know two angles and a side, or two sides and an angle opposite one of them. It also gives a quick corollary: the sides are in the ratio sin A : sin B : sin C, so the largest side faces the largest angle.

The cosine rule and projection formulas

a² = b² + c² − 2bc cos A, or equivalently cos A = (b² + c² − a²)/(2bc). The same holds cyclically for b and c.

It generalises Pythagoras, and it gives the fastest test of a triangle's type. Let a be the longest side:

The projection formulas, a = b cos C + c cos B (and cyclic), say that side a is the sum of the projections of the other two sides onto it. They are handy for proving identities.

Area formulas

The last three rows give the two radius formulas you actually use: R = abc/(4Δ) and r = Δ/s.

Half-angle formulas

• 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))] = r/(s − a)

They appear less often than the rules above, but tan(A/2) = r/(s − a) links the inradius to the angles neatly.

Special triangles worth memorising

• Right-angled with hypotenuse c: R = c/2, since the hypotenuse is a diameter; r = (a + b − c)/2.
• Equilateral with side a: R = a/√3, r = a/(2√3), so R = 2r. Area (√3/4)a².
• Angles 30°, 60°, 90°: sides in the ratio 1 : √3 : 2.
• Angles 45°, 45°, 90°: sides in the ratio 1 : 1 : √2.

Worked NDA-style MCQs

Q1. A triangle has sides 13, 14 and 15. Its inradius and circumradius are:
(a) 4 and 65/8 (b) 3 and 65/8 (c) 4 and 7 (d) 4 and 8

s = 21, and Δ = √(21 × 8 × 7 × 6) = √7056 = 84. r = Δ/s = 84/21 = 4. R = abc/(4Δ) = (13 × 14 × 15)/336 = 2730/336 = 65/8. Answer: (a).

Q2. The largest angle of a triangle with sides 3, 5 and 7 is:
(a) 90° (b) 120° (c) 135° (d) 150°

It faces the side 7. cos C = (9 + 25 − 49)/(2 × 3 × 5) = −15/30 = −1/2, so C = 120°. Answer: (b).

Q3. The angles of a triangle are in the ratio 1 : 2 : 3. The sides are in the ratio:
(a) 1 : 2 : 3 (b) √3 : 1 : 2 (c) 1 : 1 : √2 (d) 1 : √3 : 2

The angles are 30°, 60° and 90°. Sides are in the ratio sin 30° : sin 60° : sin 90° = 1/2 : √3/2 : 1 = 1 : √3 : 2. Answer: (d).

Q4. In a triangle, b² + c² − a² = bc. Angle A is:
(a) 30° (b) 45° (c) 60° (d) 120°

cos A = (b² + c² − a²)/(2bc) = bc/(2bc) = 1/2, so A = 60°. Answer: (c).

Q5. If a cos A = b cos B, the triangle is:
(a) always isosceles (b) always right-angled (c) isosceles or right-angled (d) equilateral

By the sine rule, sin A cos A = sin B cos B, so sin 2A = sin 2B. Either 2A = 2B (A = B, isosceles) or 2A = 180° − 2B (A + B = 90°, right-angled at C). Answer: (c).

Q6. In triangle ABC, A = 45°, B = 60° and a = 2. Then b equals:
(a) √3 (b) √2 (c) 2√3 (d) √6

b = a sin B/sin A = 2 × (√3/2) ÷ (1/√2) = √3 × √2 = √6. Answer: (d).

Common mistakes

• Using ½ab sin C with the wrong angle. The angle must be the one between the two sides you use.
• Testing the type with a side that is not the longest. Only the longest side can face an obtuse angle.
• Forgetting the square root in Heron's formula, or computing s as the full perimeter.
• Writing R = abc/Δ. It is abc/(4Δ).
• Assuming a unique triangle from two sides and a non-included angle. The sine rule can give two possible angles (one acute, one obtuse); check both against the angle sum.

Practice set

1. The area of a triangle with sides 7, 8 and 9 is: (a) 12√5 (b) 24 (c) 6√5 (d) 20
2. The circumradius of a right triangle with hypotenuse 10 is: (a) 10 (b) 4 (c) 5√2 (d) 5
3. If b = 5, c = 8 and A = 60°, then a equals: (a) 6 (b) √89 (c) 7 (d) 8
4. For an equilateral triangle of side 6, R and r are: (a) 2√3 and √3 (b) 3 and 1.5 (c) 6 and 3 (d) √3 and 2√3
5. If b = 6, c = 8 and A = 30°, the area is: (a) 12 (b) 24 (c) 12√3 (d) 48
6. The inradius of the 5-12-13 triangle is: (a) 1 (b) 2 (c) 3 (d) 2.5
7. A triangle with sides 2, 3 and 4 is: (a) acute-angled (b) right-angled (c) obtuse-angled (d) impossible
8. In a triangle with A = 30° and a = 5, the circumradius is: (a) 2.5 (b) 5 (c) 10 (d) 5√3

Answers:

1. (a). s = 12; √(12 × 5 × 4 × 3) = √720 = 12√5.
2. (d). R is half the hypotenuse.
3. (c). a² = 25 + 64 − 2 × 5 × 8 × ½ = 49.
4. (a). R = 6/√3 = 2√3 and r = 6/(2√3) = √3.
5. (a). ½ × 6 × 8 × sin 30° = 12.
6. (b). r = (5 + 12 − 13)/2 = 2. Check: Δ/s = 30/15 = 2.
7. (c). 4² = 16 > 2² + 3² = 13, so the largest angle is obtuse (its cosine is −1/4).
8. (b). 2R = a/sin A = 5/(1/2) = 10.

What to do next

• Fill in the area and radius tables from memory, then prove the sine rule once on paper.
• Solve 20 old NDA questions from this chapter, and for each one write which formula the given data pointed to.
• Revise the identities you used here in trigonometric identities, and practise right-triangle applications in heights and distances.