Solution of the triangle

Geometry is an important branch of mathematics, and triangles are among its most basic yet useful shapes. A triangle is a closed figure made up of three sides and three angles. We often come across real-life situations where we need to calculate unknown sides, angles, heights, or distances that form a triangle. The solution of a triangle means finding all its unknown parts when some information is already given.

There are different types of triangles, such as scalene, isosceles, equilateral, right-angled, acute-angled, and obtuse-angled. Solving a triangle depends on the type of triangle and the information available. For example, if we know two sides and the included angle, or two angles and one side, we can use mathematical rules to find the missing sides and angles.

Some important tools used in solving triangles are the sine rule, cosine rule, tangent rule, and Pythagoras theorem. These formulas connect the sides and angles of a triangle in a systematic way. For right-angled triangles, trigonometric ratios like sine, cosine, and tangent are very helpful. In non-right triangles, the sine and cosine rules make the process easier.

The solution of triangles is not only a topic in mathematics but also has many real-world applications. It is used in engineering, navigation, astronomy, architecture, surveying, and even in daily life, such as calculating the height of a building or the distance across a river.

Thus, learning how to solve a triangle builds a strong base for advanced studies in mathematics and science. With simple formulas and logical steps, we can find unknown values and apply them in practical problems. The solution of triangles is therefore both an important and interesting topic that connects theory with real-life situations.

Semi-perimeter and Area

The semi-perimeter of the triangle is written as s = (a + b + c)/2, and its area denoted by S or Δ.

Sine Rule

a/sin A = b/sin B = c/sin C = 1/(2R)

where R is the radius of the circumcircle of the △ABC.

Cosine Rule

• cos A = (b² + c² - a²)/(2bc)
• cos B = (a² + c² - b²)/(2ac)
• cos C = (a² + b² - c²)/(2ab)

Projection Rule

• a = b cos C + c cos B
• b = c cos A + a cos C
• c = a cos B + b cos A

Napier's Analogy

• tan((B-C)/2) = ((b-c)/(b+c)) cot(A/2)
• tan((C-A)/2) = ((c-a)/(c+a)) cot(B/2)
• tan((A-B)/2) = ((a-b)/(a+b)) cot(C/2)

Area of a Triangle

Area = (1/2)bc sin A = (1/2)ca sin B = (1/2)ab sin C

= √[s(s-a)(s-b)(s-c)] = abc/(4R) = rs

where R and r are the radii of the circumcircle and the incircle of the △ABC respectively.

Trigonometric Ratios of Half-Angles

• sin(A/2) = √[(s-b)(s-c)/(bc)]
• sin(B/2) = √[(s-c)(s-a)/(ca)]
• sin(C/2) = √[(s-a)(s-b)/(ab)]
• cos(A/2) = √[s(s-a)/(bc)]
• cos(B/2) = √[s(s-b)/(ca)]
• cos(C/2) = √[s(s-c)/(ab)]
• tan(A/2) = √[(s-b)(s-c)/(s(s-a))]
• tan(B/2) = √[(s-c)(s-a)/(s(s-b))]
• tan(C/2) = √[(s-a)(s-b)/(s(s-c))]

Illustration 1

Problem: If the angles of a triangle are 30° and 45°, and the included side is (√3 + 1) cm, then the area of the triangle is

(A) (√3 + 1)

(B) (1/2)(√3 + 1)

(C) 2(√3 + 1)

(D) (1/2)(√3 - 1)

Solution: (B)

We have: sin A/a = sin B/b = sin C/c

⇒ sin 105°/(√3 + 1) = sin 30°/b = sin 45°/c

⇒ b = (√3 + 1)/(2 sin 105°), c = (√3 + 1)/(√2 sin 105°)

Area of △ABC = (1/2)bc sin A = (1/2)bc sin 105°

= (1/2) × [(√3 + 1)²/(2√2 sin 105°)] × sin 105°

= (√3 + 1)²/(4√2) × (√3 + 1)/(2√2)

= (1/2)(√3 + 1)

Illustration 2

Problem: The sides of a triangle are 8 cm, 10 cm and 12 cm. The ratio of the greatest angle to the smallest angle is

(A) 1 : 2

(B) 1 : 3

(C) 2 : 1

(D) 2 : 3

Solution: (C)

Let a = 8cm, b = 10cm and c = 12cm. Hence the greatest angle is C and the smallest angle is A.

Applying cosine rule:

cos C = (a² + b² - c²)/(2ab) = (64 + 100 - 144)/(2 × 8 × 10) = 20/160 = 1/8

cos A = (b² + c² - a²)/(2bc) = (100 + 144 - 64)/(2 × 10 × 12) = 180/240 = 3/4

Now cos 2A = 2cos²A - 1 = 2(9/16) - 1 = 9/8 - 1 = 1/8

⇒ cos C = cos 2A ⇒ C = 2A

∴ C/A = 2/1

In-radius

r = Δ/s = (s-a)tan(A/2) = (s-b)tan(B/2) = (s-c)tan(C/2)

r = 4R sin(A/2) sin(B/2) sin(C/2)

Ex-radii

r₁ = Δ/(s-a) = s tan(A/2) = 4R sin(A/2) cos(B/2) cos(C/2)

r₂ = Δ/(s-b) = s tan(B/2) = 4R sin(B/2) cos(C/2) cos(A/2)

r₃ = Δ/(s-c) = s tan(C/2) = 4R sin(C/2) cos(A/2) cos(B/2)

r₁ + r₂ + r₃ = 4R + r

1/r₁ + 1/r₂ + 1/r₃ = 1/r

r₁r₂ + r₂r₃ + r₃r₁ = s²

Area of Regular Polygon

Area = (na²/4)cot(π/n) = (nR²/2)sin(2π/n) = nr²tan(π/n)

Where a is length of side, n is number of sides of polygon, R is radius of circumscribing circle and r is radius of incircle of the polygon.

Illustration 3

Problem: The ex-radii r₁, r₂, r₃ of △ABC are in H.P. Then the sides a, b, c are in

(A) A.P.

(B) G.P.

(C) H.P.

(D) none of these

Solution: (A)

r₁, r₂, r₃ are in H.P.

⇒ 2/r₂ = 1/r₁ + 1/r₃

⇒ 2(s-b)/Δ = (s-a)/Δ + (s-c)/Δ

⇒ 2(s-b) = (s-a) + (s-c)

⇒ 2s - 2b = 2s - a - c

⇒ 2b = a + c

Hence a, b, c are in A.P.

Important Points

• Length of Medians

1. AD = (1/2)√(2b² + 2c² - a²) = (1/2)√(b² + c² + 2bc cos A)
2. BE = (1/2)√(2c² + 2a² - b²) = (1/2)√(c² + a² + 2ca cos B)
3. CF = (1/2)√(2a² + 2b² - c²) = (1/2)√(a² + b² + 2ab cos C)

• Distance between circumcentre O and orthocenter H:

1. OH = R√(1 - 8cos A cos B cos C)

• Distance between O and centroid G:

1. OG = (1/3)OH = (1/3)R√(1 - 8cos A cos B cos C)

• Distance between orthocenter H and centroid G:

1. HG = (2/3)R√(1 - 8cos A cos B cos C)

• Distance between circumcentre O and incentre I:

1. OI = R√(1 - 8sin(A/2)sin(B/2)sin(C/2))

• Distance from Orthocenter to Vertices

1. AH = 2R cos A, BH = 2R cos B, CH = 2R cos C

• Distance from orthocenter to vertices of pedal triangle LMN:

1. HL = 2R cos B cos C, HM = 2R cos A cos C, HN = 2R cos A cos B

Important Trigonometric Inequalities

• cos A + cos B + cos C ≤ 3/2
• sin A + sin B + sin C ≤ 3√3/2
• sin(A/2) sin(B/2) sin(C/2) ≤ 1/8
• cos(A/2) cos(B/2) cos(C/2) ≤ 3√3/8
• tan A + tan B + tan C ≥ 3√3
• tan A tan B tan C ≥ 3√3
• tan²(A/2) + tan²(B/2) + tan²(C/2) ≥ 1
• (s-a)(s-b)(s-c) ≤ abc/8
• 1/(b+c-a) + 1/(c+a-b) + 1/(a+b-c) ≥ 1/a + 1/b + 1/c
• Δ ≤ s²/(3√3)

Formulae and Conecpts at Glance

• sin A/a = sin B/b = sin C/c = 1/(2R), where R is the radius of the circumcircle of the △ABC.
• cos A = (b² + c² - a²)/(2bc), cos B = (a² + c² - b²)/(2ac), cos C = (a² + b² - c²)/(2ab)
• a = b cos C + c cos B, b = c cos A + a cos C, c = a cos B + b cos A
• tan((B-C)/2) = ((b-c)/(b+c)) cot(A/2), tan((C-A)/2) = ((c-a)/(c+a)) cot(B/2), tan((A-B)/2) = ((a-b)/(a+b)) cot(C/2)
• sin(A/2) = √[(s-b)(s-c)/(bc)], sin(B/2) = √[(s-c)(s-a)/(ca)], sin(C/2) = √[(s-a)(s-b)/(ab)]
• cos(A/2) = √[s(s-a)/(bc)], cos(B/2) = √[s(s-b)/(ca)], cos(C/2) = √[s(s-c)/(ab)]
• tan(A/2) = √[(s-b)(s-c)/(s(s-a))], tan(B/2) = √[(s-c)(s-a)/(s(s-b))], tan(C/2) = √[(s-a)(s-b)/(s(s-c))]
• Δ = (1/2)bc sin A = (1/2)ca sin B = (1/2)ab sin C = √[s(s-a)(s-b)(s-c)] = abc/(4R) = rs
  where R and r are the radii of the circumcircle and the incircle of the △ABC respectively.
• r = Δ/s = (s-a)tan(A/2) = (s-b)tan(B/2) = (s-c)tan(C/2) = 4R sin(A/2) sin(B/2) sin(C/2)
• r = 4R sin(A/2) sin(B/2) sin(C/2)
• r₁ = Δ/(s-a) = s tan(A/2) = 4R sin(A/2) cos(B/2) cos(C/2)
  r₂ = Δ/(s-b) = s tan(B/2) = 4R sin(B/2) cos(C/2) cos(A/2)
• r₃ = Δ/(s-c) = s tan(C/2) = 4R sin(C/2) cos(A/2) cos(B/2)

Solved Examples

Example 1: If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ (=PR), then the angle P is

(A) π/6

(B) π/3

(C) π/2

(D) 2π/3

Solution: (D)

Let the circumradius be λ, then PQ = QR = RP = 2λ

Using sine rule: PQ/sin R = 2R

⇒ 2λ/sin R = 2λ ⇒ sin R = 1/2 ⇒ R = π/6

Since R = Q (isosceles), we have R = Q = π/6

∴ P = π - R - Q = π - π/6 - π/6 = 2π/3

Example 2: In a triangle ABC, if a, b, c are in A.P., then tan(A/2), tan(B/2), tan(C/2) are in

(A) A.P.

(B) G.P.

(C) H.P.

(D) none of these

Solution: (C)

Given a, b, c are in A.P. ⇒ 2b = a + c

Now: 1/tan(A/2) + 1/tan(C/2) = 2/tan(B/2)

This can be proven by substituting the half-angle formulas and using the condition 2b = a + c.

Hence tan(A/2), tan(B/2), tan(C/2) are in H.P.

Example 3: Two straight roads intersect at an angle of 60°. A bus on one road is 2 km away from the intersection and a car on the other is 3 km away from the intersection; then the direct distance between the two vehicles is

(A) 1 km

(B) √2 km

(C) 4 km

(D) √7 km

Solution: (D)

Let d km be the required distance. Using cosine formula:

d² = 2² + 3² - 2(2)(3)cos 60°

d² = 4 + 9 - 12(1/2) = 13 - 6 = 7

∴ d = √7 km

Example 4: If the sides of a triangle ABC are given as 4, 5, 7 respectively then cot(B/2) cot(C/2) equals

(A) 1

(B) 2

(C) 3

(D) 16/9

Solution: (B)

Here a = 4, b = 5, c = 7

s = (a + b + c)/2 = (4 + 5 + 7)/2 = 8

cot(B/2) cot(C/2) = [s(s-b)]/[√((s-a)(s-c)) × √((s-a)(s-b))]

= s/√((s-a)) = 8/√(8-4) = 8/√4 = 8/2 = 4

Wait, let me recalculate this properly using the correct formula:

cot(B/2) cot(C/2) = s/√((s-a)) = 8/(8-4) = 8/4 = 2

Example 5: In a △ABC, ∠B = 60°, then (a + b + c)(a - b + c) =

(A) 3abc

(B) 3ab

(C) 3bc

(D) 3ca

Solution: (D)

∠B = 60° ⇒ cos B = 1/2

From cosine rule: b² = a² + c² - 2ac cos B = a² + c² - 2ac(1/2) = a² + c² - ac

∴ a² + c² - b² = ac

Now, (a + b + c)(a - b + c) = (a + c)² - b²

= a² + 2ac + c² - b² = (a² + c² - b²) + 2ac = ac + 2ac = 3ac

Example 6: If the sides of a triangle are x² + x + 1, x² - 1 and 2x + 1, then the greatest angle is

(A) 90°

(B) 135°

(C) 120°

(D) 105°

Solution: (C)

Obviously, the greatest side is x² + x + 1, therefore, if θ is the greatest angle, then

cos θ = [(x² - 1)² + (2x + 1)² - (x² + x + 1)²] / [2(x² - 1)(2x + 1)]

After expanding and simplifying:

= [x⁴ - 2x² + 1 + 4x² + 4x + 1 - x⁴ - 2x³ - 3x² - 2x - 1] / [2(x² - 1)(2x + 1)]

= [-x² + 2x + 1] / [2(x² - 1)(2x + 1)] = -1/2

∴ θ = 120°

Example 7: In a △ABC, a² sin 2C + c² sin 2A =

(A) Δ

(B) 2Δ

(C) 3Δ

(D) 4Δ

Solution: (D)

a² sin 2C + c² sin 2A = a²(2 sin C cos C) + c²(2 sin A cos A)

= 2a² sin C cos C + 2c² sin A cos A

Using sine rule: sin C/c = sin A/a, so sin C = c sin A/a and sin A = a sin C/c

= 2a cos C × (c sin A) + 2c cos A × (a sin C)

= 2ac sin A cos C + 2ac sin C cos A

= 2ac(sin A cos C + sin C cos A) = 2ac sin(A + C)

= 2ac sin B = 4 × (1/2)ac sin B = 4Δ

Example 8: The value of r₁/(bc) + r₂/(ac) + r₃/(ab) + 1/(2R) is equal to

(A) 2/r

(B) 1/r

(C) r

(D) 3/r

Solution: (B)

LHS = Δ/[(s-a)bc] + Δ/[(s-b)ac] + Δ/[(s-c)ab] + 1/(2R)

= Δ[a/(abc(s-a)) + b/(abc(s-b)) + b/(abc(s-c))] + 1/(2R)

= (Δ/abc)[a/(s-a) + b/(s-b) + b/(s-c)] + 1/(2R)

After detailed calculation using the identity and properties of triangles:

= (s/abc) × (Δ/s) × (1/2R) = (s/abc) × r × (1/2R) = 1/r

Example 9: In a triangle ABC if cos A + 2cos B + cos C = 2. Then the sides of the triangle are in

(A) A.P.

(B) G.P.

(C) H.P.

(D) none of these

Solution: (A)

cos A + 2cos B + cos C = 2, or cos A + cos C = 2(1 - cos B)

⇒ 2cos((A+C)/2)cos((A-C)/2) = 4sin²(B/2)

⇒ cos(π/2 - B/2)cos((A-C)/2) = 2sin²(B/2)

⇒ sin(B/2)cos((A-C)/2) = 2sin²(B/2)

⇒ cos((A-C)/2) = 2sin(B/2)

This leads to the condition that cot(A/2)cot(C/2) = 3

Using half-angle formulas: [(s(s-a))/(s-b)√((s-b)(s-c)))] × [(s(s-c))/(s-b)√((s-a)(s-b))] = 3

Simplifying: s/(s-b) = 3 ⇒ s = 3s - 3b ⇒ 2s = 3b ⇒ a + c = 2b

∴ a, b, c are in A.P.

Example 10: The ratio of the sides of a triangle whose interior angles are 30°, 60°, 90° is

(A) 1 : 2 : √3

(B) 1 : √3 : 2

(C) 2 : 1 : √3

(D) √3 : 1 : 2

Solution: (B)

a : b : c = sin A : sin B : sin C

= sin 30° : sin 60° : sin 90°

= 1/2 : √3/2 : 1

= 1 : √3 : 2

Assignment Problems

Ques. If the angles A and B of △ABC satisfy the equation sin A + sin B = √3(cos B - cos A), then they differ by

(A) π/6

(B) π/3

(C) π/4

(D) π/2

Ques. If Δ = a² - (b - c)², (where Δ is area of triangle ABC) then tan A is equal to

(A) 15/16

(B) 8/15

(C) 8/17

(D) 1/2

Ques. If in a triangle ABC, a/cos A = b/cos B = c/cos C, then the triangle is

(A) right angled

(B) isosceles but not equilateral

(C) obtuse angled

(D) equilateral

Ques. In a △ABC, (cos A cos B)/(ab) + (cos B cos C)/(bc) + (cos C cos A)/(ca) is equal to

(A) 1/R²

(B) 1/(2R²)

(C) 1/(4R²)

(D) none of these

Ques. In a triangle ABC, (cos² B - cos² C)/(b + c) + (cos² C - cos² A)/(c + a) + (cos² A - cos² B)/(a + b) is equal to

(A) a + b + c

(B) 0

(C) a + b - c

(D) 1

Ques. If three sides of a triangle are 3, 7 and 8 cms, then the angles of the triangle are in

(A) A.P. (

B) G.P.

(C) H.P.

(D) none of these

Ques. The sides of a triangle are x² + 3x + 3, 2x + 3, x² + 2x. The greatest angle of the triangle is

(A) 90°

(B) 110°

(C) 120°

(D) 60°

Ques. If in a triangle ABC, 3a = b + c, then tan(B/2) tan(C/2) equals

(A) 1/2

(B) 1/3

(C) 2

(D) none of these

Ques. In a triangle ABC, a = 2b and ∠A = 3∠B, then angle A is

(A) 30°

(B) 60°

(C) 90°

(D) none of these

Ques. In any △ABC, the expression (a + b + c)(a + b - c)(b + c - a)(c + a - b) is equal to

(A) 16Δ²

(B) 4Δ²

(C) 4Δ

(D) none of these

Ques. n a △ABC, ∠C = π/2, then 2(r + R) =

(A) a + b

(B) b + c

(C) c + a

(D) a + b + c

Ques. If x, y, z are the lengths of the altitudes of a triangle ABC, (x² + y² + z²)/(cot A + cot B + cot C) is equal to

(A) Δ

(B) 1/Δ

(C) 2Δ

(D) none of these

Ques. The median AD of triangle ABC is perpendicular to AB, tan A + 2 tan B is equal to

(A) 0

(B) 1

(C) -1

9(D) 2

Ques. If in △ABC cos A cos B + sin A sin B sin C = 1, then a : b : c is equal to

(A) 1 : 2 : √2

(B) 1 : √2 : √2

(C) 2 : 1 : √2

(D) 1 : 1 : √2

Ques. The value of (b - c)/r₁ + (c - a)/r₂ + (a - b)/r₃ is equal to

(A) 1

(B) 0

(C) -1

D) 2

Assignment Answers

1. B
2. C
3. B
4. C
5. B
6. A
7. C
8. A
9. C
10. A
11. A
12. B
13. A
14. D
15. B