Solved Examples on Solution of Triangles

Illustration:

In a triangle ABC, the sides are 6 cm, 10 cm and 14 cm. Show that the triangle is obtuse angled with the obtuse angle equal to 120^o.

Solution:

Let a = 14, b = 10, c = 6

⇒ s = (a + b + c)/2 = 15.

The largest angle is opposite to the largest side.

Hence tan A/2 = √(s - b) (s - c)/s(s - a)

= √(5 X 9/15 = √3

⇒ A/2 = 60^o ⇒ A = 120^o.

Alternative Method:

cosA = (b² + c² - a²)/2bc

= 100 + 36 - 196/120

= - 1/2

⇒ A = 120^o.

Illustration:

Two sides of a triangle are √3 – 1 and √3 + 1 units and their included angle is 60^o. Solve the triangle.

Solution:

Let A = 60^o ⇒ B + C = 120^o.

Also tan (B – C)/ 2 = (b – c)/(b + c) cot A/2

= (√3 + 1 - √3 + 1) / (√3 + 1 + √3 – 1) cot 30^o

= 1

⇒ (B – C)/2 = 45^o

⇒ B – C = 90^o

⇒ B = 105^o, C = 15^o

Also a = b sin / sin B = (√3 + 1)/ sin 60° cos 45° + cos 60° sin 45°

= √2(√3 + 1). √3/(√3 + 1) = √6.

Alternatively:

a² = b² + c² – 2bc cos A

= (√3 + 1)² + (√3 – 1)² – 2(√3 + 1)(√3 – 1) cos 60^o

= 8 – 4. 1/2 = 6

⇒ a = √6.

Illustration:

If a, b and A are given in a triangle and c₁, c₂ are the possible values of the third side, prove that c₁² + c₂² – 2c₁c² cos2A = 4a² cos² A.

Solution:

We know cos A = (b² + c² - a²)/2bc

⇒ c² – 2bc cos A + b² – a² = 0.

Using the formulae of sum and product of roots, we have

c₁ + c₂ = 2bcos A and c₁c₂ = b² – a².

Now, c₁² + c₂² – 2c₁c₂cos 2A = (c₁ + c₂)² – 2c₁c₂ (1 + cos 2A)

= 4b²cos²A – 2(b² – a²)2cos²A

= 4a²cos²A.

Illustration:

Solve the triangle if B = 30^o, C = 60^o, a = 6 cm

Solution:

A = 180^o – (B + C) = 90^o.

Also a/sin A = b/sin B = c/sin C

⇒ b = a sinB = 6 × 1/2 = 3,

c = a sinC = 3√3.

Hence, these are the required lengths of the sides of the triangle.

Illustration:

Find the ratio of the sides of a triangle whose interior angles are 30^o, 60^o and 90^o.

Solution:

Let the sides of the triangle be a, b and c.

Then we know a : b : c = sin A : sin B : sin C

= sin 30^o: sin 60^o: sin 90^o

= 1/2: √3/2: 1

= 1: √3: 2.

Illustration:

The radii r₁, r₂, r₃ of escribed circle of a triangle ABC are in H.P. If its area is 24 cm²and its perimeter is 24 cm, find the length of its sides.

Solution:

Let a, b, c be the length of the sides of the triangle, s be the semi perimeter and Δ be the area.

r₁ = Δ/(s – a) = 24/(12 – a)

r₂ = Δ/(12 – b)

r₃ = Δ/s - c = 24/(12 – c)

Since r₁, r₂, r₃ are in H.P.

⇒ 1/r₁, 1/r₂, 1/r₃are in A.P.

⇒ 2/r₂, 1/r₁, 1/r₃ or 2 (12 - b )/24 = 12 - a/24 + 12 - c/24

or 2b = a + c …… (1)

Also, a + b + c = 24 …… (2)

From (1) and (2), b = 9 cm and a + c = 16 cm

Now, Δ = √s(s - a) (s - b) (s - c)

⇒ 24 = √12 (12 - a) (12 - b) (12 - c)

⇒ 24 × 24 = 12 (12 – a) × 4 × (12 – 16 + a)

⇒ 12 = (12 – a)(a – 4)

⇒ a² – 16a + 60 = 0

⇒ a = 6 or 10.

When a = 6, c = 10 and when a = 10, c = 6.

Hence sides are 6, 8, 10 or 10, 8, 6.

Illustration:

Prove that a triangle ABC is equilateral iff tan A + tan B + tan C = 3√3. (1998)

Solution:

Let us suppose that the triangle ABC is equilateral.

Then A = B = C = 60°

Hence, the required value tan A + tan B + tan C

= tan 60° + tan 60° + tan 60°

= 2 tan 60°

= 3√3

Conversely, assume that tan A + tan B + tan C = 3√3

But, in triangle ABC, A + B = 180° - C

Taking tan on both the sides we have

tan (A +B) = tan (180° - C)

(tan A + tan B)/(1 - tan A tan B) = - tan C

Hence, this gives tan A + tan B = - tan C + tan A tan B tan C

This yields tan A + tan B + tan C = tan A tan B tan C = 3√3

This means that none of tan A, tan B or tan C can be negative.

Hence, ?ABC cannot be an obtuse angled triangle.

Moreover, we know A.M. ≥ G.M.

1/3 [tan A + tan B + tan C] ≥ [tan A + tan B + tan C]^1/3

This means 1/3(3√3) ≥ (3√3)^1/3

This gives √3 ≥ √3.

Now, this equality can hold iff tan A = tan B = tan C

Hence, this means iff A = B = C.

Hence, the triangle is equilateral.

Illustration:

For a triangle ABC it is given that cos A + cos B + cos C = 3/2. Prove that the triangle is equilateral.

Solution:

Let ‘a’, ‘b’ and ‘c’ be the sides of the triangle ABC.

It is given that cos A + cos B + cos C = 3/2.

Hence, (b² + c² – a²)/2bc + (a² + c² – b²)/2ac + (a² + b² – c²)/2ab = 3/2

Simplifying this we get,

ab² + ac² – a³ + ba² + bc² – b³ + ca² + cb² – c³ = 3abc

Hence, a(b-c)² + b(c-a)² + c(a-b)² = (a+b+c)/2 [(a-b)² + (b-c)² + (c-a)²]

This gives

(a+b-c) (a-b)² + (b+c-a) (b-c)² + (c+a-b) (c-a)² = 0

(Since we know that (a+b-c) > 0, (b+c-a) > 0, (c+a-b) > 0)

On the left hand side, we have a coefficient multiplied by the square of a number. Moreover, each coefficient is positive and hence for the sum to be zero, each term separately must be equal to zero. This means we must have

(a+b-c) (a-b)² = 0 = (b+c-a) (b-c)² = (c+a-b) (c-a)²

This implies a = b = c

This implies that the triangle is equilateral.

Illustration:

If in a triangle ABC, cos A cos B + sin A sin B sin C = 1, show that a : b : c = 1 : 1 : 2.

Solution:

Given that cos A cos B + sin A sin B sin C = 1

Now, this gives sin C = (1- cos A cos B)/sin A sin B … (1)

Also, we know sin C ≤ 1, which implies that

(1- cos A cos B)/sin A sin B ≤ 1

Hence, (1- cos A cos B) ≤ sin A sin B

⇒ 1 ≤ sin A sin B + cos A cos B

⇒ 1 ≤ cos (A-B)

⇒ cos (A-B) ≥ 1

But, cos θ ≤ 1, so we have

⇒ cos (A-B) = 1.

⇒ A-B = 0.

Putting A = B in equation (1) we have,

sin C = (1-cos²A)/sin²A

⇒ sin C = 1 ⇒ C = π/2

Now, A + B + C = π

⇒ A + B = π/2

Since A = B and C = π/2, so we have A = π/4

sin A : sin B : sin C = sin π/4 : sin π/4 : sin π/4

= 1/√2 : 1/√2 : 1

Hence, a : b : c = 1 : 1 : √2.

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