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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 120o.

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 = 60o ⇒ A = 120o.

Alternative Method:

cosA = (b2 + c2 - a2)/2bc

= 100 + 36 - 196/120

= - 1/2

⇒ A = 120o.

Illustration:

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

Solution:

Let A = 60o ⇒ B + C = 120o.

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

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

= 1

⇒ (B – C)/2 = 45o

⇒ B – C = 90o

⇒ B = 105o, C = 15o

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 60o

= 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 = 30o, C = 60o, a = 6 cm

Solution:

A = 180o – (B + C) = 90o.

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 30o, 60o and 90o.

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 30o: sin 60o: sin 90o

= 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/r3 are 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.

You might like to refer some of the related resources listed below:

Look into the Revision Notes on Solution of Triangles for a quick revision.

Various Recommended Books of Mathematics are just a click away.

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