DETERMINE WHETHER THE FOLLOWING MEASUREMENTS PRODUCE A TRIANGLE

Example 1 :

Determine whether the following measurements produce one triangle, two triangles or no triangle: 

∠B = 88°, a = 23, b = 2Sine formula

Solve if solution exists.

Solution :

Sine formula :

a/sinA = b/sinB = c/sinC

23/sinA = 2/sin88° = c/sinC

23/sinA = 2/sin88°

sinA = (23/2)sin 88°

sinA = 11.5sin 88°

The maximum value of sinA is 1. So, it is not possible.

Hence the given measurements will not produce a triangle.

Example 2 :

If the sides of a triangle ABC are a = 4, b = 6 and c = 8, then show that 4cosB + 3cosC = 2.

Solution :

Cosine formula :

cosA = (b² + c² - a²)/2bc ----(1)

cosB = (a² + c² - b²)/2ac ----(2)

cosC = (a² + b² - c²)/2ab ----(3)

From (2), 

cosB = (a² + c² - b²)/2ac 

= (4² + 8² - 6²)/[2(4)(8)]

= (16 + 64 - 36)/64

= 44/64

= 11/16

From (3),

cosC = (a² + b² - c²)/2ab

= (4² + 6² - 8²)/[2(4)(6)]

= (16 + 36 - 64)/48

= -12/48

= - 1/4

4cosB + 3cosC = 4(11/16) + 3(-1/4)

= (11/4) - (3/4)

= (11 - 3)/4

= 8/4

= 2

Example 3 :

In a triangle ABC, if a = √3 − 1, b = √3 + 1 and C = 60°, find the other side and other two angles

Solution :

cosC = (a² + b² - c²)/2ab

Substitute a = √3 - 1 and b = √3 + 1.

a² = (√3 - 1)²

  = 3 + 1 - 2√3

a = 4 - 2√3

b = √3 + 1,

b² = (√3 + 1)²

 = 3 + 1 + 2√3

= 4 + 2√3

Substituting these values in cosine formula,

cos60 = (4 - 2√3 + 4 + 2√3 - c²)/[2(√3 − 1)(√3 + 1)]

1/2 = (8 - c²)/[2(2)]

1/2 = (8 - c²)/4

Multiply each side by 4.

2 = 8 - c²

c² = 8 - 2

c² = 6

c = √6

Sine formula :

a/sinA = b/sinB = c/sinC

(√3 - 1)/sinA = (√3 + 1)/sinB = √6/sin60°

(√3 + 1)/sinB = √6/(√3/2)

(√3 + 1)/sinB = 2√6/√3

 (√3 + 1)/sinB = 6√2/3

(√3 + 1)/sinB = 2√2

sinB = (√3 + 1)/2√2

sinB = (√3/2)(1/√2) + (1/2)(1/√2)

= sin60°cos45° + cos60°sin45°

= sin(60° + 45°)

= sin105°

∠B = 105°

In triangle ABC,

∠A + ∠B + ∠C = 180°

Substitute. 

∠A + 105° + 60° = 180°

∠A + 165° = 180°

∠A = 15°

Therefore, 

c = √6, ∠A = 15°, ∠B = 105°

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