Solving for x in Triangle ABC Using the Law of Cosines and Half-Angle Formula

In triangle ABC, the sides are given as AB x - 3 cm, BC x^3 cm, AC 8 cm, and the angle BAextends to C 60 degrees. To find the value of x, we can apply the Law of Cosines, a fundamental theorem used in trigonometry and geometry. Additionally, the half-angle formula can be employed to verify the solution.

Law of Cosines Approach

The Law of Cosines is expressed as:

c^2  a^2   b^2 - 2ab cdot cos(C)

In this triangle:

Substituting these values:

x^3^2  8^2   (x - 3)^2 - 2 cdot 8 cdot (x - 3) cdot cos(60^circ)

Since cos(60^circ) 1/2, the equation simplifies to:

x^6   6x^2   9  64   x^2 - 6x   9 - 8x   24

Expanding and simplifying:

x^6   6x^2   9  x^2 - 14x   101

Subtracting x^2 - 14x 101 from both sides:

6x^2 - 14x - 92  0

This is a quadratic equation in the form of ax^2 bx c 0. Solving it using the quadratic formula:

x  frac{-b pm sqrt{b^2 - 4ac}}{2a}

By substituting a 6, b -14, and c -92, we find:

x  frac{14 pm sqrt{196   2208}}{12}  frac{14 pm 50}{12}

Giving us two possible solutions:

x  frac{64}{12}  5.33 quad and quad x  frac{-36}{12}  -3

Since a negative length isn't feasible, we take x 4.4.

Half-Angle Formula

The half-angle formula for sine provides an alternative method to verify the solution:

sin frac{A}{2}  sqrt{frac{s - b - c}{bc}}

where s frac{a b c}{2} is the semi-perimeter. In triangle ABC:

The semi-perimeter s frac{x 8 x^3 - 3}{2}. Substituting into the half-angle formula:

sin 30  sqrt{frac{x^3 - 4 - 8}{8(x^3 - 3)}}

Since sin 30 1/2

frac{1}{2}  sqrt{frac{7x^3 - 12}{8x^3 - 24}}

Squaring both sides:

frac{1}{4}  frac{7x^3 - 12}{8x^3 - 24}

Cross-multiplying to find:

8x^3 - 24  28x^3 - 48

Collecting like terms:

2^3  24

Solving for x:

x  frac{24}{20}  4.4

Hence, the value of x is confirmed as 4.4 cm.

Napier’s Analogy (For Advanced Readers)

Napier’s Analogy provides another method to solve for the angles #8722; C and B in a non-spherical triangle, involving the tangent function. It can complicate the process and might not be the most straightforward method.