Question

? Help if you know how to do this

Answer 1

Answer:

$x = 2.5$

Step-by-step explanation:

Given

$AC = 2x - 1$

$AB = 2x + 1$

$BC = 2\sqrt{7$

$\angle A = 60^{\circ}$

Required

Find x

To find x, we make use of cosine formula which states:

$a^2 = b^2 + c^2 - 2bcCosA$

In this case:

$BC^2 = AB^2 + AC^2 - 2*AB*BC*CosA$

Substitute values

$(2\sqrt{7})^2 = (2x + 1)^2 + (2x-1)^2 - 2*(2x + 1)*(2x - 1)*Cos(60^{\circ})$

Evaluate all exponents

$4*7 = (4x^2 + 4x + 1)+(4x^2 - 4x + 1) - 2(4x^2 - 1)*cos(60^{\circ})$

Open brackets

$28 = 4x^2 + 4x + 1+4x^2 - 4x + 1 - (8x^2 - 2)*cos(60^{\circ})$

Collect Like Terms

$28 = 4x^2 +4x^2+ 4x - 4x+ 1 + 1 - (8x^2 - 2)*cos(60^{\circ})$

$28 = 8x^2+ 2 - (8x^2 - 2)*cos(60^{\circ})$

Substitute 0.5 for cos(60)

$28 = 8x^2+ 2 - (8x^2 - 2)*0.5$

Open bracket

$28 = 8x^2+ 2 - 4x^2 + 1$

Collect Like Terms

$8x^2- 4x^2 + 2 + 1 - 28 = 0$

$4x^2 -25 = 0$

Collect Like Terms

$4x^2 =25$

Divide both sides by 4

$x^2 =\frac{25}{4}$

$x^2 =6.25$

Take positive square root of both sides

$x = \sqrt{6.25$

$x = 2.5$