1 Answer

Eddiec · Answer 1 · 2020-09-12T03:59:29+0000

In this triangle, we have

$\cos30^\circ=(4x\sqrt3-5\sqrt3)/(5x-4)$

You should know that
$\cos30^\circ=\frac{\sqrt3}2$ , so that

$\frac{\sqrt3}2=(4x\sqrt3-5\sqrt3)/(5x-4)$

Divide both sides by
$\sqrt3$ to get

$\frac12=(4x-5)/(5x-4)$

Multiply both sides by
5x-4 to get

$\frac{5x-4}2=4x-5$

Simplify and solve for
:

$\frac{5x}2-2=4x-5$

$5-2=4x-\frac{5x}2$

$3=\frac{3x}2$

6=3x

$\implies\boxed{x=2}$

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