Sin rule

Sin rule

Postby kate » Fri Jul 04, 2008 2:00 pm

We have the triangle ABC, α = 2β, BC = 12cm and AB = 8cm. Find the length of the third side.
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Postby redmafiya » Wed Jul 23, 2008 9:07 pm

Solving it geometrically might be easier.

Draw the angle bisector of [tex]\angle A[/tex]. It meets BC at D.

[tex]\angle ADC = 2\beta[/tex]

[tex]\triangle ABC[/tex] and [tex]\triangle ACD[/tex] are similar because
[tex]\angle ABC = \angle DAC[/tex]
[tex]\angle BAC = \angle ADC[/tex]

So

[tex]\frac{AB}{AD} = \frac{AC}{DC} = \frac{BC}{AC}[/tex]

Let
[tex]BD = AD = x[/tex]
[tex]AC = y[/tex]

We have
[tex]\frac{8}{x} = \frac{y}{12 - x} = \frac{12}{y}[/tex]

Solving
[tex]\frac{8}{x} = \frac{12}{y}[/tex]
[tex]\frac{y}{12 - x} = \frac{12}{y}[/tex]

we have
[tex]8y = 12x[/tex]
[tex]y^2 = 12(12 - x)[/tex]

[tex]y^2 + 12y - 144 = 0[/tex]

[tex]y = 4\sqrt{10} - 4[/tex]

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