Sin rule

Sin rule

We have the triangle ABC, α = 2β, BC = 12cm and AB = 8cm. Find the length of the third side.
kate

Posts: 91
Joined: Mon Apr 09, 2007 3:58 pm
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Solving it geometrically might be easier.

Draw the angle bisector of $$\angle A$$. It meets BC at D.

$$\angle ADC = 2\beta$$

$$\triangle ABC$$ and $$\triangle ACD$$ are similar because
$$\angle ABC = \angle DAC$$
$$\angle BAC = \angle ADC$$

So

$$\frac{AB}{AD} = \frac{AC}{DC} = \frac{BC}{AC}$$

Let
$$BD = AD = x$$
$$AC = y$$

We have
$$\frac{8}{x} = \frac{y}{12 - x} = \frac{12}{y}$$

Solving
$$\frac{8}{x} = \frac{12}{y}$$
$$\frac{y}{12 - x} = \frac{12}{y}$$

we have
$$8y = 12x$$
$$y^2 = 12(12 - x)$$

$$y^2 + 12y - 144 = 0$$

$$y = 4\sqrt{10} - 4$$

redmafiya

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Joined: Tue Jun 17, 2008 7:05 pm
Location: New Zealand
Reputation: 1