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I am studying A level maths on my own as an interest. I have the following problem:

In $\triangle ABC, AB = 9 cm, AC = 12 cm, \angle B = 2\theta, \angle C = \theta.$ Without using tables or calculators find $\cos\theta$ and the length of BC.

Using the cosine rule I have established that:$$\cosθ = \frac{(x^2 + 63)}{24x}$$ and $$\cos2\theta = \frac{x^2 - 63}{18x}$$ where x = BC.

I can also see that $\angle A = π - 3θ$ and that using the cosine rule $\cos 3\theta = \frac{x^2 - 225}{216}$

That’s as far as I can go.

Davide Giraudo
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Steblo
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    At some stage you will probably want to use $\cos(2\theta) = 2 \cos^2(\theta)-1$ or something similar – Henry Oct 08 '19 at 10:24

1 Answers1

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Using sine rule , $$\frac{\sin {2\theta}}{12} = \frac {\sin \theta}{9}$$ $$ \frac{2\sin\theta\cos\theta}{\sin\theta} = \frac{12}9$$ $$\cos\theta = \frac 23.$$

Davide Giraudo
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The Demonix _ Hermit
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