# Quod Erat Demonstrandum

## 2015/10/07

### 三角題

Filed under: NSS — johnmayhk @ 2:58 下午
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Just come across a part of a core mathematics trigonometry problem today morning, I and my student used different ways to solve it, but actually, same results turn up:

Given

$AC=b$,

$CB=a$, where $a > b$

$\angle CAB=\theta < 90^o$,

find $c$ in terms of $a,b,\theta$.

(Note: this is an ‘A.S.S.’ given, but the triangle can be uniquely determined.)

Method 1 (cosine law)

$a^2=b^2+c^2-2bc\cos\theta$

$c^2-2bc\cos\theta+b^2-a^2=0$

$c=\frac{2b\cos\theta\pm \sqrt{(2b\cos\theta)^2-4(b^2-a^2)}}{2}$

$=b\cos\theta \pm\sqrt{a^2-b^2\sin^2\theta}$

It is easy to show $b^2\cos^2\theta < a^2-b^2\sin^2\theta$, hence $b\cos\theta-\sqrt{a^2-b^2\sin^2\theta} < 0$

thus,

$c=b\cos\theta +\sqrt{a^2-b^2\sin^2\theta}$

Method 2 (sine law)

$\sin B=\frac{b}{a}\sin \theta$, and

$c$

$=\frac{a\sin(180^o-(\theta + B))}{\sin \theta}$

$=\frac{a\sin(\theta + B)}{\sin \theta}$

$=\frac{a}{\sin\theta}(\sin\theta\cos B+\sin B\cos\theta)$

$=a(\cos B+\sin B\cot\theta)$

$=a(\sqrt{1-\sin^2B}+\frac{b}{a}\cos\theta)$ 　　　($-\sqrt{1-\sin^2B}$ is rejected)

$=b\cos\theta +\sqrt{a^2-b^2\sin^2\theta}$

as before.

Of course, values of $a,b,\theta$ are concretely given in that core math question.