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:

johnmayhk-trigonometry

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.

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