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by **Nyla** » Sun Aug 18, 2013 10:21 am

MYP Mathematics Pre-diploma – Trigonometry, Cosine Rule

How can we find the side $a$ of a triangle ABC given that $\hat{A}=60^{\circ} , c=1 \ and \ b=2$ ?

Thanks!

by **Jack** » Sun Aug 18, 2013 10:28 am

MYP Mathematics Pre-diploma – Trigonometry, Cosine Rule

The Cosine rule states

$a^2=b^2+c^2-2(bc)cosA$

So in your case the side $a$ can be calculated as follows:

$a^2=1^2+2^2-2(1(2))cos60^{\circ}$

$a^2=1+4-4\cdot \frac{1}{2}=5-2=3$

$a=\sqrt{3}$

Hope these help!!

by **Nyla** » Sun Aug 18, 2013 10:29 am

Thank you Jack!!

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MYP Math Cosine Rule

MYP Math Cosine Rule

Re: MYP Math Cosine Rule

Re: MYP Math Cosine Rule