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by **Oliver** » Wed Jan 29, 2014 7:20 am

IB Mathematics HL – Trigonometry, Cosine Rule

How can we solve the triangle $ABC$ using Cosine Rule given that $\hat{A}=\frac{\pi}{3}$ , $AC=6$ and $AB=8$ .

Thanks

by **miranda** » Wed Jan 29, 2014 7:20 am

IB Math HL – Trigonometry, Cosine Rule

Applying the Cosine Rule we have that

$(BC)^2=(AB)^2+(AC)^2-2(AB)(AC)cos \hat{A} \Rightarrow$

$(BC)^2=8^2+6^2-2\cdot 8\cdot 6 cos \frac{\pi}{3}\Rightarrow$

$(BC)^2=100-96\cdot \frac{1}{2}\Rightarrow$

$(BC)^2=100-48=52 \Rightarrow BC=\sqrt{52}$

Now, in order to find another angle, for example $\hat{C}$ , we could apply again cosine rule and solve for $cos\hat{C}$ as following

$(AB)^2=(AC)^2+(BC)^2-2(BC)(AC)cos \hat{C} \Rightarrow$

$8^2=6^2+52-2(\sqrt{52})(6)cos \hat{C} \Rightarrow$

$64=36+52-12(\sqrt{52})cos \hat{C} \Rightarrow$

$-24=-12(\sqrt{52})cos \hat{C} \Rightarrow$

$2=\sqrt{52} cos \hat{C} \Rightarrow$

$cos \hat{C}=\frac{2}{\sqrt{52}}\Rightarrow$

$\hat{C}=arcos (\frac{2}{\sqrt{52}})$

Finally, for the angle $\hat{B}$ we have that

$\hat{B}=\pi -\hat{A} -\hat{C}$

Hope these help!!

by **Oliver** » Wed Jan 29, 2014 7:21 am

Thank you miranda!!

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Trigonometry, Cosine Rule, IB Math HL

Trigonometry, Cosine Rule, IB Math HL

Re: Trigonometry, Cosine Rule, IB Math HL

Re: Trigonometry, Cosine Rule, IB Math HL