## Trigonometry, Cosine Rule, IB Math HL

Discussions for the Core part of the syllabus. Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths HL Revision Notes

### Trigonometry, Cosine Rule, IB Math HL

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
Oliver

Posts: 0
Joined: Mon Jan 28, 2013 8:08 pm

### Re: Trigonometry, Cosine Rule, IB Math HL

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!!
miranda

Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm

### Re: Trigonometry, Cosine Rule, IB Math HL

Thank you miranda!!
Oliver

Posts: 0
Joined: Mon Jan 28, 2013 8:08 pm