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AstroNerd · Nov 5, 2015

Can someone please explain how the answer is root 41

si2136 · Nov 5, 2015

View attachment 32657

Remember that you would need to be a right angled triangle for Trigonometry to work

AstroNerd · Nov 5, 2015

Thank you!!

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AstroNerd · Nov 5, 2015

AstroNerd said:
Thank you!!

InteGrand · Nov 5, 2015

$Have you learnt the cosine rule? If you have, you don't need to draw up right-angled triangles as this rule applies to arbitrary triangles, and you can do this question straightaway. Applying this rule, we have$

$\begin{align*}BC^2 &= AB^2 + CA^2 - 2\cdot AB \cdot CA \cdot \cos A \\ &= 13^2 + 8^2-2\times 13 \times 8 \times \frac{12}{13} \\ &=41 \\ \Rightarrow BC &= \sqrt{41}\text{ cm}.\end{align*}$

(Cosine rule (a.k.a. Law of cosines): https://www.mathsisfun.com/algebra/trig-cosine-law.html )

AstroNerd · Nov 6, 2015

I have and I used the exact same working u did except I had x cos12/13

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