National 5 Maths
The Cosine Rule

The SQA has never combined the cosine rule with simplifying surds, but it would be fair of them to do so.

YZ is opposite the angle whose cosine we know, so:

108° is the included angle and we are being asked to find the opposite length. So:

Rounding in questions like this needs extreme care. If possible, don't round at all until the very end. If you round too much before the square root stage, you will lose accuracy and your final answer might be wrong.

Although this question uses the letters A, B and C, they aren't in the same configuration as the formula. We need to be careful not to mix things up.

The required angle C is opposite the 12 mm. So we form the equation like this:

$$ \begin{eqnarray} 12^2 &=& 15^2+7^2-2(15)(7)\ cos\ C^\circ \\[6pt] 144 &=& 225+49-210\ cos\ C^\circ \\[6pt] \end{eqnarray} $$ $$ \begin{eqnarray} 144-225-49 &=& -210\ cos\ C^\circ \\[6pt] -130 &=& -210\ cos\ C^\circ \\[6pt] cos\ C^\circ &=& \frac{-130}{-210} = \frac{13}{21}\\[6pt] C &=& cos^{-1}\ \frac{13}{21} \\[6pt] &\approx& 51.8^\circ \\[6pt] \end{eqnarray} $$

Alternatively, we could avoid the need for all that rearrangement by using the second form of the cosine rule:

$$ \begin{eqnarray} cos\ C &=& \frac{7^2+15^2-12^2}{2 \times 7 \times 15} \\[6pt] &=& \frac{49+225-144}{210} \\[6pt] &=& \frac{130}{210} \\[6pt] C &=& cos^{-1}\ \frac{13}{21} \\[6pt] &\approx& 51.8^\circ \end{eqnarray} $$

A question like this appeared, somewhat unexpectedly, as Q7 on 2019 Paper 2.

It relies on the simple fact that the smallest angle is opposite the smallest side. So we take \(a=5.4\) and find \(A^\circ\).

$$ \begin{eqnarray} cos\ A &=& \frac{b^2+c^2-a^2}{2bc} \\[6pt] &=& \frac{7^2+9.1^2-5.4^2}{2 \times 7 \times 9.1} \\[6pt] &=& 0.8057299... \\[6pt] A &=& cos^{-1}\ 0.8057299... \\[6pt] &\approx& 36.3^\circ \end{eqnarray} $$

The straight line connecting the ends of Ruby's two lines form a triangle with her two pencil lines.

The third side is opposite the 37° angle.

So we can use the cosine rule to find the third length.

$$ \begin{eqnarray} a^2 &=& b^2+c^2-2bc\,cos\,A \\[6pt] &=& 4.8^2+6.3^2-2\times 4.8\times 6.3\,cos\,37^\circ \\[6pt] &=& 14.42852... \\[6pt] a &=& \sqrt{14.42852...} \\[6pt] &=& 3.79848... \\[6pt] &\approx& 3.8 \end{eqnarray} $$

So the ends of Ruby's lines should be 3.8 cm (or 38 mm) apart.

35° is the included angle and we are being asked to find the opposite length. So:

147° is the angle between the two given sides. So we can use the cosine rule to find the third side.

$$ \begin{eqnarray} QR^2 &=& 250^2+180^2-2(250)(180)\,cos\,147^\circ \\[6pt] &=& 62500+32400-90000\,cos\,147^\circ \\[6pt] &=& 170\,380.351...\\[6pt] QR &=& \sqrt{170\,380.351...} \\[6pt] &\approx& 412.8\ \small\textsf{m} \end{eqnarray} $$

XY is opposite the angle whose cosine we know, so:

The smallest angle in a triangle is opposite the smallest side. So we find \(Z^\circ\).

$$ \begin{eqnarray} cos\ Z &=& \frac{7.2^2+8.5^2-6.3^2}{2 \times 7.2 \times 8.5} \\[6pt] &=& \frac{211}{306} \\[6pt] Z &=& cos^{-1}\ \frac{211}{306} \\[6pt] &\approx& 46.4^\circ \end{eqnarray} $$

AB is opposite the angle whose cosine we know, so: