Question

A2012с16BFind cos(ZBAC + 30°).

Answer 1

Given:

$\begin{gathered} AC=12 \\ AB=20 \\ BC=16 \end{gathered}$

We know the indentity,

$\begin{gathered} \cos (A+B)=\cos A\cos B-\sin A\sin B \\ \cos (\angle BAC+30^{\circ})=\cos (\angle BAC)\cos 30^{\circ}-\sin (\angle BAC)\sin 30^{\circ} \\ \cos (\angle BAC+30^{\circ})=\cos (\angle BAC)\frac{\sqrt[]{3}}{2}^{}-\sin (\angle BAC)\frac{1}{2} \end{gathered}$

Now,

$\begin{gathered} \cos (\angle BAC)=\frac{AC}{AB}=\frac{12}{20}=\frac{3}{5} \\ \sin (\angle BAC)=\frac{CB}{AB}=\frac{16}{20}=\frac{4}{5} \end{gathered}$

It gives,

$\begin{gathered} \cos (\angle BAC+30^{\circ})=\cos (\angle BAC)\frac{\sqrt[]{3}}{2}^{}-\sin (\angle BAC)\frac{1}{2} \\ \cos (\angle BAC+30^{\circ})=\frac{3}{5}\times\frac{\sqrt[]{3}}{2}-\frac{4}{5}\times\frac{1}{2} \\ \cos (\angle BAC+30^{\circ})=\frac{3}{10}\sqrt[]{3}-\frac{2}{5} \end{gathered}$

Answer:

$\cos (\angle BAC+30^{\circ})=\frac{3}{10}\sqrt[]{3}-\frac{2}{5}$