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)(1)/(2) \end{gathered}`

Now,

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

It gives,

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

Answer:

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