Find $$\cos (<B A C+60^{\circ})$$ ![](http://admin.getprephub.com/media/editor/te15_20230123200717045831.png)

We know that, $$\cos (< B A C)=\frac{A C}{A B}=\frac{5}{13}$$ $$\cos 60^{\circ}=\frac{1}{2}, \sin (< B A C)=\frac{B C}{A B}=\frac{12}{13}$$ $$\sin 60^{\circ}=\frac{\sqrt{3}}{2}$$ Let , $$< B A C=\theta$$, $$\cos (\theta+60^{\circ})=\cos \theta \cos 60^{\circ}-\sin \theta \sin 60^{\circ}$$ Putting the values, $$\cos (\theta+60^{\circ})=(\frac{5}{13} \times \frac{1}{2})-(\frac{12}{13} \times \frac{\sqrt{3}}{2})=\frac{5}{26}-\frac{6 \sqrt{3}}{13}$$

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Question

Difficulty Level:

Medium

MCQ

Find $\cos (<B A C+60^{\circ})$

Select all that apply :

$\frac{5}{26}-\frac{6 \sqrt{3}}{13}$

$\frac{5}{26}+\frac{6 \sqrt{3}}{13}$

$\frac{2}{5}-\frac{\sqrt{3}}{10}$

$\frac{3}{10}-\frac{\sqrt{3}}{5}$

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