Find $$\sin (<A B C-60^{\circ})$$ ![](http://admin.getprephub.com/media/editor/te14_20230123200651165342.png)

We know that, $$\sin (< A B C)=\frac{\text{opposite leg length}}{\text{Hypotenuse}}=\frac{A C}{A B}=\frac{6}{10}=\frac{3}{5}$$ $$\cos 60^{\circ}=\frac{1}{2^{\prime}}$$ $$\cos (< A B C)=\frac{\text{Adjacent leg length}}{Hypotenuse}=\frac{B C}{A B}=\frac{8}{10}=\frac{4}{5}$$ $$\sin 60^{\circ}=\frac{\sqrt{3}}{2}$$ Let, $$< A B C=\theta$$ $$\sin (\theta-60^{\circ})=\sin \theta \cos 60^{\circ}-\cos \theta \sin 60^{\circ}$$ Putting the values, $$\sin (\theta-60^{\circ})=(\frac{3}{5} \times \frac{1}{2})-(\frac{4}{5} \times \frac{\sqrt{3}}{2})=\frac{3}{10}-\frac{4 \sqrt{3}}{10}=\frac{3}{10}-\frac{2 \sqrt{3}}{5}$$

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Question

Difficulty Level:

Medium

MCQ

Find $\sin (<A B C-60^{\circ})$

Select all that apply :

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

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

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

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

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