Question

cos, =√3/2 find sin
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Answer 1

Answer:

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$givn \: = \\ \\ \sinθ = \frac{ \sqrt{3} }{2} \\ \\ let \: us \: first \: draw \: a \: \: right \: angle \: △ABC\: \: < b \: = 90 \: degree \: and \: < A= θ</h3><h3> \\ \\ we \: know \: that \: \: \sinθ = \frac{AC}{AC} = \frac{(perpendicular) }{hypotenuse} = \frac{ \sqrt{3} }{2} \\ \\ by \: the \: pythagoras \: theorem \: = \\ \\ {AC}^{2} = {AB}^{2 } + {BC}^{2} \\ \\ \\ or \\ \\ \\ {AB}^{2} = {AC}^{2} - {BC}^{2} = {4k}^{2} - {3k}^{2} = {k}^{2} \\ \\ AB = k \\ \\ \\ find \: the \: sin \\ \\ = \\ \\ \secθ = \frac{1}{ \cosθ} = 2$

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