ANGLES IN QUADRANTS

ANGLES IN QUADRANTS
If  \(\theta=\) 30⁰ and 60⁰
    The values of trigonometric ratios are defined from equilateral triangle,
Let \(\Delta ABC\) be an equilateral triangle such that AB = BC = ca = a units

    \(AD \bot BC \Rightarrow BD = DC = \frac{a}{2}U\)
     from \(\Delta ABC\) ,(\(\angle ADB = {90^0}\))
\(\angle ADB = {90^0}\)
\(= {a^2} - \frac{{{a^2}}}{4}\)
\(\Rightarrow AD = \frac{{3{a^2}}}{4} \Rightarrow \frac{{\sqrt 3 }}{2}aU\)
From \(\Delta ABD\) or \(({\Delta ^\varphi }ADC)\)
\(\cos {30^0} = \frac{{AD}}{{AB}} = \frac{{\frac{{\sqrt 3 }}{2}a}}{a} = \frac{{\sqrt 3 }}{2} \Rightarrow \cos {30^0} = \frac{{\sqrt 3 }}{2}\)
\(\tan {30^0} = \frac{{BD}}{{AD}} = \frac{{\frac{a}{2}}}{{\frac{{\sqrt 3 }}{2}a}} = \frac{1}{{\sqrt 3 }} \Rightarrow \,\,\,\,\,\tan {30^0} = \frac{1}{{\sqrt 3 }}\)
\(\csc {30^0} = \frac{{AB}}{{BD}} = \frac{a}{{\frac{a}{2}}} = 2 \Rightarrow \,\,\,\,\,\csc {30^0} = 2\)
\(\sec {30^0} = \frac{{AB}}{{AD}} = \frac{a}{{\frac{{\sqrt 3 }}{2}a}} = \frac{2}{{\sqrt 3 }} \Rightarrow \sec {30^0} = \frac{2}{{\sqrt 3 }}\)