30°-60°-90° Triangle
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30°-60°-90° Triangle
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\(30^\circ\)–\(60^\circ\)–\(90^\circ\) Triangle
To construct the \(30^\circ\)–\(60^\circ\)–\(90^\circ\) triangle, start by constructing an equilateral triangle. Note that in an equilateral triangle, all side lengths and angle measures are equal. Specifically, all angles measure \(60^\circ\). The easiest side length to use is \(2\) units. (Note that you can choose any value, but using \(2\) will eliminate the need for future simplification.)
Next, bisect (cut in half) the base and the top \(60^\circ\) angle. The result will be a right triangle with \(60^\circ\), \(30^\circ\), and \(90^\circ\) angles. The two known lengths will be \(1\) unit and \(2\) units, as shown.
\(\begin{align}
a^2 + b^2 &= c^2 \\
1^2 + b^2 &= 2^2 \\
1 + b^2 &= 4 \\
b^2 &= 3 \\
b &= \sqrt 3 \\
\end{align}\)
a^2 + b^2 &= c^2 \\
1^2 + b^2 &= 2^2 \\
1 + b^2 &= 4 \\
b^2 &= 3 \\
b &= \sqrt 3 \\
\end{align}\)
Now, you can determine the exact value of each of the primary trigonometric ratios for \(30^\circ\) and \(60^\circ\).
Primary trigonometric ratios for \(30^\circ\):
\[\begin{align}
\sin \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\sin 30^\circ &= \frac{1}{2} \\
\end{align}\]
\sin \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\sin 30^\circ &= \frac{1}{2} \\
\end{align}\]
\[\begin{align}
\cos \theta &= \frac{{{\mathop{\rm adj}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\cos 30^\circ &= \frac{{\sqrt 3 }}{2} \\
\end{align}\]
\cos \theta &= \frac{{{\mathop{\rm adj}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\cos 30^\circ &= \frac{{\sqrt 3 }}{2} \\
\end{align}\]
\[\begin{align}
\tan \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm adj}\nolimits} }} \\
\tan 30^\circ &= \frac{1}{{\sqrt 3 }} \cdot\frac{{\sqrt 3 }}{{\sqrt 3 }} \\
\tan 30^\circ &= \frac{{\sqrt 3 }}{3} \\
\end{align}\]
\tan \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm adj}\nolimits} }} \\
\tan 30^\circ &= \frac{1}{{\sqrt 3 }} \cdot\frac{{\sqrt 3 }}{{\sqrt 3 }} \\
\tan 30^\circ &= \frac{{\sqrt 3 }}{3} \\
\end{align}\]
Primary trigonometric ratios for \(60^\circ\):
\[\begin{align}
\sin \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\sin 60^\circ &= \frac{{\sqrt 3 }}{2} \\
\end{align}\]
\sin \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\sin 60^\circ &= \frac{{\sqrt 3 }}{2} \\
\end{align}\]
\[\begin{align}
\cos \theta &= \frac{{{\mathop{\rm adj}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\cos 60^\circ &= \frac{1}{2} \\
\end{align}\]
\cos \theta &= \frac{{{\mathop{\rm adj}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\cos 60^\circ &= \frac{1}{2} \\
\end{align}\]
\[\begin{align}
\tan \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm adj}\nolimits} }} \\
\tan 60^\circ &= \frac{{\sqrt 3 }}{1} \\
&= \sqrt 3 \\
\end{align}\]
\tan \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm adj}\nolimits} }} \\
\tan 60^\circ &= \frac{{\sqrt 3 }}{1} \\
&= \sqrt 3 \\
\end{align}\]