45°-45°-90° Triangle
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45°-45°-90° Triangle
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\(45^\circ\)–\(45^\circ\)–\(90^\circ\) Triangle
To construct the \(45^\circ\)–\(45^\circ\)–\(90^\circ\) triangle, construct a right triangle with two \(45^\circ\) angles. Because this triangle has two equal angles (both \(45^\circ\)), it is an isosceles triangle. Isosceles triangles not only have two equal angles, they also have two equal side lengths opposite the equal angles. For simplicity, give the equal sides a length of \(1\) unit. (Note that you can choose any value, but using \(1\) will eliminate the need for future simplification.)
Next, determine the length of the hypotenuse using the Pythagorean Theorem.
\(\begin{align}
a^2 + b^2 &= c^2 \\
1^2 + 1^2 &= c^2 \\
1 + 1 &= c^2 \\
2 &= c^2 \\
\sqrt 2 &= c \\
\end{align}\)
a^2 + b^2 &= c^2 \\
1^2 + 1^2 &= c^2 \\
1 + 1 &= c^2 \\
2 &= c^2 \\
\sqrt 2 &= c \\
\end{align}\)
\[\begin{align}
\sin \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\sin 45^\circ &= \frac{1}{{\sqrt 2 }} \\
\sin 45^\circ &= \frac{1}{{\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} \\
\sin 45^\circ &= \frac{{\sqrt 2 }}{2} \\
\end{align}\]
\sin \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\sin 45^\circ &= \frac{1}{{\sqrt 2 }} \\
\sin 45^\circ &= \frac{1}{{\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} \\
\sin 45^\circ &= \frac{{\sqrt 2 }}{2} \\
\end{align}\]
\[\begin{align}
\cos \theta &= \frac{{{\mathop{\rm adj}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\cos 45^\circ &= \frac{1}{{\sqrt 2 }} \\
\cos 45^\circ &= \frac{1}{{\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} \\
\cos 45^\circ &= \frac{{\sqrt 2 }}{2} \\
\end{align}\]
\cos \theta &= \frac{{{\mathop{\rm adj}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\cos 45^\circ &= \frac{1}{{\sqrt 2 }} \\
\cos 45^\circ &= \frac{1}{{\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} \\
\cos 45^\circ &= \frac{{\sqrt 2 }}{2} \\
\end{align}\]
\[\begin{align}
\tan \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm adj}\nolimits} }} \\
\tan 45^\circ &= \frac{1}{1} \\
\tan 45^\circ &= 1 \\
\end{align}\]
\tan \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm adj}\nolimits} }} \\
\tan 45^\circ &= \frac{1}{1} \\
\tan 45^\circ &= 1 \\
\end{align}\]