Special Right Triangles

Trigonometry: Special Right Triangles

Tip: This isn't the place to ask a question because the teacher can't reply.

Key Questions

• There are 2 types of special right triangles.
Type 1. Triangle that is half of a equilateral triangle. Its 3 angle measures are: 30, 60 and 90 deg. Its side measures are : a, a/2; and (a*sqr.3)/2.
Type 2. Triangle that has its side measures in the ratio of 3:4:5. The proof is given by the Pythagor theorem: c^2 = b^2 + a^2.
Use of special right triangles.
In the old time, people use the special right triangles with sides ratio 3:4:5 to figure out, in the field, a right angle or a rectangular, or square, shape.
Now, students just use the properties of special right triangle to find, by computing, the unknown sides or angles.

• The basic properties are: An interior angle of ${45}^{\circ}$, an exterior angle of ${135}^{\circ}$, an opposite angleof ${45}^{\circ}$, a right angle (${90}^{\circ}$), opposite side length: $\frac{\sqrt{2}}{2}$, and adjacent side length of $\frac{\sqrt{2}}{2}$, and a hypotenuse of 1. (side lengths for a standard triangle inside of the Unit Circle)

Explanation

All standard right triangles inside of the unit circle have a hypotenouse of 1. This is because the unit circle is defined as a circle with a radius of 1. Thus we create Unit Circle Triangles by picking a radius, and drawing its vector components.

In this case, we know we have a triangle with a primary angle of ${45}^{\circ}$ and, as mentioned before, a hypotneouse of 1. As such we can use $\sin \theta$ and $\cos \theta$ to determine the other two lengths.

$\sin \theta = \frac{o}{h} \to h \sin \theta = o \to 1 \cdot \sin \left({45}^{\circ}\right) = \frac{\sqrt{2}}{2}$

$\cos \theta = \frac{a}{h} \to h \cos \theta = a \to 1 \cdot \cos \left({45}^{\circ}\right) = \frac{\sqrt{2}}{2}$

• $m a t h b f \left\{{30}^{\circ} \text{-"60^circ"-} {90}^{\circ}\right\}$ Triangle

The ratios of three sides of a ${30}^{\circ} \text{-"60^circ"-} {90}^{\circ}$ triangle are:

$1 : \sqrt{3} : 2$

I hope that this was helpful.

• This key question hasn't been answered yet.

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