Special right triangles Knowledge-Based Study

Study notes

1. Overview of Special Right Triangles

   In trigonometry and geometry, special right triangles are right-angled triangles whose interior angles yield fixed, predictable ratios among their side lengths. Because these geometric relationships stem directly from regular polygons—specifically the square and the equilateral triangle—they allow us to determine exact side lengths and exact trigonometric values without relying on decimal approximations from a calculator.

   The two primary special right triangles studied in trigonometry are:

   • The $45^{\circ}\text{--}45^{\circ}\text{--}90^{\circ}$ triangle (an isosceles right triangle)
   • The $30^{\circ}\text{--}60^{\circ}\text{--}90^{\circ}$ triangle (half of an equilateral triangle)

   Mastering these two configurations is essential for understanding the unit circle and evaluating exact trigonometric functions of reference angles such as $30^{\circ}$, $45^{\circ}$, and $60^{\circ}$ (or $\frac{\pi}{6}$, $\frac{\pi}{4}$, and $\frac{\pi}{3}$ radians).

   Properties and Side Ratios of the $45^{\circ}\text{--}45^{\circ}\text{--}90^{\circ}$ Triangle

   A $45^{\circ}\text{--}45^{\circ}\text{--}90^{\circ}$ triangle is formed by cutting a square in half along its diagonal. Because the two acute angles are equal, the triangle is isosceles, meaning its two legs are congruent.

   Derivation of the Side Lengths

   Let each leg have length $x$, where $x > 0$. By the Pythagorean theorem, the hypotenuse $h$ satisfies:

   $h^2 = x^2 + x^2 = 2x^2$

   Taking the positive square root of both sides gives:

   $h = x\sqrt{2}$

   Key Relationships

   • Legs: Both legs are equal in length ($\text{leg}_1 = \text{leg}_2 = x$).
   • Hypotenuse: The hypotenuse is always equal to the leg length multiplied by $\sqrt{2}$ ($\text{hypotenuse} = x\sqrt{2}$).
   • Finding a leg from the hypotenuse: If the hypotenuse $h$ is known, each leg is given by $x = \frac{h}{\sqrt{2}} = \frac{h\sqrt{2}}{2}$.

   The standard side ratio for any $45^{\circ}\text{--}45^{\circ}\text{--}90^{\circ}$ triangle is therefore $1 : 1 : \sqrt{2}$.

   Properties and Side Ratios of the $30^{\circ}\text{--}60^{\circ}\text{--}90^{\circ}$ Triangle

   A $30^{\circ}\text{--}60^{\circ}\text{--}90^{\circ}$ triangle is formed by dropping an altitude from any vertex of an equilateral triangle to the opposite base, bisecting both the vertex angle and the opposite side.

   Derivation of the Side Lengths

   Consider an equilateral triangle with side length $2x$. Bisecting one angle produces two congruent right triangles, each with acute angles of $30^{\circ}$ and $60^{\circ}$:

   • The side opposite the $30^{\circ}$ angle (the shorter leg) has length $x$.
   • The hypotenuse retains the original side length, $2x$.

   Using the Pythagorean theorem to solve for the longer leg $y$ (opposite the $60^{\circ}$ angle):

   $x^2 + y^2 = (2x)^2$

   $x^2 + y^2 = 4x^2$

   $y^2 = 3x^2$

   $y = x\sqrt{3}$

   Key Relationships

   • Shorter Leg (opposite $30^{\circ}$): Length $x$
   • Longer Leg (opposite $60^{\circ}$): Length $x\sqrt{3}$
   • Hypotenuse (opposite $90^{\circ}$): Length $2x$

   The standard side ratio for any $30^{\circ}\text{--}60^{\circ}\text{--}90^{\circ}$ triangle is therefore $1 : \sqrt{3} : 2$.

   Trigonometric Functions of Special Angles

   Using the standard definitions of sine, cosine, and tangent in a right triangle ($\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$, and $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$), we can determine the exact trigonometric values for $30^{\circ}$, $45^{\circ}$, and $60^{\circ}$.

   Values for $30^{\circ}$ ($\frac{\pi}{6}$)

   • $\sin 30^{\circ} = \frac{x}{2x} = \frac{1}{2}$
   • $\cos 30^{\circ} = \frac{x\sqrt{3}}{2x} = \frac{\sqrt{3}}{2}$
   • $\tan 30^{\circ} = \frac{x}{x\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

   Values for $45^{\circ}$ ($\frac{\pi}{4}$)

   • $\sin 45^{\circ} = \frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
   • $\cos 45^{\circ} = \frac{x}{x\sqrt{2}} = \frac{\sqrt{2}}{2}$
   • $\tan 45^{\circ} = \frac{x}{x} = 1$

   Values for $60^{\circ}$ ($\frac{\pi}{3}$)

   • $\sin 60^{\circ} = \frac{x\sqrt{3}}{2x} = \frac{\sqrt{3}}{2}$
   • $\cos 60^{\circ} = \frac{x}{2x} = \frac{1}{2}$
   • $\tan 60^{\circ} = \frac{x\sqrt{3}}{x} = \sqrt{3}$

   Applying Special Right Triangle Relationships

   Example 1: Finding Missing Sides in a $45^{\circ}\text{--}45^{\circ}\text{--}90^{\circ}$ Triangle

   Problem: A $45^{\circ}\text{--}45^{\circ}\text{--}90^{\circ}$ triangle has a hypotenuse of length 10. Find the exact length of each leg.

   Solution: In this triangle, $\text{hypotenuse} = x\sqrt{2}$. Setting up the equation:

   $x\sqrt{2} = 10$

   $x = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$

   Each leg measures $5\sqrt{2}$.

   Example 2: Finding Missing Sides in a $30^{\circ}\text{--}60^{\circ}\text{--}90^{\circ}$ Triangle

   Problem: In a right triangle, one acute angle measures $60^{\circ}$, and the leg opposite this angle has length $6$. Find the lengths of the shorter leg and the hypotenuse.

   Solution: The side opposite $60^{\circ}$ is the longer leg, so $x\sqrt{3} = 6$:

   $x = \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$

   The shorter leg has length $2\sqrt{3}$. The hypotenuse is twice the shorter leg:

   $\text{hypotenuse} = 2x = 2(2\sqrt{3}) = 4\sqrt{3}$