Special Right Triangles Study Notes: Section 9.2

Section 9.2: Special Right Triangles Overview

• Objectives:     * Learn and apply methods to find side lengths in special right triangles ($30^{\circ}-60^{\circ}-90^{\circ}$ and $45^{\circ}-45^{\circ}-90^{\circ}$).     * Develop skills to solve real-life application problems involving special right triangles.

• Date of Record: Monday, April 27, 2026.

• Context: Unit 8: Right Triangles & Trigonometry, Homework 2.

Properties of the $30^{\circ}-60^{\circ}-90^{\circ}$ Triangle

In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, the sides exist in a specific ratio relative to the "short leg," which is the side opposite the $30^{\circ}$ angle. The three sides are categorized as follows:

• Short Leg (SL): The side opposite the $30^{\circ}$ angle.

• Long Leg (LL): The side opposite the $60^{\circ}$ angle.

• Hypotenuse (Hyp): The side opposite the $90^{\circ}$ angle.

Mathematical Formulas for Side Ratios:

• Hypotenuse Calculation:     * $\text{hypotenuse} = 2 \times \text{short leg}$

• Long Leg Calculation:     * $\text{long leg} = \text{short leg} \times \sqrt{3}$

• Short Leg Extraction (from Hypotenuse):     * $\text{short leg} = \frac{\text{hypotenuse}}{2}$

• Short Leg Extraction (from Long Leg):     * $\text{short leg} = \frac{\text{long leg}}{\sqrt{3}} = \frac{\text{long leg} \times \sqrt{3}}{3}$

Properties of the $45^{\circ}-45^{\circ}-90^{\circ}$ Triangle

A $45^{\circ}-45^{\circ}-90^{\circ}$ triangle is an isosceles right triangle. The two legs are congruent, and the hypotenuse is the longest side.

• Leg (L): The sides opposite the $45^{\circ}$ angles.

• Hypotenuse (H): The side opposite the $90^{\circ}$ angle.

Mathematical Formulas for Side Ratios:

• Hypotenuse Calculation:     * $\text{hypotenuse} = \text{leg} \times \sqrt{2}$

• Leg Calculation (from Hypotenuse):     * $\text{leg} = \frac{\text{hypotenuse}}{\sqrt{2}} = \frac{\text{hypotenuse} \times \sqrt{2}}{2}$

Step-by-Step Computational Examples

Example 1: Given Hypotenuse in a $30^{\circ}-60^{\circ}-90^{\circ}$ Triangle

• Given Hypotenuse = $46$.

• Required: Find side $x$ (long leg) and side $y$ (short leg).

• Solution for y:     * $46 = 2 \times y$     * $y = 23$

• Solution for x:     * $x = y \times \sqrt{3}$     * $x = 23\sqrt{3}$

Example 2: Given Long Leg in a $30^{\circ}-60^{\circ}-90^{\circ}$ Triangle

• Given Long Leg = $20$.

• Required: Find side $x$ (hypotenuse) and side $y$ (short leg).

• Solution for y:     * $20 = y \times \sqrt{3}$     * $y = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3}$

• Solution for x:     * $x = 2 \times y$     * $x = 2 \times \left(\frac{20\sqrt{3}}{3}\right) = \frac{40\sqrt{3}}{3}$

Example 3: Given Variations in Expression

• Given Long Leg = $9\sqrt{3}$.

• Required: Find side $x$ (short leg) and side $y$ (hypotenuse).

• Solution for x:     * $9\sqrt{3} = x \times \sqrt{3}$     * $x = 9$

• Solution for y:     * $y = 2 \times 9 = 18$

Real-World Applications

Application 1: Estimating the Area of a Road Sign

• Scenario: A Yield sign is shaped like an equilateral triangle with a side length of $36\,\text{in}$.

• Conceptual Approach: Dropping an altitude from one vertex of an equilateral triangle creates two congruent $30^{\circ}-60^{\circ}-90^{\circ}$ triangles.

• Dimensions identified:     * Side (Hypotenuse of the right triangle) = $36\,\text{in}$.     * Base of the right triangle (Short leg) = $\frac{36}{2} = 18\,\text{in}$.     * Height $h$ (Long leg) = $18\sqrt{3}\,\text{in}$.

• Area Calculation:     * Formula: $A = \frac{1}{2}bh$     * $A = \frac{1}{2} \times (36) \times (18\sqrt{3})$     * $A = 18 \times 18\sqrt{3} = 324\sqrt{3}$     * Final Estimate: $A \approx 561.18\,\text{in}^2$

Application 2: Tipping Platform Ramp Height

• Scenario: An $80\text{-foot}$ ramp is used for unloading trucks. Find the height ($h$) of the end of the ramp at two different tipping angles.

• Case 1: Tipping Angle = $30^{\circ}$     * The height is the leg opposite the $30^{\circ}$ angle (the short leg).     * ${\text{Ramp Length (Hyp)}} = 80\,\text{ft}$.     * $h = \frac{80}{2} = 40\,\text{ft}$.

• Case 2: Tipping Angle = $45^{\circ}$     * The height is the leg of an isosceles right triangle ($45^{\circ}-45^{\circ}-90^{\circ}$).     * $80 = h \times \sqrt{2}$     * $h = \frac{80}{\sqrt{2}} = 40\sqrt{2} \approx 56.57\,\text{ft}$.

Comprehensive Problem Analysis

From the "Homework 2: Special Right Triangles" worksheet, various orientations and values are explored to reinforce side-length relationships:

• Standard Hipotenuse Given:     * Problem 1: $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with hyp = $34$. $SL = 17$, $LL = 17\sqrt{3}$.     * Problem 2: $45^{\circ}-45^{\circ}-90^{\circ}$ triangle with hyp = $13$. $L = \frac{13\sqrt{2}}{2}$.

• Radical Values in Side Lengths:     * Problem 7: $45^{\circ}-45^{\circ}-90^{\circ}$ triangle with hyp = $2\sqrt{14}$. $L = \frac{2\sqrt{14}}{\sqrt{2}} = 2\sqrt{7}$.     * Problem 8: $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with hyp = $10\sqrt{2}$. $SL = 5\sqrt{2}$, $LL = 5\sqrt{2} \times \sqrt{3} = 5\sqrt{6}$.     * Problem 9: $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with LL = $22\sqrt{3}$. $SL = 22$, $Hyp = 44$.     * Problem 11: $45^{\circ}-45^{\circ}-90^{\circ}$ triangle with leg = $\sqrt{6}$. $Hyp = \sqrt{6} \times \sqrt{2} = \sqrt{12} = 2\sqrt{3}$.

• Integer Long Legs:     * Problem 4: $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with LL = $45$. $SL = \frac{45}{\sqrt{3}} = 15\sqrt{3}$, $Hyp = 30\sqrt{3}$.     * Problem 12: $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with LL = $27$. $SL = \frac{27}{\sqrt{3}} = 9\sqrt{3}$, $Hyp = 18\sqrt{3}$.