Sunday, July 14, 2013

# What are vertical angles? GRE Math Fact

2:22 AM
When two lines intersect, two pairs of angles are formed. The opposite angles in such an intersection are called vertical angles.

Vertical angles are always equal to each other.
In the above figure, $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \angle x_1$  and  $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \angle x_2$ are vertical angles. Such that $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \angle x_1$$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \angle x_2$. Similarly $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \angle y_1$ and $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \angle y_2$ are vertical angles such that $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \angle y_1$$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \angle y_2$
Let’s look at an example where this concept plays a role:
Example:
In the above figure, lines $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} l_1$ $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} l_2$ and $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} l_3$ intersect at point O. If $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle p$ =100º and $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle h$ =30º, what is the measure of $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle t$ ?
A) 30º
B) 50º
C) 80º
D) 100º
E) 150º
Solution:

Step 1: First draw the vertical angle $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle x$ as shown in the following figure:
Since $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle x$ and $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle h$ are vertical angles, $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle x$ $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle h$ = 30º
Step 2: $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle p$ $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle x$ and $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle t$ are angles along the straight line $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} l_1$ and therefore the sum of their measures should be equal to 180º
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle p + \angle x + \angle t$ = 180º
Since $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle p$ = 100º and $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle x$ = 30º,  $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} \angle t = 180 - 100 - 30 = 50$