Triangles - Right Triangles

SAT Questions that focus on Right Triangles require knowledge of the following topics.

Pythagorean Theorem

In the right triangle ABC (figure below):
-"b" and "c" are the catheti, and "a" is the hypotenuse ($B\hat{A}C=90º$),
- "h" is the altitude relative to side "a",
- "m" is the projection of side "b" on side "a",
- "n" is the projection of side "c" on side "a":

Triangles ABC, HBA and HAC are similar (three congruent interior angles). Therefore, corresponding sides are proportional:

Equation 1: $\frac{h}{n}=\frac{m}{h}$. Thus, $h^2=mn$

Equation 2:  $\frac{a}{b}=\frac{c}{h}$. Thus, $b.c=ah$

Equation 3: $\frac{a}{b}=\frac{b}{m}$. Thus, $b^2=am$

Equation 4: $\frac{a}{c}=\frac{c}{n}$. Thus, $c^2=an$

Adding equations 1 and 2:

$b^2+c^2=am+an=a(m+n)=a(a)$

Equation 5: $b^2+c^2=a^2$ (Pythagorean Theorem)

Metric Relation between Catheti and Altitude

Consider now the following relation:

$\frac{1}{b^2}+\frac{1}{c^2}=\frac{b^2+c^2}{b^2c^2}=\frac{a^2}{(bc)^2}$

Since, by Equation 2, $bc=ah$,

$\frac{a^2}{(bc)^2}=\frac{a^2}{(ah)^2}=\frac{1}{h^2}$

Therefore,

$\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{h^2}$

Fundamental Trigonometric Equation

By definition:
Sine = Opposite Cathetus / Hypotenuse
Cosine = Adjacent Cathetus / Hypotenuse
Tangent = Opposite Cathetus / Adjacent Cathetus

In the right triangle in the figure below:

$sin(a)=\frac{C1}{H}=cos(b)$

$cos(a)=\frac{C2}{H}=sen(b)$

$tan(a)=\frac{C1}{C2}=\frac{C1/H}{C2/H}=\frac{sin(a)}{cos(a)}$


Consider now the following equation:

$sin²(a)+cos²(a)=(\frac{C1}{H})^2+(\frac{C2}{H})^2$=

$\frac{C1^2}{H^2}+\frac{C2^2}{H^2}=\frac{C1^2+C2^2}{H^2}$

By the Pythagoras Theorem $C1^2+C2^2=H^2$. Thus:

$\frac{C1^2+C2^2}{H^2}=\frac{H^2}{H^2}=1$

$sin²(a)+cos²(a)=1$ (Fundamental Trigonometric Equation)

Important Angles in Trigonometry

30º:
$sin(30º)=\frac{1}{2}$
$cos(30º)=\frac{\sqrt{3}}{2}$
$tan(30º)=\frac{\sqrt{3}}{3}$

45º:
$sin(45º)=\frac{\sqrt{2}}{2}$
$cos(45º)=\frac{\sqrt{2}}{2}$
$tan(45º)=1$

60º:
$sin(60º)=\frac{\sqrt{3}}{2}$
$cos(60º)=\frac{1}{2}$
$tan(60º)=\sqrt{3}$

Considere the equilateral triangle and the square in the figure below, both with sides that measure 1:

The height (h) of the triangle measures $\frac{\sqrt{3}}{2}$, and the diagonal (d) of the square measures $\sqrt{2}$.
With these measures, and using the catheti and the hypotenuses of the triangles formed, one can compute the values of the sine, cosine and tangent of 30º, 45º and 60º.

Solved SAT Practice Tests

Find Practice Tests in the following link:

SAT Practice Tests - Right Triangles and

Additional Practice Tests - Right Triangles