Right Triangles and Trigonometry Pattern - Direct Calculation

This pattern tests your ability to use the Pythagorean theorem and SOHCAHTOA to find a missing side length or trigonometric ratio in a right triangle.

 

The Pythagorean Theorem

In any right triangle with legs $a$ and $b$ and hypotenuse $c$:

$a^2 + b^2 = c^2$

The hypotenuse is always the longest side — it's the one opposite the right angle. The other two sides are the legs.

 

Finding the Hypotenuse

If you know both legs, square them, add, and take the square root.

In the right triangle shown, the lengths of the two legs are given. Which of the following is closest to the length of the hypotenuse?

A) $4.0$
B) $12.1$
C) $13.0$
D) $16.0$

$c = \sqrt{5^2 + 11^2} = \sqrt{25 + 121} = \sqrt{146} \approx 12.1$. The answer is B. Option D ($16$) comes from adding $5 + 11$ — that's not how the Pythagorean theorem works (you must square first, then add, then take the root).

 

Finding a Leg

If you know the hypotenuse and one leg, rearrange: $a = \sqrt{c^2 - b^2}$.

In the right triangle shown, the length of leg $q$ is $7$ and the length of the hypotenuse $r$ is $12$. Which expression gives the value of $p$?

A) $\sqrt{12^2 + 7^2}$
B) $\sqrt{12^2 - 7^2}$
C) $12 - 7$
D) $\sqrt{12 \cdot 7}$

We need a leg, so we subtract: $p = \sqrt{r^2 - q^2} = \sqrt{12^2 - 7^2}$. The answer is B. Option A adds the squares — that would give the hypotenuse if $12$ and $7$ were both legs. Option C just subtracts, which doesn't apply here.

Gotcha — Adding vs. Subtracting: - Finding the hypotenuse: $c = \sqrt{a^2 + b^2}$ (add the squares) - Finding a leg: $a = \sqrt{c^2 - b^2}$ (subtract the squares)

The key: ask yourself "Am I looking for the longest side or a shorter side?" If the longest, add. If a shorter, subtract.

 

Common Pythagorean Triples

These integer sets satisfy $a^2 + b^2 = c^2$ and show up constantly on the SAT:

• $3$-$4$-$5$ (and multiples: $6$-$8$-$10$, $9$-$12$-$15$, $15$-$20$-$25$, ...)
• $5$-$12$-$13$ (and multiples: $10$-$24$-$26$, ...)
• $8$-$15$-$17$
• $7$-$24$-$25$

Recognizing these saves time — you can skip the calculation entirely.

 

Word Problems: Ladders, Diagonals, Braces

Real-world problems almost always create a right triangle. Identify the three sides:

A ladder that is $13$ feet long leans against a vertical wall. The top of the ladder touches the wall at a point $7$ feet above the ground. How far is the base of the ladder from the base of the wall, in feet?

A) $6$
B) $120$
C) $\sqrt{120}$
D) $\sqrt{218}$

The ladder is the hypotenuse ($13$), the wall height is one leg ($7$), and the distance from the wall is the other leg.
$d = \sqrt{13^2 - 7^2} = \sqrt{169 - 49} = \sqrt{120}$
The answer is C. Option A ($6 = 13 - 7$) just subtracts, which is wrong. Option D ($\sqrt{218}$) adds the squares — that would apply if you were finding the hypotenuse.

A rectangular garden gate is reinforced with a diagonal brace. The height of the gate is $12$ feet and the diagonal brace is $13$ feet. Which expression represents the width $w$ of the gate?

A) $13^2 - 12^2$
B) $13 - 12$
C) $\sqrt{13^2 - 12^2}$
D) $\sqrt{13^2 + 12^2}$

The diagonal is the hypotenuse, so: $w = \sqrt{13^2 - 12^2}$. The answer is C. Option A forgot the square root. Option D adds instead of subtracting. (Incidentally, $\sqrt{169 - 144} = \sqrt{25} = 5$, a classic $5$-$12$-$13$ triple.)

Tip for word problems: The longest measurement (diagonal, ladder, brace, distance "as the crow flies") is almost always the hypotenuse.

 

SOHCAHTOA: Trigonometric Ratios

For an acute angle $\theta$ in a right triangle:

$\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$, $\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$, $\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$

"Opposite" and "adjacent" are always relative to the angle you're looking at. The hypotenuse is always the same side regardless of the angle.

In the right triangle shown, what is the value of $\cos y°$?

A) $\dfrac{17}{15}$
B) $\dfrac{15}{17}$
C) $\dfrac{2}{17}$
D) $\dfrac{32}{17}$

For $\cos y°$: adjacent over hypotenuse. The side adjacent to angle $y$ is $15$, and the hypotenuse is $17$.
$\cos y° = \dfrac{15}{17}$
The answer is B. Option A ($\dfrac{17}{15}$) flips the fraction — that's $\sec y$, not $\cos y$. Cosine is always $\leq 1$ for an acute angle.

In right triangle $PQR$, the right angle is at $Q$. The sides are $a = 5$, $b = 12$, $c = 13$. The measure of $\angle PRQ$ is $y°$. What is $\sin y°$?

A) $\dfrac{12}{13}$
B) $\dfrac{5}{12}$
C) $\dfrac{5}{13}$
D) $\dfrac{13}{5}$

For $\sin y°$: opposite over hypotenuse. The side opposite $\angle PRQ$ is $PQ = 5$ (across from the angle), and the hypotenuse is $PR = 13$.
$\sin y° = \dfrac{5}{13}$
The answer is C. Option A ($\dfrac{12}{13}$) gives the cosine of $y$, not the sine. Option B ($\dfrac{5}{12}$) gives the tangent.

Gotcha — Picking the Right Sides:

Students commonly mix up "opposite" and "adjacent." Here's a foolproof method: 1. Point to your angle. Now look across the triangle — the side directly opposite is the "opposite" side. 2. The side touching your angle (that isn't the hypotenuse) is the "adjacent" side. 3. The hypotenuse is always the longest side, opposite the $90°$ angle.

 

When You Need the Pythagorean Theorem AND Trig

Some problems give you two sides of a right triangle and ask for a trig ratio. You may need to find the missing third side first.

In a right triangle, the legs are $12$ and $5$. What is $\cos$ of the angle opposite the side of length $5$?

First find the hypotenuse: $c = \sqrt{12^2 + 5^2} = \sqrt{169} = 13$.
The angle opposite the side of length $5$ has adjacent side $12$ and hypotenuse $13$.
$\cos \theta = \dfrac{12}{13}$
The answer is $\dfrac{12}{13}$.

 

What to Do on Test Day

• Pythagorean theorem: $a^2 + b^2 = c^2$. The hypotenuse $c$ is always the longest side (opposite the right angle)
• Finding hypotenuse → add squares. Finding a leg → subtract squares. Getting this backward is the #1 error
• Memorize the common triples: $3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$, $7$-$24$-$25$. Also recognize multiples (like $6$-$8$-$10$)
• SOHCAHTOA: $\sin = \dfrac{\text{opp}}{\text{hyp}}$, $\cos = \dfrac{\text{adj}}{\text{hyp}}$, $\tan = \dfrac{\text{opp}}{\text{adj}}$
• "Opposite" and "adjacent" depend on which angle you're examining. Always ask: "Opposite to which angle? Adjacent to which angle?"
• Sine and cosine of an acute angle are always between 0 and 1. If your answer is greater than 1, you've flipped the fraction
• Tangent can be greater than 1 (when the opposite side is longer than the adjacent side)
• In word problems: ladder/diagonal/brace = hypotenuse. Height and base distance = legs
• "Closest to" problems require computing $\sqrt{\text{something}}$. Estimate by finding which perfect squares it falls between