PSAT 10 Right Triangles & Trigonometry: Pythagorean, Special Triangles, Sine-Cosine

What You Need to Know

Right-triangle geometry shows up constantly on the PSAT 10 because it connects algebra, geometry, and basic trigonometry in quick, solvable setups. The test mainly expects you to:

• Use the Pythagorean Theorem to find missing side lengths.
• Recognize and use special right triangles (no calculator-style trig tables needed).
• Use sine and cosine in right triangles to relate an angle to side ratios.

Core ideas (the “must-know” backbone)

• Right triangle: one angle is $90^\circ$.
• Hypotenuse: the side opposite the right angle; it’s always the longest side.
• Legs: the two non-hypotenuse sides.

Theorem + trig definitions (in one place)

• Pythagorean Theorem (only for right triangles):

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

where $c$ is the hypotenuse.

• Sine and cosine (right triangle, with respect to an acute angle $\theta$):

$\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$

$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$

Use these when you know an angle and one side and need another side (or when you need to build an equation from a word problem).

Critical reminder: On PSAT 10, trig is almost always right-triangle trig with acute angles (less than $90^\circ$).

Step-by-Step Breakdown

A) Using the Pythagorean Theorem to find a missing side

1. Confirm it’s a right triangle (given a right angle, or implied by perpendicular lines/axes).
2. Label the hypotenuse $c$ (opposite the $90^\circ$ angle).
3. Plug into:

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

4. Solve for the missing variable.
5. If the side length must be positive, reject negative roots.

Mini example (missing leg):

• Hypotenuse $c=13$, leg $a=5$, find leg $b$.

$5^2+b^2=13^2$

$25+b^2=169$

$b^2=144$

$b=12$

B) Recognizing special right triangles fast

1. Look for angles $30^\circ$, $45^\circ$, $60^\circ$, or words like isosceles right triangle.
2. Match to the correct ratio set (see table below).
3. Apply a scale factor to go from the ratio triangle to the actual triangle.

Mini example (scale factor):

• A $30\text{-}60\text{-}90$ triangle has short leg $=7$. Then:
  • hypotenuse $=14$
  • long leg $=7\sqrt{3}$

C) Using sine/cosine to find an unknown side

1. Choose the reference angle $\theta$ (the angle given).
2. Identify sides relative to $\theta$:
   • Opposite: across from $\theta$
   • Adjacent: touches $\theta$ (but is not the hypotenuse)
   • Hypotenuse: across from $90^\circ$
3. Pick the right ratio:
   • if you have/need opposite and hypotenuse, use $\sin(\theta)$
   • if you have/need adjacent and hypotenuse, use $\cos(\theta)$
4. Write an equation and solve.

Mini example (cosine):

• Adjacent to $\theta$ is $12$ and hypotenuse is $20$. Find $\cos(\theta)$:

$\cos(\theta)=\frac{12}{20}=\frac{3}{5}$

D) Using trig to find an angle (when ratios are given)

1. Identify which ratio is given (opposite/hypotenuse or adjacent/hypotenuse).
2. Set it equal to $\sin(\theta)$ or $\cos(\theta)$.
3. If the ratio matches a special-angle value, recognize it (common on PSAT 10).

Mini example (recognize special angle):

$\sin(\theta)=\frac{1}{2} \Rightarrow \theta=30^\circ$

If an answer choice uses special angles, the problem often expects recognition, not a calculator.

Key Formulas, Rules & Facts

Essential formulas and when to use them

Special right triangles (memorize the ratios)

$45\text{-}45\text{-}90$

• Angles: $45^\circ,45^\circ,90^\circ$
• Side ratio:

$1:1:\sqrt{2}$

If legs are $x$ and $x$, then hypotenuse is:

$x\sqrt{2}$

$30\text{-}60\text{-}90$

• Angles: $30^\circ,60^\circ,90^\circ$
• Side ratio (short leg opposite $30^\circ$):

$1:\sqrt{3}:2$

If short leg is $x$ (opposite $30^\circ$), then:

• long leg (opposite $60^\circ$) is $x\sqrt{3}$
• hypotenuse is $2x$

Special-angle trig values you should know

Common Pythagorean triples (recognize them quickly)

These are right triangles with integer side lengths.

Scaling works too (multiply all sides by the same factor). Example: $3\text{-}4\text{-}5$ becomes $6\text{-}8\text{-}10$.

Examples & Applications

Example 1: Pythagorean Theorem (algebraic side)

A right triangle has legs $x$ and $x+6$ and hypotenuse $x+12$. Find $x$.

Set up:

$x^2+(x+6)^2=(x+12)^2$

Key insight: Expand carefully and simplify.

$x^2+(x^2+12x+36)=x^2+24x+144$

$2x^2+12x+36=x^2+24x+144$

$x^2-12x-108=0$

Factor:

$(x-18)(x+6)=0$

Reject negative length, so:

$x=18$

Example 2: Distance on a coordinate grid (built from Pythagorean)

Find the distance between points $(-2,3)$ and $(4,-5)$.

Horizontal change:

$\Delta x=4-(-2)=6$

Vertical change:

$\Delta y=-5-3=-8$

Distance:

$d=\sqrt{6^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=10$

Key insight: This is just the Pythagorean Theorem on a coordinate plane.

Example 3: Special triangle identification

A right triangle has one acute angle $45^\circ$ and hypotenuse $10\sqrt{2}$. Find each leg.

Key insight: $45\text{-}45\text{-}90$ ratio is $1:1:\sqrt{2}$, so:

$\text{leg}=\frac{10\sqrt{2}}{\sqrt{2}}=10$

So both legs are:

$10 \text{ and } 10$

Example 4: Using sine/cosine in a word-style setup

A right triangle has hypotenuse $20$ and angle $\theta=30^\circ$. Find the length of the side opposite $\theta$.

Use sine:

$\sin(30^\circ)=\frac{\text{opp}}{20}$

$\frac{1}{2}=\frac{\text{opp}}{20}$

$\text{opp}=10$

Variation to expect: If they ask for the adjacent side instead, you’d use $\cos(30^\circ)=\frac{\sqrt{3}}{2}$.

Common Mistakes & Traps

1. Mixing up hypotenuse vs a leg: You plug the wrong side in as $c$ in $a^2+b^2=c^2$. The hypotenuse is always opposite $90^\circ$ and is the longest side. **Fix**: Circle the right angle; the opposite side is $c$.

2. Forgetting to take the square root: You solve for $b^2$ and stop there. **Fix**: Your final side length must be $b$, so take $\sqrt{\,}$ and keep only the positive value.

3. Using special-triangle ratios backwards: In a $30\text{-}60\text{-}90$, students often think the short leg is $\sqrt{3}$ times the long leg. It’s the other way: $\text{long}=\text{short}\times\sqrt{3}$. **Fix**: Always anchor: short leg is opposite $30^\circ$.

4. Calling the wrong side “opposite” or “adjacent”: Opposite/adjacent depends on the chosen angle $\theta$, not on the triangle itself. **Fix**: Point at $\theta$: the side across is opposite; the side touching (not hypotenuse) is adjacent.

5. Using sine when you need cosine (or vice versa): You pick based on which sides you have, not on what “feels right.” Fix: Write the fraction you need first (like $\frac{\text{adj}}{\text{hyp}}$), then match it to $\cos(\theta)$.

6. Assuming all right triangles are special triangles: If the angle is not $30^\circ$, $45^\circ$, or $60^\circ$, do not force special ratios. Fix: Use Pythagorean (if two sides given) or trig ratios (if an angle is given).

7. Dropping radicals incorrectly: Example: simplifying $\sqrt{50}$ as $5\sqrt{2}$ is correct, but many write $\sqrt{50}=\sqrt{25}+\sqrt{25}$ (wrong). **Fix**: Factor perfect squares: $\sqrt{50}=\sqrt{25\cdot 2}=5\sqrt{2}$.

8. Rounding too early (if a calculator is allowed in practice): Early rounding can change answer choices. Fix: Keep exact values like $\frac{\sqrt{2}}{2}$ and $\frac{\sqrt{3}}{2}$ as long as possible.

Memory Aids & Quick Tricks

Quick Review Checklist

• You can instantly label hypotenuse as the side opposite $90^\circ$.
• You can apply:

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

• You know special triangles:
  • $45\text{-}45\text{-}90$: $1:1:\sqrt{2}$
  • $30\text{-}60\text{-}90$: $1:\sqrt{3}:2$ (short leg opposite $30^\circ$)
• You know:

$\sin(\theta)=\frac{\text{opp}}{\text{hyp}}$

$\cos(\theta)=\frac{\text{adj}}{\text{hyp}}$

• You remember special-angle values for $30^\circ,45^\circ,60^\circ$.
• You avoid traps: don’t mix up opposite/adjacent, don’t forget square roots, don’t force special triangles.

You’ve got this, just stay disciplined about labels and ratios.