Plane Geometry and Trigonometry

📐 Plane Geometry & Trigonometry

Part 1 of 5 — Angles, Lines & Triangle Sums

Topics in This Part

🔑 Why this matters: Plane geometry and trig make up about 35–40% of the ACT Math section. Almost every figure on the test starts with a few angle facts — master these and you unlock the rest.

Angle Pairs You Must Know

Vertical angles are always equal. When two lines intersect, the angle pairs across from each other match:

$\text{if one angle is } 70^\circ,\ \text{the one across from it is also } 70^\circ$

The other two angles are each $180^\circ - 70^\circ = 110^\circ$ (supplementary, on a straight line).

💡 ACT shortcut: angles around a single point on one side of a line always total $180^\circ$; angles all the way around a point total $360^\circ$.

Concept Check 🎯

Parallel Lines Cut by a Transversal

When a straight line (a transversal) crosses two parallel lines, only two angle sizes appear — and they pair up:

Example

Two parallel lines are cut by a transversal. One interior angle is $110^\circ$.

• The alternate interior angle is also $110^\circ$.
• The co-interior angle on the same side is $180^\circ - 110^\circ = 70^\circ$.

⚠️ These rules only work when the lines are parallel. The ACT will tell you (or mark the figure with arrows) when they are.

Parallel-Line Angles 🔽

Lines $\ell$ and $m$ are parallel, cut by a transversal. An angle of $65^\circ$ is marked. Choose each related angle.

The Triangle Angle-Sum

The three interior angles of any triangle add to $180^\circ$:

$\angle A + \angle B + \angle C = 180^\circ$

Example: a triangle has angles $50^\circ$ and $60^\circ$. The third is $180^\circ - 50^\circ - 60^\circ = 70^\circ.$

Exterior Angle Theorem

An exterior angle of a triangle equals the sum of the two non-adjacent (remote) interior angles:

$\text{exterior} = 45^\circ + 70^\circ = 115^\circ$

💡 Quick check: that exterior angle ($115^\circ$) plus its neighbor interior angle ($65^\circ$) makes $180^\circ$ — a straight line. ✓

Find the Missing Angle 🧮

Enter each missing angle in degrees (just the number).

1) A triangle has angles $40^\circ$ and $75^\circ$. The third angle $= \,?$ 2) A right triangle has one acute angle of $32^\circ$. The other acute angle $= \,?$ 3) An exterior angle of a triangle equals the sum of remote interior angles $52^\circ$ and $61^\circ$. The exterior angle $= \,?$

📐 Plane Geometry & Trigonometry

Part 2 of 5 — Triangles: Pythagoras, Special Triangles & Similarity

🔑 The big three: the Pythagorean theorem, the two special right triangles, and similar-triangle proportions. These appear on virtually every ACT.

The Pythagorean Theorem

In a right triangle with legs $a$, $b$ and hypotenuse $c$ (the side opposite the right angle, always the longest):

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

Pythagorean Triples to Memorize

Multiples count too: $3\text{-}4\text{-}5$ scaled by $3$ gives $9\text{-}12\text{-}15$.

Example: find a leg

Hypotenuse $= 10$, one leg $= 6$. Find the other leg: $6^2 + b^2 = 10^2 \;\Rightarrow\; b^2 = 100 - 36 = 64 \;\Rightarrow\; b = 8$

⚠️ The hypotenuse is always $c$ — never put the longest side in place of a leg.

Pythagorean Drill 🧮

Enter each missing side length (a whole number).

1) Legs $9$ and $12$. Hypotenuse $= \,?$ 2) Leg $5$, hypotenuse $13$. Other leg $= \,?$ 3) Legs $8$ and $15$. Hypotenuse $= \,?$

The Two Special Right Triangles

These side ratios are fixed — the ACT loves them because no calculator is needed.

$45^\circ$-$45^\circ$-$90^\circ$ (isosceles right)

$\text{leg} : \text{leg} : \text{hypotenuse} \;=\; 1 : 1 : \sqrt{2}$

If each leg is $5$, the hypotenuse is $5\sqrt{2}$.

$30^\circ$-$60^\circ$-$90^\circ$

$\text{short leg} : \text{long leg} : \text{hypotenuse} \;=\; 1 : \sqrt{3} : 2$

• The short leg is opposite the $30^\circ$ angle.
• The hypotenuse is twice the short leg.
• The long leg (opposite $60^\circ$) is the short leg times $\sqrt{3}$.

Example: short leg $= 4 \Rightarrow$ long leg $= 4\sqrt{3}$, hypotenuse $= 8$.

💡 In a $30$-$60$-$90$, the hypotenuse is the biggest number ($2$), so it is double the smallest side — a fast sanity check.

Special Triangles 🎯

Similar Triangles

Two triangles are similar when their angles match. Corresponding sides are then proportional — same shape, scaled size.

$\frac{\text{side in triangle 1}}{\text{matching side in triangle 2}} = \text{constant scale factor}$

Example

Triangle $ABC$ is similar to triangle $DEF$. Side $AB = 6$ matches $DE = 9$; side $BC = x$ matches $EF = 12$. Set up the proportion:

$\frac{6}{9} = \frac{x}{12} \;\Rightarrow\; x = \frac{6 \cdot 12}{9} = \frac{72}{9} = 8$

🔑 Set-up rule: keep the same triangle on top of both fractions. Cross-multiply, then solve.

Similar-Triangle Proportions 🧮

Solve for the missing length (a whole number).

1) $\dfrac{4}{6} = \dfrac{x}{15}$. Find $x$. 2) A $5$-ft pole casts a $4$-ft shadow. At the same time a tree casts a $20$-ft shadow. Using $\dfrac{5}{4} = \dfrac{h}{20}$, find the tree height $h$ (ft).

📐 Plane Geometry & Trigonometry

Part 3 of 5 — Polygons, Circles, Area & Perimeter

🔑 Formula-heavy part: the ACT does not give you a formula sheet, so the area, circumference, and polygon-angle rules here must be memorized cold.

Area & Perimeter Formulas

Example: triangle area

Base $= 10$, height $= 6$: $A = \tfrac{1}{2}(10)(6) = 30$

⚠️ In a triangle, the height is the perpendicular distance to the base — not a slanted side. Use the side that meets the base at $90^\circ$.

Circles

For a circle with radius $r$ (and diameter $d = 2r$):

$\text{Area} = \pi r^2 \qquad \text{Circumference} = 2\pi r = \pi d$

Example: $r = 5$

$\text{Area} = \pi (5)^2 = 25\pi \qquad \text{Circumference} = 2\pi(5) = 10\pi$

Sectors (pie slices)

A sector spanning a central angle of $\theta$ degrees is the fraction $\dfrac{\theta}{360^\circ}$ of the whole circle:

$\text{Sector area} = \frac{\theta}{360^\circ}\cdot \pi r^2 \qquad \text{Arc length} = \frac{\theta}{360^\circ}\cdot 2\pi r$

Example: a $90^\circ$ sector of a circle with $r = 6$ is $\frac{90}{360} = \frac14$ of the circle: $\text{Sector area} = \tfrac14 \cdot 36\pi = 9\pi, \qquad \text{Arc length} = \tfrac14 \cdot 12\pi = 3\pi.$

Circle Check 🎯

Polygon Interior Angles

The interior angles of an $n$-sided polygon sum to:

$\text{Sum} = (n - 2)\cdot 180^\circ$

For a regular polygon (all angles equal), each interior angle is that sum divided by $n$:

Example (hexagon): $(6-2)\cdot 180^\circ = 720^\circ$, and $720^\circ \div 6 = 120^\circ$ each.

💡 The exterior angles of any convex polygon always sum to $360^\circ$ — a handy backup formula.

Polygons & Areas 🔽

Putting the Area Formulas to Work

Before the drill, lock in the three you will use most:

$\text{Rectangle: } A = lw, \qquad \text{Triangle: } A = \tfrac12 bh, \qquad \text{Circle: } A = \pi r^2$

On the ACT, leave a circle's area in terms of $\pi$ (e.g. $9\pi$) unless the question asks for a decimal — it is faster and avoids rounding mistakes.

💡 If a figure is a composite shape (a rectangle with a half-circle on top, say), find each piece's area separately and add them.

Area Drill 🧮

Enter each answer as a number (use $\pi$ as the symbol, not a decimal — e.g. write the coefficient only).

1) Rectangle $8$ by $5$: area $= \,?$ 2) Triangle, base $12$, height $7$: area $= \,?$ 3) Circle with radius $3$: area is $?\,\pi$ (enter the coefficient)

📐 Plane Geometry & Trigonometry

Part 4 of 5 — Right-Triangle Trigonometry (SOH-CAH-TOA)

🔑 The whole game: the three basic trig ratios connect an acute angle of a right triangle to its sides. Memorize SOH-CAH-TOA and you can find any missing side or angle.

SOH-CAH-TOA

Relative to a chosen acute angle $\theta$, label the sides:

• Opposite — the leg across from $\theta$
• Adjacent — the leg next to $\theta$ (not the hypotenuse)
• Hypotenuse — opposite the right angle (longest)

$\sin\theta = \frac{\text{Opp}}{\text{Hyp}}, \qquad \cos\theta = \frac{\text{Adj}}{\text{Hyp}}, \qquad \tan\theta = \frac{\text{Opp}}{\text{Adj}}$

Example: the $3$-$4$-$5$ triangle

Let $\theta$ be the angle with opposite leg $3$, adjacent leg $4$, hypotenuse $5$:

$\sin\theta = \frac{3}{5}, \qquad \cos\theta = \frac{4}{5}, \qquad \tan\theta = \frac{3}{4}$

💡 SOH-CAH-TOA = Sin Opp/Hyp, Cos Adj/Hyp, Tan Opp/Adj.

Identify the Ratio 🎯

Trig of the Special Angles

From the special right triangles, these exact values come up constantly:

Solving for a side

If you know an angle and one side, multiply or divide to get another.

Example: the hypotenuse is $10$ and one acute angle is $30^\circ$. The side opposite that angle is $\text{Opp} = \text{Hyp}\cdot \sin 30^\circ = 10 \cdot \tfrac{1}{2} = 5.$

⚠️ Notice $\sin 30^\circ = \frac12$ but $\sin 60^\circ = \frac{\sqrt3}{2}$ — they are not the same. Match the ratio to the angle carefully.

Special-Angle Values 🔽

Turning a Ratio Into a Side

To find a side when you know an angle and another side, set up the matching ratio and solve:

• Know the hypotenuse, want the opposite side → use $\sin$: $\;\text{Opp} = \text{Hyp}\cdot\sin\theta$.
• Know the hypotenuse, want the adjacent side → use $\cos$: $\;\text{Adj} = \text{Hyp}\cdot\cos\theta$.
• Know one leg, want the other leg → use $\tan$.

🔑 Pick the ratio that contains the side you know and the side you want — then it is just one multiplication or one cross-multiply away.

Solve the Triangle 🧮

Enter each answer as a whole number.

1) Right triangle, hypotenuse $20$, one acute angle $30^\circ$. The leg opposite that angle $= 20\sin 30^\circ = \,?$ 2) A right triangle has $\tan\theta = \frac{3}{4}$ with opposite leg $6$. Using $\frac{6}{\text{adj}} = \frac34$, the adjacent leg $= \,?$

📐 Plane Geometry & Trigonometry

Part 5 of 5 — Coordinate Geometry & Mastery Check

The ACT also tests geometry on the $xy$-plane: distance, midpoint, and slope. Then we put everything together.

Distance, Midpoint & Slope

For two points $(x_1, y_1)$ and $(x_2, y_2)$:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\text{Midpoint} = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)$

$\text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}$

Worked Example

Points $(1, 2)$ and $(4, 6)$: $\text{Distance} = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ $\text{Slope} = \frac{6 - 2}{4 - 1} = \frac{4}{3}$

💡 The distance formula is just the Pythagorean theorem in disguise: the horizontal and vertical gaps are the legs, the distance is the hypotenuse.

Coordinate Drill 🧮

Use the points $(2, 3)$ and $(8, 11)$.

1) Midpoint $x$-coordinate $= \,?$ 2) Midpoint $y$-coordinate $= \,?$ 3) Slope of the line through them $= \,?$ (reduce to a fraction or whole number)

Quick Reference

⚠️ The ACT gives no formula sheet. Drill these until they are automatic.

Mixed Practice 🎯

One Last Look Before the Exit Quiz

You have covered the full ACT plane-geometry-and-trig toolkit:

1. Angles & lines — complementary/supplementary/vertical, parallel-line pairs, $180^\circ$ triangle sum.
2. Triangles — Pythagoras, the $1{:}1{:}\sqrt2$ and $1{:}\sqrt3{:}2$ special triangles, similarity proportions.
3. Polygons & circles — $(n-2)\cdot 180^\circ$, $A=\pi r^2$, sectors.
4. Trig — SOH-CAH-TOA and the $30^\circ$/$45^\circ$/$60^\circ$ exact values.
5. Coordinate geometry — distance, midpoint, slope.

🔑 On test day, redraw and label every figure. Most ACT geometry errors come from mis-reading the picture, not from the math itself.

Exit Quiz ✅

Answer all three to finish the lesson.