Geometry & Trigonometry — 700-800

Geometry & Trigonometry: The 700-800 Patterns

Part 1 of 3 — The Archetypes Hard-Tier Items Are Built From

This is the largest hard-tier bank on the test, and it splits cleanly into two halves: right-triangle trigonometry and coordinate geometry. Learn six archetypes and you have covered nearly all of it.

Archetype 1: The Two-Observation Height Problem

A surveyor sights the top of a tower at $27^{\circ}$, walks $35$ m closer, and sights it at $44^{\circ}$. A drone hovers between two ground points $600$ m apart. A lighthouse keeper watches a boat approach.

Every one of these has the same skeleton. Let $h$ be the unknown height. The two horizontal distances are $\frac{h}{\tan\theta_{1}}$ and $\frac{h}{\tan\theta_{2}}$, and the given distance is either their difference (observer moves toward the object) or their sum (object is between the two observers). Factor out $h$:

$h\left(\frac{1}{\tan 27^{\circ}} - \frac{1}{\tan 44^{\circ}}\right) = 35$

One equation, one unknown, no law of sines needed. Angle of depression from a height equals the angle of elevation from the ground — they are alternate interior angles.

The planted answers are the two horizontal distances, plus the value $35\tan 27^{\circ}$ from treating the $35$ m as the whole distance to the base.

Archetype 2: A Trig Ratio Names a Pythagorean Triple

$\tan A = \frac{7}{24}$ is not an invitation to use $\arctan$. It says the legs are $7k$ and $24k$ and the hypotenuse is $25k$. Likewise $\sin D = \frac{8}{17}$ gives the $8$-$15$-$17$ family, and $\cos A = 0.6 = \frac{3}{5}$ gives $3$-$4$-$5$.

Find $k$ from the given side, scale all three, then answer the perimeter or area question. The two standing distractors: using the ratio numbers themselves as side lengths, and reporting the sum of the legs when the perimeter was asked.

Worth memorizing: $3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$, $7$-$24$-$25$, $20$-$21$-$29$, $9$-$40$-$41$.

Archetype 3: Complementary-Angle Identities

In a right triangle the two acute angles are complementary, which produces three facts hard items lean on constantly:

• $\sin\theta = \cos(90^{\circ} - \theta)$, so $\sin(2a) = \cos(a+15^{\circ})$ means $2a + (a+15) = 90$. Pure algebra, no calculator.
• $\cos R = \sin P$ when $P$ and $R$ are the two acute angles.
• $\tan B = \frac{1}{\tan A}$, so $\tan A \cdot \tan B = 1$ for the two acute angles of any right triangle.

The distractors are always the right angle with the wrong ratio ($\sin 50^{\circ}$ offered next to $\tan 50^{\circ}$) and the right ratio of the other angle ($\tan 40^{\circ}$).

Archetype 4: Similar Figures and the Squared Ratio

Whenever a line is drawn parallel to a side, or a figure is enlarged, or two triangles are declared similar: lengths scale by $k$, areas by $k^{2}$.

If $AD:DB = 4:3$ then $AD:AB = 4:7$ — converting a part-to-part ratio into a part-to-whole ratio is where most of the errors live. The area ratio is then $\frac{16}{49}$, and the quadrilateral is the big triangle minus the small one, never a direct ratio.

Running it backwards works too: an enlargement with $1.69$ times the area has linear factor $\sqrt{1.69} = 1.3$.

Archetype 5: Area as a Bridge (the Altitude Trick)

A triangle has one area but three bases and three altitudes. That gives a two-line route to any altitude:

1. Compute the area using the convenient base (a horizontal or vertical side, an isosceles triangle's natural altitude, or, in the coordinate plane, an enclosing rectangle minus its corner triangles).
2. Set $\text{Area} = \frac{1}{2}(\text{other base})(h)$ and solve.

The distractors write themselves: the area and the base are both choices, and so is the answer that divided the area by the base without doubling first.

Archetype 6: Coordinate Geometry Is Three Formulas and One Constraint

Distance, midpoint, slope — plus one condition that resolves an ambiguity.

• Unknown coordinate + given distance: $(a-1)^{2} + 8^{2} = 17^{2}$ gives two values of $a$, and a condition like "$a > 0$" or "$k > 3$" picks one. The rejected root's answer is always a choice.
• Moving a set distance along a line: a slope of $\frac{3}{4}$ means each step of $4$ right and $3$ up covers exactly $5$ units. To travel $15$, take three steps. Never use the distance formula with a variable here.
• Perpendicular bisector = the set of points equidistant from two points. "Equidistant from $A$ and $B$" and "on line $\ell$" is a two-line system.
• Shortest distance from a point to a line is the path that meets the line at a right angle. To a horizontal line $y = c$ it is just $|y_{0} - c|$; to a vertical line $x = c$ it is $|x_{0} - c|$. For a slanted line, give the path the negative-reciprocal slope, solve the system to find where it meets the line, then use the distance formula between the two points.
• Area from vertices: if a side is horizontal or vertical, use it as the base and read the height off the coordinates. Otherwise, enclose the figure in a rectangle and subtract the right triangles cut off in its corners.

The final trap in this half is almost always which number was asked for: the $x$-coordinate when the $y$ was wanted, the intercept on the wrong axis, the area when the cost was wanted.

Geometry & Trigonometry: Traps & Speed

Part 2 of 3 — Distractor Autopsy

Hard-tier geometry rarely punishes a bad theorem. It punishes a correct calculation that stopped one step early or answered a neighbouring quantity. Here is the full catalogue of what the wrong options actually are.

Distractor Species 1: The Intermediate Leg

You are asked for a perimeter, an area, or a difference, and one option is the leg you found on the way. In a $\tan A = \frac{7}{24}$ problem with hypotenuse $75$, the legs are $21$ and $72$ — and $72$ will be sitting right there in the option list. Before you bubble, reread the last six words of the question.

Distractor Species 2: The Un-Scaled Triple

Given $\sin A = \frac{8}{17}$, your instinct correctly says $8\text{-}15\text{-}17$. But the triangle in the question has hypotenuse $34$, not $17$. The scale factor is $34 \div 17 = 2$, so the legs are $16$ and $30$, not $8$ and $15$. The area of the unscaled triangle is always planted as an option.

Memorize the triples so the recognition is free: $3\text{-}4\text{-}5$, $5\text{-}12\text{-}13$, $7\text{-}24\text{-}25$, $8\text{-}15\text{-}17$, $9\text{-}40\text{-}41$, $20\text{-}21\text{-}29$.

Distractor Species 3: Equal Instead of Complementary

When a stem says $\sin(\text{something}) = \cos(\text{something else})$, the relationship is complementary: the two angle expressions sum to $90^{\circ}$. The trap answer comes from setting the expressions equal to each other, and a second trap comes from summing them to $180^{\circ}$. Both produce clean integers, which is exactly why they are believable.

Distractor Species 4: Vertical Distance Masquerading as Perpendicular Distance

"How far is the tower from the road?" means the perpendicular distance. Measuring straight up or straight across to the line gives a larger number that is always an option. Build the right angle instead: the path's slope is the negative reciprocal of the line's slope, so write that path through the point, solve the two-line system to find where it meets the line, and finish with the distance formula. Using the plain reciprocal (dropping the sign change) is the planted slip, and it lands on the wrong point of the line.

Distractor Species 5: Linear Ratio Where Area Ratio Belongs

Lengths scale by $k$; areas scale by $k^{2}$. An item that gives you a difference of areas is testing exactly this: if $k = 3$, then $\text{big} - \text{small} = 9A - A = 8A$, and the trap divides by $2$ instead.

---

Speed Techniques

Ratio → triangle. Given $\sin\theta = \frac{2}{5}$ or $\tan\theta = \frac{2}{5}$, immediately draw the triangle with those two sides and fill the third by Pythagoras. Every other ratio is then free — no calculator needed.

The two-observation formula. Two elevation angles from points $d$ apart, same side: $h = \frac{d}{\frac{1}{\tan\theta_{\text{far}}} - \frac{1}{\tan\theta_{\text{near}}}}$ Each $\frac{1}{\tan\theta}$ is the horizontal distance per unit of height, so subtract the reciprocals of the tangents, never the tangents themselves. That single sign-of-approach error is the most common wrong answer in the bank.

Area in the coordinate plane: box it, then subtract. If the figure has a horizontal or vertical side, use that side as the base and read the height straight off the coordinates. If it has none, draw the smallest rectangle whose sides pass through the vertices, then subtract the right triangles cut off in the corners. Each corner triangle's legs are just differences of coordinates, so no distance formula is needed. The planted errors are forgetting to halve the corner triangles and reporting the rectangle itself.

Midpoint runs backwards. If $M$ is the midpoint of $\overline{AB}$ and you know $A$, then $B = 2M - A$. Doing $M - A$ instead is a planted option.

Geometry & Trigonometry: Timed Drill

Part 3 of 3 — Four Items at Test Pace

Give yourself about 90 seconds per question. These are built from the four highest-frequency hard-tier skeletons: the two-observation height, the reversed midpoint, the similar-figure area difference, and the boxed-in polygon with a cost step.

Before each answer, run the two-second checklist:

1. What quantity was asked? Height or distance? Area or cost? The small triangle or the quadrilateral?
2. Did I scale? A ratio gives a shape, not a size.
3. Is my answer the right order of magnitude? A perpendicular distance is shorter than a vertical one; a part is smaller than its whole.