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Geometry & Trigonometry — 700-800

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Geometry & Trigonometry — 700-800 - Complete Interactive Lesson

Part 1: The 700-800 Patterns

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 $32^{\circ}$ , walks $40$ m closer, and sights it at $48^{\circ}$ . A drone hovers between two ground points $500$ 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 , and the given distance is either their (observer moves toward the object) or their (object is between the two observers). Factor out :

$h\left(\frac{1}{\tan 32^{\circ}} - \frac{1}{\tan 48^{\circ}}\right) = 40$

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 $40\tan 32^{\circ}$ from treating the $40$ 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 and the hypotenuse is . Likewise gives the -- family, and gives --.

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: $3$ - $4$ - $5$ , $5$ - $12$ - $13$ , $8$ --, --, --, --.

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 $\sin(3a) = \cos(a+10^{\circ})$ means . Pure algebra, no calculator.

The distractors are always the right angle with the wrong ratio ( $\sin 60^{\circ}$ offered next to $\tan 60^{\circ}$ ) and the right ratio of the other angle ( $\tan 30^{\circ}$ ).

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 $AD:DB = 3:2$ then — converting a part-to-part ratio into a part-to-whole ratio is where most of the errors live. The area ratio is then , and the is the big triangle minus the small one, never a direct ratio.

Part 2: Traps & Speed

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 , the legs are and — and will be sitting right there in the option list. Before you bubble, reread the last six words of the question.

Part 3: Timed Drill

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 shoelace-plus-cost polygon.

Before each answer, run the two-second checklist:

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

\frac{h}{\tan\theta_{2}}

\frac{h}{t a n θ _{2}}

\sin D = \frac{8}{17}

\cos A = 0.6 = \frac{3}{5}

\cos R = \sin P

when

P

and

R

are the two acute angles.

\tan B = \frac{1}{\tan A}

, so

\sin A \cdot \tan B = \sin A \cdot \frac{\cos A}{\sin A} = \cos A

Running it backwards works too: an enlargement with $2.25$ times the area has linear factor $\sqrt{2.25} = 1.5$ .

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:

Compute the area using the convenient base (a horizontal side, or an isosceles triangle's natural altitude, or the shoelace formula in the coordinate plane).
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: $(a+2)^{2} + 5^{2} = 13^{2}$ 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 $\frac{5}{12}$ means each step of $12$ right and $5$ up covers exactly $13$ units. To travel , take two 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.
Distance from a point to a line is $\frac{|Ax_{0}+By_{0}+C|}{\sqrt{A^{2}+B^{2}}}$ — the shortest path, and the only one the item means by "shortest."
Shoelace handles any polygon's area from its vertices, and the vertices must be listed in order around the figure.

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.

Distractor Species 2: The Un-Scaled Triple

Given $\sin A = \frac{8}{17}$ , your instinct correctly says "8-15-17." But the triangle in the question has hypotenuse $68$ , not $17$ . The scale factor is $68 \div 17 = 4$ , so the legs are $32$ and $60$ , 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. Use $d = \frac{|Ax_{0} + By_{0} - C|}{\sqrt{A^{2} + B^{2}}}$ for a line written $Ax + By = C$ .

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 $k = 2.5$ , then $\text{big} - \text{small} = 6.25A - A = 5.25A$ , and the trap divides by $1.5$ instead.

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, no inverse trig.

The two-observation formula. Two elevation angles from points $d$ apart, same side: $h = \frac{d}{\cot\theta_{\text{far}} - \cot\theta_{\text{near}}} = \frac{d}{\frac{1}{\tan\theta_{\text{far}}} - \frac{1}{\tan\theta_{\text{near}}}}$ Subtract the cotangents, never the tangents. That single sign-of-approach error is the most common wrong answer in the bank.

Shoelace for any polygon. List the vertices in order, repeat the first at the bottom, cross-multiply down-right minus down-left, halve the absolute value. It beats decomposition every time and never miscounts a region.

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.