Mistake Master
Stand at the angle, then read the sides
Three ratios and one identity cover all of SAT trig. The arithmetic is a fraction. The points slip away when the ratio is built from the wrong sides, when sine-equals-cosine gets solved as an equality instead of through the $90^\circ$ sum, and when the other angle's ratio, a flipped fraction, or an unscaled numerator stands in for the answer. Locate opposite and adjacent first.
§1
What this topic is about
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Right-triangle trig is three ratios and one identity. The points ride on three skills: locating opposite and adjacent from the named angle before writing any fraction, using the complementary identity $\sin x^\circ = \cos(90 - x)^\circ$ when sine meets cosine, and reporting the ratio (or side) that belongs to the angle and quantity actually asked.
§2
Stand at the angle, then read the sides
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Opposite and adjacent are not fixed labels; they depend on which acute angle you stand at. The hypotenuse never moves, and it appears only in sine and cosine.
- $\sin = \dfrac{\text{opposite}}{\text{hypotenuse}}$, $\cos = \dfrac{\text{adjacent}}{\text{hypotenuse}}$, $\tan = \dfrac{\text{opposite}}{\text{adjacent}}$.
- Opposite is the side ACROSS from the named angle; adjacent is the leg touching it.
- Sine and cosine of an acute angle are always less than $1$; a bigger-than-$1$ answer flipped.
Worked example. In a right triangle, the side opposite angle $A$ is $6$, the adjacent leg is $8$, the hypotenuse $10$. Find $\sin A$, $\cos A$, $\tan A$.
$\sin A = \dfrac{6}{10} = \dfrac{3}{5}$, $\cos A = \dfrac{8}{10} = \dfrac{4}{5}$, $\tan A = \dfrac{6}{8} = \dfrac{3}{4}$.
Each wrong choice on this pattern is one of the OTHER ratios; the sides get swapped, not miscomputed.
§3
Sine meets cosine across complements
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In any right triangle the two acute angles sum to $90^\circ$, so one angle's opposite is the other's adjacent. That geometry is the identity: $\sin x^\circ = \cos(90 - x)^\circ$.
- $\sin A = \cos B$ whenever $A + B = 90$; the angles are complements, not equals.
- Given $\sin(\text{expr}_1) = \cos(\text{expr}_2)$ with acute angles, set $\text{expr}_1 + \text{expr}_2 = 90$.
- The pairing runs through $90$, never $180$.
Worked example. Acute angles satisfy $\sin(2x + 10)^\circ = \cos(3x - 5)^\circ$. Find $x$.
$(2x + 10) + (3x - 5) = 90$, so $5x + 5 = 90$ and $x = 17$.
Setting $2x + 10 = 3x - 5$ gives $x = 15$; equal function VALUES do not make equal angles when the functions differ.
§4
From ratio to side length
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A trig ratio is a shape statement: $\sin A = \dfrac{4}{5}$ names a family of similar triangles. To get an actual side, scale the ratio up to the triangle's real dimensions.
- Opposite $= \text{hypotenuse} \cdot \sin$; adjacent $= \text{hypotenuse} \cdot \cos$.
- The ratio's numerator is NOT the side; it must be scaled to this triangle.
- Familiar triples hide inside given ratios: $\sin A = \dfrac{4}{5}$ is the $3$-$4$-$5$ family.
Worked example. $\sin A = \dfrac{4}{5}$ and the hypotenuse is $35$. Find the opposite leg.
$35 \cdot \dfrac{4}{5} = 28$.
$4$ is the unscaled numerator, and $21$ is the ADJACENT leg (the triple's other member scaled by $7$).
§5
Which angle owns the ratio
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Every right triangle offers two acute angles, and each ratio belongs to one of them. $\sin B$ is $\cos A$'s twin, and $\tan B$ is $\tan A$ flipped. Asked about $B$, re-seat yourself before reading sides.
Locate opposite and adjacent from the angle actually named, send sine-equals-cosine problems through the $90^\circ$ sum, scale ratio numerators to real triangles, and re-seat before answering for the other angle.
§6
Three patterns that cost real points
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Three patterns recur on trig questions. They are the same ones the diagnostic routes on.
The ratio is built from the wrong sides.
The adjacent leg lands in a sine, the hypotenuse sneaks into a tangent, or opposite and adjacent trade places outright.
Fix. Stand at the named angle first: across is opposite, touching is adjacent. Then let the function pick its pair, hypotenuse only in sine and cosine.
Sine-equals-cosine gets mishandled.
The two angle expressions are set equal, or their sum is sent to $180$, or the identity swaps the function without swapping the angle.
Fix. Equal sine and cosine of acute angles means COMPLEMENTARY angles: their sum is $90$. Write that equation and solve.
The answer belongs to something else.
The other angle's ratio gets reported, the fraction arrives flipped, or a ratio's bare numerator stands in for a real side length.
Fix. Reread which angle and which quantity the question names. Re-seat for that angle, keep sine and cosine below $1$, and scale ratios up to the triangle's actual size.
Ten quick checks across the patterns: building each ratio from the named angle, the complementary identity numerically and with algebra, ratio-to-side scaling, and which angle owns the ratio. Pick or type your answer, then check. Progress is saved.