Apply sine, cosine, and tangent ratios in right triangles, and recall exact values at 30°, 45°, and 60° from memory — the standard angles are what let you skip the calculator entirely. Every question comes with a written walkthrough of exactly which ratio applied.
"Opposite" and "adjacent" only make sense relative to a specific angle — the same side can be opposite one angle and adjacent to another in the same triangle.
sin(30°) = 1/2 while cos(30°) = √3/2, and sin(60°) = √3/2 while cos(60°) = 1/2 — the two are mirror images of each other, and mixing them up under time pressure is the single most common error here.
Once a ratio like sin(x) = 0.5 is set up, x itself requires applying arcsin, not just reading off the ratio value as if it were the angle.
Straight from the bank — one per difficulty tier. Reveal the answer to see the explanation you'd get in a real session.
In a right triangle, the angle is 30°, and the hypotenuse is 12. Find the length of the side opposite the 30° angle.
A — 6. Sine relates the opposite side to the hypotenuse: sin(30°) = opposite/12. Since sin(30°) = 1/2, the opposite side is 12 × 1/2 = 6. 6√3 (B) comes from using cosine instead of sine — computing 12 × cos(30°) = 12 × (√3/2) = 6√3, mixing up which ratio applies to the opposite side.
A right triangle has legs of length 8 and 15, with the right angle between them. Find the measure of the angle opposite the side of length 8, to the nearest degree.
A — 28°. The angle opposite the side of length 8 satisfies tan(θ) = 8/15 (opposite over adjacent), so θ = arctan(8/15) ≈ 28°. 62° (B) is the other acute angle in the triangle — the one opposite the side of length 15 — a common mix-up about which angle a given ratio actually describes.
A surveyor stands 50 meters from the base of a building and measures the angle of elevation to the top of the building as 40°. From the same spot, she measures the angle of elevation to a flagpole on top of the building as 48°. Find the height of the flagpole itself, not including the building, to the nearest tenth of a meter.
A — 13.6 m. Find each height separately using tan(angle) = height/50, then subtract. The building's height: 50·tan(40°) ≈ 41.95 m. The total height to the top of the flagpole: 50·tan(48°) ≈ 55.53 m. The flagpole alone is the difference: 55.53 − 41.95 ≈ 13.6 m. 97.5 m (D) comes from adding the two heights instead of subtracting them, treating the two elevation readings as separate objects instead of one stacked on the other.
SOHCAHTOA problems are quick once the ratio and the standard-angle values are automatic — the setup is naming which side and angle you actually have, not the trig itself.
Before writing a ratio, mark which side is opposite and which is adjacent to the specific angle in question — don't assume it carries over from a different angle in the same triangle. Untimed practice is where that habit sticks.
Move to timed sessions once ratio setup and standard-angle recall are instant. Multi-step problems like angles of elevation run a bit longer.
Pair this with Law of Sines and Law of Cosines in a mock — SOHCAHTOA is the right-triangle special case both laws generalize beyond.
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