12.1 Right-Triangle Trigonometry and Special Triangles

12.1 Right-Triangle Trigonometry and Special Triangles

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

SOH-CAH-TOA in a right triangle

Label a right triangle ABC with the right angle at C. Acute angle A has three sides relative to A:

• Opposite is the side that does not touch A (here, side BC).
• Adjacent is the leg that does touch A but is not the hypotenuse (here, side AC).
• Hypotenuse is the side opposite the right angle (here, side AB). It is always the longest side.

The three primary ratios are:

sin A = opposite / hypotenuse

cos A = adjacent / hypotenuse

tan A = opposite / adjacent

The mnemonic SOH-CAH-TOA is the whole skill: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent. AAF will not grade you on reciting the letters; it will grade you on picking the correct two sides for the named acute angle. Switch the named angle from A to B and opposite and adjacent trade places. The hypotenuse never moves.

Reciprocal identities

The three reciprocal functions flip the primary ratios:

csc A = 1 / sin A = hypotenuse / opposite

sec A = 1 / cos A = hypotenuse / adjacent

cot A = 1 / tan A = adjacent / opposite

If you already have sin A, csc A is its reciprocal — do not rebuild it from a different pair of sides. Two more compact identities sit on top of the same triangle: tan A = sin A / cos A and cot A = cos A / sin A. Reciprocal does not mean complementary. csc A is the flip of sin A; it is not cos B.

Worked example: sin A = 5/13, then every other ratio

Right triangle ABC has a right angle at C, and sin A = 5/13. Find cos A, tan A, csc A, and sin B.

Because sin A = opposite / hypotenuse = 5/13, the opposite side may be taken as 5 and the hypotenuse as 13 (any positive multiple of that pair gives the same ratio). The missing adjacent side comes from the Pythagorean theorem:

adjacent^2 + 5^2 = 13^2

adjacent^2 + 25 = 169

adjacent^2 = 144

adjacent = 12

(A length is positive, so discard −12.) This is the 5-12-13 triple. Now every ratio is a fraction of those three numbers:

• cos A = 12/13
• tan A = 5/12
• csc A = 13/5
• sec A = 13/12
• cot A = 12/5

Train on this triple. Do not drill a published sample that starts from a cosine of five-eighths; AAF will reuse the method, not that particular pair of sides.

Complementary angles: sin A = cos B

In right triangle ABC with right angle C, the two acute angles are complementary: A + B = 90°. The side opposite A is adjacent to B, and the side opposite B is adjacent to A. Therefore the cofunction identities are not extra formulas — they are the same two legs, read from the other angle:

sin A = cos B

cos A = sin B

tan A = cot B

From the 5-12-13 example, sin A = 5/13, so cos B = 5/13 as well — not 12/13. The trap is treating complementary as “the same angle.” Complementary means the angles add to 90°; the cofunction identity swaps sine with cosine. The same identity in special-angle clothing is sin 30° = cos 60° = 1/2 and sin 45° = cos 45° = √2/2.

sin B uses the side opposite B, which is the adjacent to A, so sin B = 12/13. That matches cos A = 12/13, as the cofunction rule promised.

Finding the third side first

AAF often gives two sides and asks for a trig ratio that needs the third. Always sketch, label opposite/adjacent/hypotenuse relative to the named acute angle, then apply Pythagoras if a side is missing. Pythagoras for right triangle ABC with right angle C is a^2 + b^2 = c^2, the same identity taught with the distance formula in Distance Formula and Pythagorean Theorem.

Worked: Legs 8 and 15, hypotenuse unknown, angle A opposite the side of length 8. Then hypotenuse = √(8^2 + 15^2) = √(64 + 225) = √289 = 17. So sin A = 8/17, cos A = 15/17, tan A = 8/15. If the stem had instead said angle A is adjacent to the 8, you would swap: sin A = 15/17. The number 17 does not tell you which ratio — the words opposite versus adjacent do.

Keep 7-24-25 and 20-21-29 as extra triples you can finish quickly, but you do not need a catalog. You need to put the hypotenuse in the c slot and not in a leg slot. A stem that already gives a hypotenuse of 25 and a leg of 7 is one subtraction away from 24: adjacent^2 = 25^2 − 7^2 = 625 − 49 = 576 = 24^2.

Special triangle: 45-45-90

An isosceles right triangle has acute angles 45° and 45°. If each leg is x, the hypotenuse is x√2. The two legs are equal, so tan 45° = x/x = 1 at every scale.

Worked: Each leg is 7. Hypotenuse = 7√2. Then sin 45° = 7 / (7√2) = 1/√2 = √2/2 after rationalizing. If the hypotenuse is 10√2, each leg is 10, not 10√2. Divide by √2: leg = hyp / √2 = (10√2)/√2 = 10.

Worked from a hypotenuse without a radical already showing: Hypotenuse 18. Then x√2 = 18, so x = 18/√2 = 9√2. Each leg is 9√2, and sin 45° is still √2/2 because the scale cancels.

Trap: treating a 45-45-90 like a 30-60-90 and putting √3 on a side. There is no √3 in a 45-45-90 triangle. A second trap: writing the hypotenuse as 2x because “the hypotenuse is twice a leg.” That doubling rule belongs only to 30-60-90.

Special triangle: 30-60-90

A 30-60-90 triangle has sides in the ratio x : x√3 : 2x.

• Side opposite 30° is the shortest leg, x.
• Side opposite 60° is the longer leg, x√3.
• Hypotenuse is opposite 90°, 2x.

Worked from the short leg: The side opposite 30° is 5. Then x = 5, hypotenuse 2x = 10, and the side opposite 60° is 5√3. So sin 30° = 5/10 = 1/2, cos 30° = (5√3)/10 = √3/2, sin 60° = (5√3)/10 = √3/2, cos 60° = 5/10 = 1/2.

Worked from the hypotenuse: Hypotenuse 14. Then 2x = 14, so x = 7. Opposite 30° is 7; opposite 60° is 7√3.

Worked from the long leg: Opposite 60° is 9√3. Then x√3 = 9√3, so x = 9. Opposite 30° is 9; hypotenuse is 18. If a stem instead gives the long leg as 9 with no radical, then x√3 = 9, so x = 9/√3 = 3√3, hypotenuse 6√3, and opposite 30° is 3√3.

Special-angle value table (exact)

Memorize the exact values. AAF will mix them with algebra — 2 sin 30° + cos 60° is 2(1/2) + 1/2 = 3/2, not a calculator decimal. tan 30° · tan 60° = (√3/3) · √3 = 1, which is also tan 45°.

Sine increases through 0, 1/2, √2/2, √3/2, 1 while cosine decreases through the same five numbers in reverse. That is the complementary pattern again: sin 30° = cos 60°. Tangent is sine over cosine, which is why tan 60° = (√3/2) / (1/2) = √3 and why tan 90° is undefined (division by cosine zero).

AAF traps for right-triangle trig

• Wrong pair of sides. tan never uses the hypotenuse. If you divide a leg by the hypotenuse you have sine or cosine, not tangent.
• Wrong reference angle. Opposite and adjacent switch when you move from angle A to angle B. Compute sin A and sin B separately after you label.
• Leaving Pythagoras unfinished. A stem that gives two legs and asks for sin A is not a two-side ratio until you find the hypotenuse.
• Exact versus decimal. sin 45° = √2/2, not 0.7. If every option is exact, do not pick a rounded neighbor.
• Reciprocal mix-up. csc A = 13/5 is the flip of sin A = 5/13. 13/5 is not cos A.
• 30-60-90 assignment. The √3 always sits opposite 60°, never opposite 30°. Opposite 30° is half the hypotenuse.
• Complementary mix-up. sin A = cos B, not sin A = cos A (unless A = 45°).

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Mini mixed set (work these before the quizzes)

1. Right triangle, right angle C, a = 9, c = 15. Then b = 12 (3-4-5 scaled by 3). If a is opposite A, sin A = 9/15 = 3/5, cos A = 12/15 = 4/5, tan A = 9/12 = 3/4, and csc A = 5/3.
2. 45-45-90 with hypotenuse 6√2. Legs = 6. tan 45° = 1 regardless of the scale, and sec 45° = √2.
3. 30-60-90 with longer leg 10√3. Then x√3 = 10√3, x = 10, hypotenuse 20, sin 60° = (10√3)/20 = √3/2, tan 30° = 10 / (10√3) = √3/3.
4. Right triangle ABC, right angle C, sin A = 20/29. Adjacent = √(29^2 − 20^2) = √(841 − 400) = √441 = 21. Then cos A = 21/29 and cos B = 20/29 by the cofunction rule.

That is the whole 250–262 skill: name the sides, finish Pythagoras, quote an exact special-angle value when the triangle is 30-60-90 or 45-45-90, and remember that complementary acute angles swap sine with cosine.