Right triangles and trigonometry

Learning objectives

Identify the hypotenuse; use the Pythagorean theorem and special right-triangle ratios; select sine, cosine, or tangent relative to a named angle; exploit complementary angles; solve geometric models without unnecessary angle calculations.

The right triangle is the organizing structure

A right triangle contains a 90° angle. The side opposite that angle is the hypotenuse; it is always the longest side. The other two sides are the legs. If the legs are a and b and the hypotenuse is c, then

$a 2 + b 2 = c 2 .$

To find a leg, rearrange before taking the square root: $\sqrt{c^{2} − b^{2}}$ To find the hypotenuse, use $\sqrt{a^{2} + b^{2}}$ A right-angle condition must be stated or established before using either formula.

“Opposite” and “adjacent” depend on the chosen acute angle. The hypotenuse does not.

Common integer side patterns include 3–4–5, 5–12–13, 8–15–17, and 7–24–25, plus their positive multiples. Recognizing a pattern saves arithmetic, but the theorem remains your check. A set of three numbers is not a right-triangle pattern just because the numbers are consecutive or familiar.

Two recurring constructions: a rectangle’s diagonal makes two right triangles; an altitude can split an isosceles or equilateral triangle into right triangles. In a general triangle, an altitude need not bisect the base. In an isosceles triangle, the altitude from the vertex between the equal sides does bisect the base.

Special right triangles and exact values

In a 45°–45°–90° triangle, the legs are equal. If each leg is s, the Pythagorean theorem gives $c 2 = 2s 2,$ so the side ratio is $\sqrt{2}$ A 30°–60°–90° triangle is half an equilateral triangle. If the short leg is s, the hypotenuse is 2s, and the long leg is $\sqrt{({2s)}^{2} − s^{2}}$ $\sqrt{3}$

The short leg is opposite 30°, the long leg opposite 60°, and the hypotenuse opposite 90°. Identify the given side’s role before applying a multiplier.

Sine, cosine, and tangent

For an acute angle θ in a right triangle,

$\frac{opposite}{hypotenuse}$ $\frac{adjacent}{hypotenuse}$ $\frac{opposite}{adjacent}$

“SOH–CAH–TOA” is a memory aid, not a substitute for labeling the sides relative to the correct angle. The adjacent side in the cosine and tangent ratios is the adjacent leg, not the hypotenuse.

Why ratios work. Any two right triangles sharing the same acute angle are similar by AA. Their side lengths may differ, but the ratios are identical. This lets you use a convenient representative triangle: if $tan θ = 3/4,$ choose legs 3 and 4, then derive hypotenuse 5. The actual triangle might be larger or smaller.

Angle	sin θ	cos θ	tan θ
30°	1/2	$\sqrt{3}$	$\sqrt{3}$
45°	$\sqrt{2}$	$\sqrt{2}$	1
60°	$\sqrt{3}$	1/2	$\sqrt{3}$

Two useful identities follow directly from the definitions:

$sin 2 θ + cos 2 θ = 1,$ $\frac{sin θ}{cos θ}$ $(cos θ \neq 0).$

The first means $(sin θ) 2 + (cos θ) 2 = 1,$ not $sin(θ 2) + cos(θ 2) = 1.$

Complementary angles swap sine and cosine

The acute angles of a right triangle sum to 90°. A side opposite one acute angle is adjacent to the other, so

$sin θ = cos(90° - θ),$ $cos θ = sin(90° - θ).$

For acute angles u and v, an equation $sin u = cos v$ implies $u + v = 90°.$ Without a restriction on the angles, periodicity can allow other possibilities; do not apply the acute-angle conclusion indiscriminately.

Calculator mode and inverse functions

When an angle is given in degrees, use degree mode; when it is in radians, use radian mode. The same input number has different meanings in the two modes. Use arcsin, arccos, or arctan only when the angle itself is requested. The notation $sin -1 (u)$ for inverse sine does not mean $1/sin(u).$ If a side ratio is requested, a representative triangle is often cleaner than finding an angle and then evaluating another function.

15 worked examples

3.01. Find the hypotenuse

A right triangle has legs of lengths 9 and 12. Find its hypotenuse.

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Recognize the structure. The unknown side is opposite the right angle, so add the leg squares.

Work it through. Let the hypotenuse be c. Then

$c 2 = 9 2 + 12 2 = 81 + 144 = 225,$

so $c = 15.$ Only the positive square root represents a length.

Answer: 15 units.

Another efficient route. The triangle is three times a 3–4–5 triangle.

Avoid the trap. $9 + 12 = 21$ is not the hypotenuse. The theorem relates the squares of the side lengths.

3.02. Find a missing leg

A right triangle has hypotenuse 17 and one leg 8. Find the other leg.

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Recognize the structure. The hypotenuse square equals the sum of the leg squares, so subtract to isolate a leg.

Work it through. If the missing leg is b, then

$82 + b 2 = 17 2 b 2 = 289 - 64 = 225 b = 15.$

Answer: 15 units.

Key takeaway. The result must be shorter than 17. A calculation such as $\sqrt{17^{2} + 8^{2}}$ would produce a side longer than the stated hypotenuse and immediately fail that check.

3.03. A rectangle hides the right triangle

A rectangle has diagonal 26 and width 10. What is its area?

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Recognize the structure. The diagonal is the hypotenuse of a right triangle formed by the rectangle’s sides.

Work it through. If the length is l, then

$l 2 + 10 2 = 26 2 l 2 = 576 l = 24.$

The area is $lw = 24(10) = 240.$

Answer: 240 square units.

Avoid the trap. The diagonal is not a side of the rectangle for purposes of the area formula. Multiplying 26 by 10 uses the wrong dimensions.

3.04. Work backward in a 45–45–90 triangle

A 45°–45°–90° triangle has hypotenuse 14. What is its perimeter? A) 28 B) 14 + 7 $\sqrt{2}$ C) 14 + 14 $\sqrt{2}$ D) 28 $\sqrt{2}$

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Recognize the structure. Each leg is the hypotenuse divided by $\sqrt{2}$

Work it through. If each leg is s, then $\sqrt{2}$ giving $\sqrt{2}$ The perimeter is

$\sqrt{2}$

Answer: C, $\sqrt{2}$ units.

Avoid the trap. The formula $\sqrt{2}$ goes from a leg to the hypotenuse. Starting with the hypotenuse requires division, not multiplication.

3.05. Identify the long leg in a 30–60–90 triangle

The side opposite 60° in a 30°–60°–90° triangle has length 9 $\sqrt{3}$ . Find the hypotenuse.

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Recognize the structure. The given side is the long leg, which equals $\sqrt{3}$ when the short leg is s.

Work it through. Set $\sqrt{3}$ so $s = 9.$ The hypotenuse is $2s = 18.$

Answer: 18 units.

Key takeaway. The role of the side matters more than the appearance of the radical. A side involving $\sqrt{3}$ is not automatically the long leg unless the angle information establishes that role.

3.06. An equilateral triangle: altitude, then area

An equilateral triangle has side length 12. What is its area?

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Recognize the structure. An altitude splits the triangle into two 30°–60°–90° triangles.

Work it through. Each half-base is 6, so the altitude is $\sqrt{3}$ Therefore

$\frac{1}{2}$

Answer: $\sqrt{3}$ square units. Another efficient route. Deriving the general formula in the same way gives $\sqrt{3}$ Substituting $s = 12$ yields the same result.

Avoid the trap. Using 12 as both the base and the perpendicular height overstates the area. The altitude is shorter than a slanted side.

3.07. Ratios depend on the named angle

In right triangle ABC, ∠C = 90°, AC = 12, and BC = 5. What is tan A? A) 5/13 B) 12/13 C) 12/5 D) 5/12

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Recognize the structure. Relative to angle A, the opposite leg is BC and the adjacent leg is AC.

Work it through. By definition, $\frac{BC}{AC}$

The hypotenuse is 13, so $sin A = 5/13$ and $cos A = 12/13,$ but neither is the requested ratio.

Answer: D, 5/12.

Key takeaway. You did not need to calculate the hypotenuse to find tangent. Selecting the right ratio can eliminate unnecessary work.

3.08. Convert tangent to sine without finding the angle

For an acute angle θ, tan θ = 3/4. What is sin θ?

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Recognize the structure. Build a representative right triangle with opposite and adjacent legs in the ratio 3 : 4.

Work it through. Choose legs 3 and 4. The hypotenuse is 5 by the Pythagorean theorem. Therefore

$\frac{opposite}{hypotenuse}$

Answer: 3/5.

Another efficient route. With actual legs 3k and 4k, the hypotenuse is 5k, and the factor k cancels. This explains why the convenient smaller triangle is legitimate.

3.09. A trigonometric ratio determines a side length

In a right triangle, an acute angle θ has sin θ = 5/13. The hypotenuse is 39. Find the leg adjacent to θ.

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Recognize the structure. The ratio describes a scaled 5–12–13 triangle.

Work it through. The scale factor is $39/13 = 3.$ The opposite leg is 15, and the adjacent leg is $12(3) = 36.$

Answer: 36 units.

Another efficient route. Find the opposite leg from $o/39 = 5/13,$ then compute $\sqrt{39^{2} − 15^{2}}$

Avoid the trap. $39(5/13) = 15$ gives the opposite leg, not the adjacent leg. The named side is part of the target.

3.10. Complementary-angle algebra

Angles (2x + 5)° and (3x + 10)° are both acute. If sin((2x + 5)°) = cos((3x + 10)°), what is x?

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Recognize the structure. For acute angles, equal sine and cosine values in this form imply that the angles are complementary.

Work it through. Write

$(2x + 5) + (3x + 10) = 90 5x + 15 = 90 x = 15.$

The angles are 35° and 55°, which satisfy the stated acute-angle conditions.

Answer: 15.

Avoid the trap. Do not set $2x + 5 = 3x + 10.$ The functions differ, so the angle expressions are complements, not equal angles.

3.11. Similarity preserves trigonometric ratios

Right triangles ABC and DEF have right angles at C and F, and ∠A = ∠D. If AB = 20, BC = 12, and DE = 35, what is cos D?

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Recognize the structure. The triangles are similar, so $cos D = cos A;$ the second hypotenuse is unnecessary.

Work it through. In ABC, the adjacent leg to A is

$\sqrt{20^{2} − 12^{2}}$

Thus $cos D = cos A = 16/20 = 4/5.$

Answer: 4/5.

Key takeaway. The extra length 35 does not force you to scale the whole triangle. If you do scale, $DF = 28,$ and 28/35 still simplifies to 4/5.

3.12. Model an angle of elevation

From a point on level ground 28 meters from the base of a vertical pole, the angle of elevation to the top is 38°. The angle is measured at ground level. What is the pole’s height, to the nearest tenth of a meter?

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Recognize the structure. The height is opposite the angle; the horizontal distance is adjacent. Use tangent.

Work it through. With height h,

$\frac{h}{28}$

Answer: 21.9 meters.

Avoid the trap. Use degree mode. Also note the ground-level measurement: if the observation point were at eye height, that vertical offset would need to be added to the computed rise.

Key takeaway. Because $38° < 45°,$ the height should be less than the horizontal distance. The answer passes this check.

3.13. Use a ratio and a perimeter to recover area

A right triangle has an acute angle θ with tan θ = 3/4. Its perimeter is 36. What is its area?

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Recognize the structure. Use one common scale factor for all three sides.

Work it through. The legs can be written 3k and 4k, making the hypotenuse 5k. The perimeter condition gives $3k + 4k + 5k = 36 k = 3.$

The legs are 9 and 12, so the area is $(1/2)(9)(12) = 54.$

Answer: 54 square units.

Avoid the trap. The ratio 3 : 4 fixes shape, not size. Treating the legs as literally 3 and 4 would ignore the perimeter condition.

3.14. Target the hypotenuse squared directly

A right triangle has area 60 square units. For one acute angle A, sin A = 5/13. If the hypotenuse has length c, what is c²?

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Recognize the structure. The sine ratio fixes the side proportions, and the question asks for a square rather than a length.

Work it through. Write the opposite leg as 5t and the hypotenuse as 13t. The other leg is 12t. Then

$\frac{1}{2}$

Therefore $c 2 = (13t) 2 = 169t 2 = 338.$

Answer: 338.

Key takeaway. There is no need to calculate $\sqrt{2}$ or $\sqrt{2}$ Solving for the expression actually requested keeps the work shorter and exact.

3.15. Two observation points, one hidden height

Two points on level ground lie on the same side of a vertical tower and on a straight line through its base. The farther point is 40 meters farther from the base than the nearer point. The angles of elevation to the top are 30° from the farther point and 60° from the nearer point. Find the tower’s height.

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Recognize the structure. Let the nearer horizontal distance be x; both right triangles have the same height.

Work it through. From the nearer point, $\sqrt{3}$ From the farther point,

$\frac{x + 40}{\sqrt{3}}$

Equate the heights: $\sqrt{3}$ so $3x = x + 40$ and $x = 20.$ Thus $\sqrt{3}$

Answer: $\sqrt{3}$ meters.

Avoid the trap. The 40 meters separates the observation points; it is not either point’s distance to the tower. Naming x prevents that modeling error.

Mastery check and error prevention

You should be able to switch the reference angle in one triangle and relabel opposite and adjacent correctly; derive a missing trigonometric ratio from a representative triangle; and decide when no angle calculation is necessary.

Useful magnitude checks: for an acute angle, $0 < sin θ < 1$ and $0 < cos θ < 1,$ while tangent can exceed 1. The hypotenuse must be the longest side. A claimed sine of 5/3, a leg longer than the hypotenuse, or a height exceeding the horizontal distance at an elevation below 45° signals a setup error.