Right Triangle Trigonometry — Free Trigonometry Review Games.

SOH: sine = opposite over hypotenuse.

CAH: cosine = adjacent over hypotenuse.

TOA: tangent = opposite over adjacent.

sin = opposite/hypotenuse = 3/5.

Adjacent = \(\sqrt{169-25} = 12\). \(\cos = \frac{12}{13}\).

Angle of elevation is measured from the horizontal looking up.

height = 50 * tan(60) = 50 * 1.732 = 86.6 ft.

\(\sin = \sqrt{1-0.36} = \sqrt{0.64} = 0.8\).

Cotangent is the reciprocal of tangent: adjacent/opposite.

Cosecant is the reciprocal of sine: hypotenuse/opposite.

height = 200 * tan(35) = 200 * 0.70 = 140 ft.

cos = 3/5, sin = 4/5, tan = sin/cos = 4/3.

opposite = hyp * sin(A) = 10 * 0.643 = 6.43.

distance = 100/tan(25) = 100/0.466 = 214.6 ft.

Hypotenuse = \(\sqrt{49+576} = 25\). \(\sin = \frac{7}{25}\).

The hypotenuse is defined as the side opposite the \(90°\) angle, and it is always the longest side of a right triangle. The choice "Opposite side" is wrong because that term is relative to a chosen acute angle, not the right angle itself. Remembering that the hypotenuse is fixed relative to the right angle helps students correctly set up SOH-CAH-TOA ratios.

Tangent is defined by the TOA part of SOH-CAH-TOA as opposite over adjacent, since it compares the two legs relative to the reference angle. The choice \(\frac{\text{opposite}}{\text{hypotenuse}}\) is wrong because that ratio defines sine, not tangent. Keeping the SOH-CAH-TOA mnemonic straight prevents mixing up which sides belong to which ratio.

By the Pythagorean theorem, \(c=\sqrt{6^2+8^2}=\sqrt{100}=10\), since the hypotenuse squared equals the sum of the squares of the legs. The choice \(14\) is wrong because it simply adds the legs instead of using the Pythagorean relationship. Always confirm right-triangle side lengths satisfy \(a^2+b^2=c^2\) before trusting a numeric answer.

In a 30-60-90 triangle the side opposite the \(30°\) angle is half the hypotenuse, giving \(\sin(30°)=\frac{1}{2}\) exactly. The choice \(\frac{\sqrt{3}}{2}\) is wrong because that value corresponds to \(\sin(60°)\), not \(\sin(30°)\). Memorizing the exact trig values for \(30°\), \(45°\), and \(60°\) speeds up problem solving without a calculator.

In a 30-60-90 triangle, the side adjacent to the \(60°\) angle is half the hypotenuse, so \(\cos(60°)=\frac{1}{2}\). The choice \(\frac{\sqrt{3}}{2}\) is wrong because that is the value of \(\cos(30°)\), illustrating the complementary relationship \(\sin(\theta)=\cos(90°-\theta)\). Keeping these standard angle values memorized avoids errors on non-calculator sections.

In a 45-45-90 triangle the two legs are equal, so opposite over adjacent equals \(1\), making \(\tan(45°)=1\). The choice \(\sqrt{2}\) is wrong because that value equals the ratio of the hypotenuse to a leg, not the tangent ratio. Recognizing that equal legs produce a tangent of \(1\) is a quick check for isosceles right triangles.

An angle of depression is formed when an observer looks downward from a horizontal reference line to an object positioned below their eye level. The choice "Above the observer" is wrong because that scenario describes an angle of elevation, not depression. Distinguishing elevation from depression is essential since the two angles are alternate interior angles equal in measure when the sight lines are parallel to the horizontal.

Using the Pythagorean theorem, \(\text{adjacent}=\sqrt{13^2-5^2}=\sqrt{144}=12\), since the legs and hypotenuse satisfy \(a^2+b^2=c^2\). The choice \(18\) is wrong because it incorrectly adds the given lengths instead of applying the Pythagorean relationship. This 5-12-13 triple is a common Pythagorean triple worth memorizing for quick right-triangle problems.

Cosecant is defined as \(\csc(\theta)=\frac{1}{\sin(\theta)}\), making it the reciprocal of the sine function by definition. The choice "Secant" is wrong because secant is the reciprocal of cosine, not sine. Knowing the three reciprocal pairs, sine-cosecant, cosine-secant, and tangent-cotangent, prevents confusion on exam problems involving reciprocal identities.

Secant is defined as \(\sec(\theta)=\frac{1}{\cos(\theta)}\), so it is by definition the reciprocal of cosine. The choice "Cosecant" is wrong because cosecant is the reciprocal of sine, not cosine. Pairing each primary function with its correct reciprocal is a foundational skill for simplifying trig expressions.

Cotangent is defined as \(\cot(\theta)=\frac{1}{\tan(\theta)}=\frac{\text{adjacent}}{\text{opposite}}\), making it the reciprocal of tangent. The choice "Secant" is wrong because secant is the reciprocal of cosine, unrelated to the tangent ratio. Recognizing reciprocal pairs quickly helps when converting between equivalent trig expressions on tests.

At \(90°\) the opposite side equals the hypotenuse in the limiting right-triangle configuration, so \(\sin(90°)=1\) exactly. The choice \(0\) is wrong because that is the value of \(\cos(90°)\), not \(\sin(90°)\). Knowing the boundary values at \(0°\) and \(90°\) helps confirm whether a computed trig ratio is reasonable.

At \(0°\) the adjacent side equals the hypotenuse, so \(\cos(0°)=1\) by the CAH definition of cosine. The choice \(0\) is wrong because that is the value of \(\sin(0°)\), not \(\cos(0°)\). Checking these limiting values is a fast way to verify trig calculations near the axes.

Since \(\tan(35°)=\frac{\text{opposite}}{10}\), solving gives opposite \(=10\tan(35°)\approx 7.0\), applying the TOA relationship directly. The choice \(8.5\) is wrong because it does not match the tangent computation for a \(35°\) angle. Always identify which sides are known and which trig ratio links them before solving for a missing side.

Since \(\sin(25°)=\frac{\text{opposite}}{20}\), solving gives opposite \(=20\sin(25°)\approx 8.45\), using the SOH relationship. The choice \(18.13\) is wrong because it corresponds to \(20\cos(25°)\), the adjacent side, not the opposite side. Matching the correct ratio, sine for opposite over hypotenuse, to the given information avoids swapping sides.

Since \(\tan(\theta)=\frac{7}{24}\), taking the inverse tangent gives \(\theta=\arctan\left(\frac{7}{24}\right)\approx 16.3°\). The choice \(73.7°\) is wrong because that is the complementary angle, obtained if the ratio were flipped to \(\frac{24}{7}\). Always check whether opposite/adjacent or adjacent/opposite is being computed before taking the inverse function.

Using \(\tan(32°)=\frac{h}{50}\), solving gives \(h=50\tan(32°)\approx31.2\) ft, since the height is opposite the elevation angle. The choice \(42.4\) ft is wrong because it results from using \(\sin(32°)\) with \(50\) as if it were the hypotenuse, which it is not. In elevation problems, the horizontal distance is typically the adjacent side, so tangent should be used to relate it to the height.

Because the angle of depression equals the angle of elevation from the boat, \(\tan(18°)=\frac{120}{d}\), giving \(d=\frac{120}{\tan(18°)}\approx369\) ft. The choice \(114\) ft is wrong because it results from computing \(120\tan(18°)\) instead of dividing, misplacing the height and distance in the ratio. Depression problems rely on alternate interior angles making the depression angle equal to the elevation angle from the ground observer.

Since \(\tan(55°)=\frac{\text{opposite}}{12}\), solving gives opposite \(=12\tan(55°)\approx17.1\), applying the TOA ratio to the given adjacent side. The choice \(6.9\) is wrong because it results from computing \(12\cos(55°)\) instead of using tangent to relate the two legs. Correctly labeling which side is adjacent to the given angle is critical before choosing sine, cosine, or tangent.

The two acute angles in a right triangle are complementary because all three angles must sum to \(180°\) and one angle is already \(90°\), so \(90°-50°=40°\). The choice \(130°\) is wrong because it would make the triangle's angle sum exceed \(180°\) when combined with the right angle and the given angle. Remembering that acute angles in a right triangle always sum to \(90°\) is essential for solving triangles quickly.

Since the base distance is adjacent to the \(65°\) angle and the ladder is the hypotenuse, \(\cos(65°)=\frac{5}{L}\), giving \(L=\frac{5}{\cos(65°)}\approx11.8\) ft. The choice \(5.5\) ft is wrong because it is smaller than the adjacent side itself, which is impossible since the hypotenuse must be the longest side. When the hypotenuse is unknown but an angle and one leg are given, division rather than multiplication by the trig ratio is required.

Using \(\tan(27°)=\frac{h}{40}\), solving gives \(h=40\tan(27°)\approx20.4\) ft, since the height is opposite the elevation angle and the distance is adjacent. The choice \(44.9\) ft is wrong because it results from dividing by \(\tan(27°)\) instead of multiplying, which would be appropriate only if the height were the known side. Always confirm whether the unknown side is opposite or adjacent before deciding whether to multiply or divide by the tangent ratio.

By the Pythagorean theorem, \(x=\sqrt{15^2-9^2}=\sqrt{144}=12\), since the hypotenuse squared equals the sum of the squares of both legs. The choice \(6\) is wrong because it does not satisfy \(9^2+6^2=15^2\), since \(81+36=117\neq225\). This is another instance of a common Pythagorean triple, \(9\)-\(12\)-\(15\), a multiple of the \(3\)-\(4\)-\(5\) triangle.

With opposite \(=3\) and adjacent \(=4\), the Pythagorean theorem gives hypotenuse \(=5\), so \(\sec(\theta)=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{5}{4}\). The choice \(\frac{5}{3}\) is wrong because that ratio represents \(\csc(\theta)\), using the opposite side in the denominator instead of the adjacent side. Building the full 3-4-5 triangle from a given ratio is a reliable method for finding any other trig function of the same angle.

Since the horizontal distance is adjacent to the \(8°\) angle and the rise is opposite, \(\tan(8°)=\frac{h}{500}\), giving \(h=500\tan(8°)\approx70.2\) ft. The choice \(495.1\) ft is wrong because it results from using cosine instead of tangent, confusing the rise with the road's actual length. Grade or slope angle problems are modeled the same way as elevation problems, with tangent linking the rise to the horizontal run.

Using \(\tan(60°)=\frac{h}{30}\), solving gives \(h=30\tan(60°)=30\sqrt{3}\approx51.96\) ft, since tangent relates the opposite height to the adjacent distance. The choice \(34.64\) ft is wrong because it results from computing \(30\sqrt{3}/\sqrt{3}\) incorrectly, effectively using \(\tan(30°)\) instead of \(\tan(60°)\). Being careful with exact values like \(\tan(60°)=\sqrt{3}\) avoids arithmetic slips on non-calculator problems.

Taking the inverse sine, \(\theta=\arcsin(0.8)\approx53.1°\), which is the standard angle from the 3-4-5 triangle scaled so the opposite side is \(0.8\) of the hypotenuse. The choice \(36.9°\) is wrong because that is the complementary angle, corresponding to \(\arccos(0.8)\) instead of \(\arcsin(0.8)\). Recognizing the 3-4-5 triangle's angles, approximately \(36.9°\) and \(53.1°\), is a useful shortcut for many trigonometry problems.

Since the angle of depression equals the angle of elevation from the airport, \(\tan(15°)=\frac{3000}{d}\), giving \(d=\frac{3000}{\tan(15°)}\approx11,196\) ft. The choice \(3106\) ft is wrong because it results from computing \(3000\tan(15°)\), which would only be correct if the horizontal distance were the known side and altitude were unknown. When the small angle is opposite a much smaller side compared to the adjacent side, the adjacent distance grows large, so results near \(11{,}000\) ft are reasonable for a \(15°\) angle.

Since \(\sin(28°)=\frac{a}{14}\), solving gives \(a=14\sin(28°)\approx6.6\), using the SOH relationship between the opposite side and hypotenuse. The choice \(12.4\) is wrong because it corresponds to \(14\cos(28°)\), which gives the adjacent side rather than the opposite side. Labeling triangle sides relative to the given angle before choosing sine, cosine, or tangent prevents mixing up opposite and adjacent.

Because the two acute angles of a right triangle are complementary, the side opposite one angle is adjacent to the other, giving \(\sin(\theta)=\cos(90°-\theta)\). The choice \(\sin(\theta)=\cos(180°-\theta)\) is wrong because \(180°-\theta\) is not the complementary angle in a right triangle context and produces a sign change instead. This co-function identity is a direct consequence of the complementary acute angles in every right triangle.

Since the horizontal distance is adjacent to the \(22°\) angle and the drop is opposite, \(\tan(22°)=\frac{h}{200}\), giving \(h=200\tan(22°)\approx80.8\) ft. The choice \(185.4\) ft is wrong because it results from using cosine instead of tangent, which would find the slope length rather than the vertical drop. Depression problems on inclines use the same opposite-adjacent tangent relationship as elevation problems on flat ground.

Equal legs mean \(\tan(\theta)=1\) for both acute angles, and since \(\arctan(1)=45°\), each acute angle must measure \(45°\). The choice \(60°\) is wrong because a \(60°\)-\(30°\) pair, not two equal \(60°\) angles, describes an unequal-leg right triangle. Recognizing an isosceles right triangle instantly gives \(45°\)-\(45°\)-\(90°\) without further calculation.

Setting \(d=h\) (from \(\tan45°=1\)) and \(d+100=h\sqrt{3}\) (from \(\tan30°=\frac{1}{\sqrt3}\)), solving simultaneously gives \(h(\sqrt3-1)=100\), so \(h=\frac{100}{\sqrt3-1}\approx136.6\) ft. The choice \(100.0\) ft is wrong because it ignores the change in angle entirely and simply reuses the distance moved. Two-observation elevation problems require setting up a system of equations from two right triangles sharing the same height.

Since the height up the wall is opposite the \(70°\) angle and the ladder is the hypotenuse, \(\sin(70°)=\frac{h}{15}\), giving \(h=15\sin(70°)\approx14.1\) ft. The choice \(5.1\) ft is wrong because it results from using \(\cos(70°)\), which gives the horizontal base distance instead of the vertical height. Correctly matching the trig function, sine for the vertical reach and cosine for the horizontal base, is essential in ladder problems.

The horizontal distance between buildings is \(d=\frac{50}{\tan(35°)}\approx71.4\) ft, and since the nearby building's top rises above the observer's eye level by \(d\tan(20°)\approx26.0\) ft, its total height is \(50+26.0\approx76.0\) ft. The choice \(50.0\) ft is wrong because it ignores the additional height above the observer indicated by the angle of elevation. Multi-angle building problems require finding a shared horizontal distance first, then applying it to each vertical segment separately.

The hill's height is \(250\tan(22°)\approx101.0\) ft and the height to the top of the tower is \(250\tan(25°)\approx116.6\) ft, so the tower's own height is the difference, \(116.6-101.0\approx15.6\) ft. The choice \(101.0\) ft is wrong because that value is the hill's height alone, not the additional height contributed by the tower. When a structure sits atop another elevation, subtracting the two computed heights isolates the added structure's height.

Each observer's distance from the base is found separately: \(\frac{80}{\tan(50°)}\approx67.1\) ft and \(\frac{80}{\tan(35°)}\approx114.3\) ft, and since they stand on opposite sides, the total distance is their sum, \(\approx181.4\) ft. The choice \(67.1\) ft is wrong because it only accounts for one observer's distance, not the combined separation between both. When observers are on opposite sides of a shared vertical object, their individual horizontal distances must be added, not compared directly.

Since the height is opposite the \(50°\) angle and the string is the hypotenuse, \(\sin(50°)=\frac{h}{200}\), giving \(h=200\sin(50°)\approx153.2\) ft. The choice \(128.6\) ft is wrong because it results from using \(\cos(50°)\), which gives the horizontal distance from the flyer rather than the vertical height. Assuming a straight string turns the kite problem into a standard SOH application with the string as the hypotenuse.

Letting \(h\) be the cliff height, the two distances satisfy \(\frac{h}{\tan(20°)}-\frac{h}{\tan(35°)}=100\), and solving gives \(h\approx\frac{100}{2.747-1.428}\approx75.8\) ft. The choice \(100.0\) ft is wrong because it merely restates the distance sailed rather than solving the resulting equation for the height. Whenever elevation angles change after moving a known distance, the two right-triangle expressions for the horizontal distance must be set up and subtracted to isolate the height.

Since the pole height is opposite the \(40°\) angle and the wire is the hypotenuse, \(\sin(40°)=\frac{60}{L}\), giving \(L=\frac{60}{\sin(40°)}\approx93.3\) ft. The choice \(46.0\) ft is wrong because it is shorter than the pole itself, which is impossible since the hypotenuse must exceed either leg of a right triangle. When the opposite side and an angle are known but the hypotenuse is unknown, dividing by the sine ratio is required rather than multiplying.

The shadow lengths satisfy \(\frac{h}{\tan(25°)}-\frac{h}{\tan(55°)}=30\), and solving gives \(h\approx\frac{30}{2.145-0.700}\approx20.8\) ft. The choice \(30.0\) ft is wrong because it simply restates the shadow's change in length rather than solving for the pole's fixed height. Shadow-shortening problems mirror the two-angle elevation setup, requiring a difference of two cotangent-based expressions equal to the known change in distance.

Since \(\sin(\theta)=\frac{10}{26}\), taking the inverse sine gives \(\theta=\arcsin\left(\frac{10}{26}\right)\approx22.6°\). The choice \(67.4°\) is wrong because that is the complementary angle, which is opposite the other leg of length \(24\), not the leg of length \(10\). Finding the other leg first, \(\sqrt{26^2-10^2}=24\), can help confirm which angle corresponds to which side before computing the inverse trig function.

Since the rise is opposite the \(5°\) angle and the ramp is the hypotenuse, \(\sin(5°)=\frac{3}{L}\), giving \(L=\frac{3}{\sin(5°)}\approx34.4\) ft as the minimum length that keeps the angle at or below \(5°\). The choice \(17.2\) ft is wrong because a shorter ramp would force a steeper incline angle greater than \(5°\), violating the accessibility requirement. Small angle constraints like ramp codes require dividing by sine to find the necessary hypotenuse length for a fixed rise.

The two horizontal distances to the airport are \(\frac{5000}{\tan(12°)}\approx23,520\) ft and \(\frac{5000}{\tan(18°)}\approx15,389\) ft, so the distance traveled is their difference, \(23,520-15,389\approx8131\) ft. The choice \(23,520\) ft is wrong because that is only the distance at the first sighting, not the change in position between the two observations. Comparing two depression angles at a fixed altitude requires computing both horizontal distances separately before subtracting to find distance traveled.

The opposite leg is \(9\tan(63°)\approx17.66\), and since the area of a right triangle equals half the product of its legs, area \(=\frac{1}{2}(9)(17.66)\approx79.5\). The choice \(40.5\) is wrong because it uses only the given leg of \(9\) squared and halved, ignoring the actual computed opposite side from the tangent ratio. Multi-step area problems require first solving for any missing leg using trig ratios before applying the standard triangle area formula.