Right Triangles and Trigonometry — Free Geometry Review Games.

By the Pythagorean theorem: \(\sqrt{9+16} = \sqrt{25} = 5\).

a^2 + b^2 = c^2, where c is the hypotenuse.

A 45-45-90 triangle is isosceles, so both legs are equal.

Sine = Opposite over Hypotenuse (SOH).

5^2 + 12^2 = 25 + 144 = 169 = 13^2. Yes, it is right.

The sides opposite \(30\), \(60\), \(90\) are in ratio \(1 : \sqrt{3} : 2\).

Hypotenuse \(= \text{leg} \cdot \sqrt{2} = 6\sqrt{2}\).

\(\cos(60) = 1/2\) from the 30-60-90 triangle ratios.

\(\sin(30) = 1/2\), the side opposite 30 degrees over the hypotenuse.

\(\cos(60) = \frac{5}{\text{ladder}}\), \(\frac{1}{2} = \frac{5}{\text{ladder}}\), ladder \(= 10\) ft.

leg^2 = 13^2 - 5^2 = 169 - 25 = 144, leg = 12.

Shorter leg \(= \frac{\text{hypotenuse}}{2} = \frac{14}{2} = 7\).

In a 45-45-90 triangle, opposite = adjacent, so \(\tan(45) = 1\).

tan(35) = height/20, height = 20 * 0.70 = 14 ft.

If sin A = 3/5, then opp=3, hyp=5, adj=4 (by Pythagorean triple). cos A = 4/5.

SOH-CAH-TOA: Cosine = Adjacent over Hypotenuse (CAH), so \(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\). Choice A describes sine (SOH), and Choice C describes tangent (TOA). Choice D is the reciprocal of cosine, known as secant.

SOH-CAH-TOA: Tangent = Opposite over Adjacent (TOA), so \(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}\). Choice B describes cosine (CAH), and Choice D describes sine (SOH). Tangent is unique in that it does not involve the hypotenuse.

Using the Pythagorean theorem: \(6^2 + 8^2 = 36 + 64 = 100 = 10^2\), so the hypotenuse is \(10\). The triple \(6, 8, 10\) is simply \(2 \times (3, 4, 5)\). Choice A fails because \(36 + 64 = 100 \ne 81 = 9^2\).

The hypotenuse is defined as the side opposite the \(90°\) angle, and it is always the longest side of a right triangle. Choice B is a common error — both legs are adjacent to the right angle; the hypotenuse is not. Choice D is false: the Pythagorean theorem gives \(c^2 = a^2 + b^2\), not \(c = a + b\).

By the Pythagorean theorem: \(c^2 = 1^2 + 1^2 = 2\), so \(c = \sqrt{2}\). In general, a \(45°\)-\(45°\)-\(90°\) triangle has legs \(a\) and hypotenuse \(a\sqrt{2}\). Choice D (\(\sqrt{3}\)) is the long leg ratio from a \(30°\)-\(60°\)-\(90°\) triangle, not a \(45°\)-\(45°\)-\(90°\) triangle.

Check: \(7^2 + 24^2 = 49 + 576 = 625 = 25^2\). Since the Pythagorean theorem is satisfied exactly, this is a right triangle. The \(7\)-\(24\)-\(25\) triple is its own primitive Pythagorean triple — not every right triangle is a multiple of \(3\)-\(4\)-\(5\), so Choice C is false.

All three interior angles of a triangle sum to \(180°\). In a right triangle, one angle equals \(90°\), so the two acute angles must sum to \(180° - 90° = 90°\). They are equal only in a \(45°\)-\(45°\)-\(90°\) triangle, which is a special case. Choice B (\(180°\)) is the full angle sum, not just the acute pair.

In a \(30°\)-\(60°\)-\(90°\) triangle the sides are in ratio \(1 : \sqrt{3} : 2\) (short leg : long leg : hypotenuse). The short leg (opposite \(30°\)) is \(5\), so the scale factor is \(5\), and the long leg (opposite \(60°\)) is \(5\sqrt{3}\). Choice B (\(10\)) is the hypotenuse; Choice C (\(5\sqrt{2}\)) is the hypotenuse ratio from a \(45°\)-\(45°\)-\(90°\) triangle.

In a \(45°\)-\(45°\)-\(90°\) triangle with legs \(1\) and hypotenuse \(\sqrt{2}\): \(\cos 45° = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\). Choice A (\(\tfrac{1}{2}\)) equals \(\cos 60°\), and Choice B (\(\tfrac{\sqrt{3}}{2}\)) equals \(\cos 30°\) — memorizing the full trig table prevents these mix-ups.

In a \(30°\)-\(60°\)-\(90°\) triangle with sides \(1, \sqrt{3}, 2\): \(\tan 30° = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\) after rationalizing the denominator. Choice B (\(\sqrt{3}\)) equals \(\tan 60°\), which is the reciprocal relationship — a common reversal error.

By the Pythagorean theorem: \(c^2 = 5^2 + 12^2 = 25 + 144 = 169\), so \(c = \sqrt{169} = 13\). The triple \(5\)-\(12\)-\(13\) is a fundamental Pythagorean triple worth memorizing. Choice C (\(\sqrt{17}\)) results from incorrectly adding rather than squaring: \(\sqrt{5 + 12} \ne c\).

\(\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{8}{10} = \dfrac{4}{5}\). The adjacent leg is \(\sqrt{10^2 - 8^2} = \sqrt{36} = 6\), so \(\cos A = \dfrac{6}{10} = \dfrac{3}{5}\), which is Choice B — a frequent sine/cosine swap error.

The horizontal distance (adjacent) is \(30\) ft and the building height (opposite) is \(h\). Using TOA: \(\tan 60° = \dfrac{h}{30}\), so \(h = 30\tan 60° = 30\sqrt{3}\) ft. Choice A results from dividing \(30\) by \(\tan 60°\) instead of multiplying — that would give the distance if the angle were \(30°\), not \(60°\).

In a \(45°\)-\(45°\)-\(90°\) triangle, hypotenuse \(= a\sqrt{2}\) where \(a\) is the leg length. Setting \(a\sqrt{2} = 8\sqrt{2}\) gives \(a = 8\). Choice C (\(4\sqrt{2}\)) is tempting but wrong: \((4\sqrt{2}) \cdot \sqrt{2} = 8\), which is too small. Choice D (\(16\)) would give a hypotenuse of \(16\sqrt{2}\).

Check: \(8^2 + 15^2 = 64 + 225 = 289 = 17^2\). The Pythagorean theorem is satisfied, confirming a right triangle. The \(8\)-\(15\)-\(17\) triple is its own primitive triple. Choices C and D describe false general rules — odd hypotenuses and multiples of \(3\)-\(4\)-\(5\) are not defining criteria.

First find the hypotenuse: \(AB = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\). For angle \(A\), the adjacent side is \(AC = 12\) and the hypotenuse is \(AB = 13\), so \(\cos A = \dfrac{12}{13}\). Choice A (\(\tfrac{5}{13}\)) equals \(\sin A\) since \(BC = 5\) is opposite angle \(A\) — a common sine-cosine swap.

In a \(30°\)-\(60°\)-\(90°\) triangle with short leg \(1\), long leg \(\sqrt{3}\), and hypotenuse \(2\): \(\tan 60° = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{\sqrt{3}}{1} = \sqrt{3}\). Choice A (\(\tfrac{\sqrt{3}}{2}\)) equals \(\sin 60°\), and Choice D (\(\tfrac{\sqrt{2}}{2}\)) equals \(\sin 45°\) or \(\cos 45°\).

With hypotenuse \(= 10\) and side ratio \(1 : \sqrt{3} : 2\): the short leg \(= \dfrac{10}{2} = 5\) and the long leg \(= 5\sqrt{3}\). Perimeter \(= 5 + 5\sqrt{3} + 10 = 15 + 5\sqrt{3}\). Choice B omits adding the hypotenuse; Choice D uses \(10\sqrt{3}\) for the long leg, which would require a short leg of \(10\) and hypotenuse of \(20\).

If \(\tan A = \dfrac{3}{4}\), label opposite \(= 3\) and adjacent \(= 4\). Find the hypotenuse: \(c = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\). Then \(\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{3}{5}\). Choice C (\(\tfrac{4}{5}\)) is \(\cos A\), not \(\sin A\). The key step is recognizing that you must find the hypotenuse before computing sine or cosine.

Apply the Pythagorean theorem: \(x^2 + (x+7)^2 = (x+8)^2\). Expanding: \(x^2 + x^2 + 14x + 49 = x^2 + 16x + 64\). Simplifying: \(x^2 - 2x - 15 = 0\), which factors as \((x-5)(x+3) = 0\). Since lengths must be positive, \(x = 5\). Verify: legs \(5\) and \(12\), hypotenuse \(13\) — the \(5\)-\(12\)-\(13\) Pythagorean triple.

The leg opposite \(30°\) has length \(20\sin 30° = 20 \cdot \dfrac{1}{2} = 10\). The leg opposite \(60°\) has length \(20\sin 60° = 20 \cdot \dfrac{\sqrt{3}}{2} = 10\sqrt{3}\). Area \(= \dfrac{1}{2} \cdot 10 \cdot 10\sqrt{3} = 50\sqrt{3}\). Choice B (\(100\sqrt{3}\)) is the result of forgetting the \(\dfrac{1}{2}\) factor in the area formula.

The angle of depression equals the angle of elevation from the boat, both \(30°\). In the right triangle: opposite \(=\) cliff height \(= 80\) ft, adjacent \(=\) horizontal distance \(d\). Using TOA: \(\tan 30° = \dfrac{80}{d}\), so \(d = \dfrac{80}{\tan 30°} = \dfrac{80}{\tfrac{\sqrt{3}}{3}} = \dfrac{240}{\sqrt{3}} = 80\sqrt{3}\) ft. Choice B results from computing \(80 \cdot \tan 30°\) (multiplying instead of dividing).

If \(\cos A = \dfrac{5}{13}\), then adjacent \(= 5\) and hypotenuse \(= 13\). By the Pythagorean theorem: opposite \(= \sqrt{13^2 - 5^2} = \sqrt{144} = 12\). Therefore \(\tan A = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{12}{5}\). Choice A (\(\tfrac{12}{13}\)) is \(\sin A\) — confusing tangent with sine is the most common error here.

With the right angle at \(R\), hypotenuse \(= PQ\), the side opposite \(P\) is \(QR\), and the side adjacent to \(P\) is \(PR = 9\). Since \(\sin P = \dfrac{3}{5}\), the \(3\)-\(4\)-\(5\) ratio gives \(\cos P = \dfrac{4}{5}\). Then \(\cos P = \dfrac{PR}{PQ} = \dfrac{9}{PQ} = \dfrac{4}{5}\), so \(PQ = \dfrac{45}{4}\). Finally, \(QR = PQ \cdot \sin P = \dfrac{45}{4} \cdot \dfrac{3}{5} = \dfrac{27}{4}\). Choice D (\(12\)) would require \(PR = 9\) to serve as the full hypotenuse, ignoring the given \(\sin P\).

Let each leg have length \(a\); then hypotenuse \(= a\sqrt{2}\) and perimeter \(= a + a + a\sqrt{2} = a(2 + \sqrt{2})\). Setting \(a(2 + \sqrt{2}) = 10 + 5\sqrt{2} = 5(2 + \sqrt{2})\) gives \(a = 5\). The hypotenuse \(= 5\sqrt{2}\). Choice B (\(10\)) is the sum of the two legs, not the hypotenuse — a careless reading error.

The Pythagorean theorem states that the sum of the squares of the two legs equals the square of the hypotenuse: \(a^2 + b^2 = c^2\). Choice A confuses the theorem with a linear relationship. Choice C would mean one leg is found by subtracting squares, which does not describe a right triangle in general.

In a \(30°\)-\(60°\)-\(90°\) triangle, the side opposite the \(30°\) angle is half the hypotenuse, so \(\sin 30° = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{1}{2}\). Choice A, \(\dfrac{\sqrt{3}}{2}\), is \(\sin 60°\) (equivalently \(\cos 30°\)), a very common mix-up.

In a \(45°\)-\(45°\)-\(90°\) triangle with legs of length \(1\) and hypotenuse \(\sqrt{2}\), \(\cos 45° = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\). Choice A, \(\dfrac{1}{2}\), is \(\cos 60°\), not \(\cos 45°\).

\(\tan 45° = \dfrac{\sin 45°}{\cos 45°} = \dfrac{\sqrt{2}/2}{\sqrt{2}/2} = 1\). Equivalently, in a \(45°\)-\(45°\)-\(90°\) triangle the two legs are equal, so \(\tan 45° = \dfrac{\text{opposite}}{\text{adjacent}} = 1\). Choice C, \(\sqrt{3}\), is \(\tan 60°\).

By the Pythagorean theorem, hypotenuse \(= \sqrt{s^2 + s^2} = \sqrt{2s^2} = s\sqrt{2}\). Choice A, \(2s\), is the hypotenuse of a \(30°\)-\(60°\)-\(90°\) triangle whose shorter leg is \(s\). Choice B, \(s\sqrt{3}\), is the longer leg of that same \(30°\)-\(60°\)-\(90°\) triangle, not the hypotenuse of this one.

In a \(30°\)-\(60°\)-\(90°\) triangle, the side lengths follow the ratio \(1 : \sqrt{3} : 2\) (short leg : long leg : hypotenuse). If the short leg is \(n\), the long leg is \(n\sqrt{3}\). Choice A, \(2n\), is the hypotenuse, not the long leg.

A Pythagorean triple satisfies \(a^2 + b^2 = c^2\). For choice B: \(5^2 + 12^2 = 25 + 144 = 169 = 13^2\). ✓ For choice A: \(3^2 + 4^2 = 25 \neq 36 = 6^2\). For choice C: \(6^2 + 8^2 = 100 \neq 121 = 11^2\).

By the SOH-CAH-TOA mnemonic, \(\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}\) (SOH). Choice A is the definition of \(\cos\theta\) (CAH). Choice B is the definition of \(\tan\theta\) (TOA). Choice C is \(\csc\theta\), the reciprocal of sine.

Using the Pythagorean theorem: \(c = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41\). Note that \((9, 40, 41)\) is a Pythagorean triple. Choice A, \(49\), results from incorrectly adding the legs (\(9 + 40\)) instead of squaring them.

In a \(30°\)-\(60°\)-\(90°\) triangle the sides are in the ratio \(n : n\sqrt{3} : 2n\). Setting the long leg equal to \(n\sqrt{3} = 6\sqrt{3}\) gives \(n = 6\), so the hypotenuse is \(2n = 12\). Choice D, \(18\), results from multiplying \(6\sqrt{3} \cdot \sqrt{3} = 18\) instead of first dividing by \(\sqrt{3}\) to find \(n\).

Using SOH: \(\sin 30° = \dfrac{\text{opposite}}{\text{hypotenuse}}\), so opposite \(= 10 \cdot \sin 30° = 10 \cdot \dfrac{1}{2} = 5\). Choice A, \(5\sqrt{3}\), is the leg adjacent to the \(30°\) angle (opposite the \(60°\) angle), found via \(10 \cdot \cos 30° = 10 \cdot \dfrac{\sqrt{3}}{2} = 5\sqrt{3}\).

Using CAH: \(\cos 45° = \dfrac{\text{adjacent}}{\text{hypotenuse}}\), so adjacent \(= 8 \cdot \cos 45° = 8 \cdot \dfrac{\sqrt{2}}{2} = 4\sqrt{2}\). Choice A, \(4\), would be correct only if \(\cos 45° = \dfrac{1}{2}\), but that is \(\cos 60°\), not \(\cos 45°\).

Using the Pythagorean theorem: \(b = \sqrt{c^2 - a^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\). The triple \((5, 12, 13)\) is a classic Pythagorean triple. Choice A, \(8\), comes from incorrectly subtracting: \(13 - 5 = 8\).

The ground distance is adjacent to the \(60°\) angle and equals \(6\) ft. Using cosine: \(\cos 60° = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{6}{L}\), so \(L = \dfrac{6}{\cos 60°} = \dfrac{6}{1/2} = 12\) feet. Choice C, \(6\sqrt{3}\), is the height the ladder reaches on the wall (the side opposite the \(60°\) angle).

Angle \(D\) is at vertex \(D\). The side opposite \(D\) is \(EF = 8\), and the hypotenuse is \(DE = 17\), so \(\sin D = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{8}{17}\). (The third side is \(DF = \sqrt{17^2 - 8^2} = \sqrt{225} = 15\).) Choice B, \(\dfrac{15}{17}\), is \(\cos D\) (adjacent over hypotenuse).

With hypotenuse \(8\), the short leg is \(\dfrac{8}{2} = 4\) and the long leg is \(4\sqrt{3}\). Area \(= \dfrac{1}{2} \cdot 4 \cdot 4\sqrt{3} = 8\sqrt{3}\). Choice B, \(16\sqrt{3}\), doubles the correct answer — a common mistake when the \(\dfrac{1}{2}\) factor in the area formula is forgotten.

\(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{7}{7} = 1\). Since \(\tan 45° = 1\), we conclude \(\theta = 45°\). This is consistent with the fact that equal legs define a \(45°\)-\(45°\)-\(90°\) triangle.

First find \(MP\): \(MP = \sqrt{MN^2 - NP^2} = \sqrt{676 - 100} = \sqrt{576} = 24\). Then \(\sin M = \dfrac{NP}{MN} = \dfrac{10}{26} = \dfrac{5}{13}\) and \(\cos M = \dfrac{MP}{MN} = \dfrac{24}{26} = \dfrac{12}{13}\). So \(\sin M + \cos M = \dfrac{5}{13} + \dfrac{12}{13} = \dfrac{17}{13}\). Choice D gives only \(\sin M\), omitting the \(\cos M\) term entirely.

Applying the Pythagorean theorem: \((3x)^2 + (4x)^2 = 15^2 \Rightarrow 9x^2 + 16x^2 = 225 \Rightarrow 25x^2 = 225 \Rightarrow x^2 = 9 \Rightarrow x = 3\). The legs are \(9\) and \(12\) with hypotenuse \(15\), a scaled \((3, 4, 5)\) triple. Choice A (\(x = 2\)) gives legs \(6\) and \(8\), which form a valid right triangle with hypotenuse \(10\), but not the one with hypotenuse \(15\).

By the geometric mean altitude theorem, the altitude to the hypotenuse is the geometric mean of the two segments it creates: \(CD = \sqrt{AD \cdot DB} = \sqrt{4 \cdot 9} = \sqrt{36} = 6\). Choice A, \(\sqrt{13}\), incorrectly uses \(\sqrt{AD + DB} = \sqrt{13}\). Choice C, \(6.5\), is the arithmetic mean \(\dfrac{4 + 9}{2}\), not the geometric mean.

The angle of depression is \(30°\), so the angle of elevation from the boat to the lighthouse top is also \(30°\) (alternate interior angles). The horizontal distance \(d\) satisfies \(\tan 30° = \dfrac{120}{d}\), giving \(d = \dfrac{120}{\tan 30°} = \dfrac{120}{1/\sqrt{3}} = 120\sqrt{3}\) feet. Choice B, \(60\sqrt{3}\), results from incorrectly using half the height or confusing \(\tan 30°\) with \(\tan 60°\).

Using the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\): \(\cos^2 A = 1 - \sin^2 A = 1 - \left(\dfrac{2}{3}\right)^2 = 1 - \dfrac{4}{9} = \dfrac{5}{9}\). Choice C, \(\dfrac{4}{9}\), is \(\sin^2 A\) itself. Choice D, \(\dfrac{\sqrt{5}}{3}\), is \(\cos A\) rather than \(\cos^2 A\).

Since \(\sin\theta = \dfrac{5}{13}\), the opposite side is \(5\) and the hypotenuse is \(13\). The adjacent side is \(\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\). Therefore \(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{5}{12}\). Choice B, \(\dfrac{12}{13}\), is \(\cos\theta\), not \(\tan\theta\).

The altitude of an equilateral triangle bisects the base, creating two \(30°\)-\(60°\)-\(90°\) right triangles. Each has a short leg of \(\dfrac{10}{2} = 5\). The altitude is the long leg: \(5 \cdot \sqrt{3} = 5\sqrt{3}\). Choice C, \(10\sqrt{3}\), incorrectly uses the full side length \(10\) as the short leg instead of the half-base \(5\).

By the geometric mean (leg) theorem, each leg is the geometric mean of the entire hypotenuse and the segment of the hypotenuse adjacent to that leg: \(PR^2 = PS \cdot PQ = 3 \cdot 12 = 36\), so \(PR = 6\). Note that \(SQ = PQ - PS = 12 - 3 = 9\). Choice D, \(9\), is the length of segment \(SQ\), not the leg \(PR\).

By the Pythagorean theorem, \(c^2 = 6^2 + 8^2 = 36 + 64 = 100\), so \(c = \sqrt{100} = 10\). This is the familiar \(3\)-\(4\)-\(5\) Pythagorean triple scaled by \(2\). Choice \(\sqrt{48} \approx 6.93\) is actually less than one of the legs, which is impossible for a hypotenuse. Choice \(12\) incorrectly adds the legs directly.

Using the mnemonic SOH-CAH-TOA: \(\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}\) (SOH). Choice A defines \(\cos\theta\) (CAH). Choice C defines \(\tan\theta\) (TOA). Choice D is the reciprocal of sine, known as cosecant (\(\csc\theta\)), which is not one of the three primary ratios.

In a \(45°\)-\(45°\)-\(90°\) triangle the hypotenuse equals a leg multiplied by \(\sqrt{2}\), so hypotenuse \(= 5\sqrt{2}\). Choice A (\(5\sqrt{3}\)) mistakenly applies the \(30°\)-\(60°\)-\(90°\) long-leg ratio. Choice B (\(10\)) doubles the leg instead of multiplying by \(\sqrt{2} \approx 1.414\). Choice D is the leg length when the hypotenuse equals \(5\), not the other way around.

Cosine equals \(\dfrac{\text{adjacent}}{\text{hypotenuse}}\). With the right angle at \(C\), the hypotenuse is \(AB\), the side adjacent to angle \(A\) is \(AC\), and the side opposite angle \(A\) is \(BC\). Therefore \(\cos A = \dfrac{AC}{AB}\). Choice A gives \(\sin A\) (opposite over hypotenuse). Choice C gives \(\tan A\) (opposite over adjacent).

A set forms a right triangle if and only if \(a^2 + b^2 = c^2\) where \(c\) is the largest value. Testing \(6, 8, 10\): \(6^2 + 8^2 = 36 + 64 = 100 = 10^2\) ✓. For \(5, 11, 13\): \(25 + 121 = 146 \neq 169\). For \(4, 6, 8\): \(16 + 36 = 52 \neq 64\). For \(5, 7, 9\): \(25 + 49 = 74 \neq 81\). The set \(6, 8, 10\) is the \(3\)-\(4\)-\(5\) triple scaled by \(2\).

In a \(30°\)-\(60°\)-\(90°\) triangle the hypotenuse is exactly twice the short leg (the side opposite \(30°\)), so hypotenuse \(= 2 \times 7 = 14\). Choice A (\(7\sqrt{3}\)) is the length of the side opposite \(60°\), not the hypotenuse. Choice B (\(7\sqrt{2}\)) incorrectly applies the \(45°\)-\(45°\)-\(90°\) ratio.

From SOH-CAH-TOA, tangent is the TOA ratio: \(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}\). Choice C is \(\sin\theta\) (SOH). Choice B is the cotangent, the reciprocal of tangent. Choice A is the cosecant, the reciprocal of sine.

Side \(BC\) is opposite angle \(A\), so \(\sin A = \dfrac{BC}{AB}\). Thus \(BC = AB \cdot \sin 35° = 12 \times 0.574 \approx 6.9\). Choice B (\(9.8\)) results from using cosine instead of sine: \(12 \times 0.819 \approx 9.8\), which gives the adjacent side \(AC\). Choice D (\(20.9\)) comes from dividing rather than multiplying: \(\dfrac{12}{0.574} \approx 20.9\).

The height \(h\) is opposite the \(40°\) angle and the \(50\) ft distance is adjacent, so \(\tan 40° = \dfrac{h}{50}\). Thus \(h = 50 \times 0.839 \approx 42.0\) ft. Choice A (\(32.2\) ft) uses sine (\(50 \times 0.643\)) instead of tangent. Choice D (\(38.3\) ft) uses cosine (\(50 \times 0.766\)). Choice C (\(59.6\) ft) inverts the ratio: \(\dfrac{50}{0.839} \approx 59.6\).

The diagonal of a square with side \(s\) divides it into two \(45°\)-\(45°\)-\(90°\) triangles, giving diagonal \(= s\sqrt{2}\). Setting \(s\sqrt{2} = 10\sqrt{2}\) yields \(s = 10\) in. Choice A (\(5\sqrt{2}\)) is what you would get if the diagonal were \(10\) (not \(10\sqrt{2}\)). Choice C (\(5\)) halves the diagonal instead of dividing by \(\sqrt{2}\).

After \(3\) hours, the northbound ship has traveled \(15 \times 3 = 45\) miles and the eastbound ship \(20 \times 3 = 60\) miles. Because north and east are perpendicular, the distance between them is the hypotenuse: \(\sqrt{45^2 + 60^2} = \sqrt{2025 + 3600} = \sqrt{5625} = 75\) miles. Choice D (\(105\)) incorrectly adds the two distances (\(45 + 60 = 105\)) instead of applying the Pythagorean theorem.

If both legs equal \(a\), then \(\tan\theta = \dfrac{a}{a} = 1\). Since \(\tan 45° = 1\), each acute angle is \(45°\), confirming a \(45°\)-\(45°\)-\(90°\) isosceles right triangle. A \(30°\)-\(60°\)-\(90°\) triangle has legs in ratio \(1:\sqrt{3}\), not \(1:1\), so equal legs cannot produce \(30°\) or \(60°\) acute angles.

The leg opposite \(30°\) is \(20\sin 30° = 20 \cdot \dfrac{1}{2} = 10\). The leg adjacent to \(30°\) is \(20\cos 30° = 20 \cdot \dfrac{\sqrt{3}}{2} = 10\sqrt{3}\). Area \(= \dfrac{1}{2}(10)(10\sqrt{3}) = 50\sqrt{3}\). Choice A (\(50\)) omits the \(\sqrt{3}\) factor from the longer leg. Choice C (\(100\)) forgets the \(\dfrac{1}{2}\) in the area formula.

By the Pythagorean theorem: \(x^2 + (x+2)^2 = 52\). Expanding: \(x^2 + x^2 + 4x + 4 = 52\), so \(2x^2 + 4x - 48 = 0\), or \(x^2 + 2x - 24 = 0\). Factoring: \((x + 6)(x - 4) = 0\), giving \(x = 4\) (taking the positive root). Check: \(4^2 + 6^2 = 16 + 36 = 52\) ✓. Choice C (\(x = 5\)) gives \(25 + 49 = 74 \neq 52\).

With \(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{5}{13}\), set adjacent \(= 5\) and hypotenuse \(= 13\). The opposite side is \(\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\). Therefore \(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{12}{5}\). Choice B (\(\dfrac{12}{13}\)) gives \(\sin\theta\), not \(\tan\theta\). Choice A (\(\dfrac{5}{12}\)) inverts the correct answer.

In a \(30°\)-\(60°\)-\(90°\) triangle with hypotenuse \(2\) and short leg \(1\), the side adjacent to the \(60°\) angle has length \(1\). So \(\cos 60° = \dfrac{1}{2}\). Choice A (\(\dfrac{\sqrt{3}}{2}\)) is \(\cos 30°\) (equivalently \(\sin 60°\)) — a common mix-up. Choice D (\(\dfrac{\sqrt{2}}{2}\)) is \(\cos 45°\).

The angle \(\theta\) satisfies \(\tan\theta = \dfrac{\text{rise}}{\text{run}} = \dfrac{3}{15} = 0.2\), so \(\theta = \tan^{-1}(0.2) \approx 11.3°\). Choice A (\(78.7°\)) results from swapping rise and run: \(\tan^{-1}\!\left(\dfrac{15}{3}\right) = \tan^{-1}(5) \approx 78.7°\), which is actually the complementary angle.

In a \(45°\)-\(45°\)-\(90°\) triangle, leg \(= \dfrac{\text{hypotenuse}}{\sqrt{2}} = \dfrac{18}{\sqrt{2}} = 9\sqrt{2}\). The perimeter is \(9\sqrt{2} + 9\sqrt{2} + 18 = 18\sqrt{2} + 18 = 18 + 18\sqrt{2}\). Choice C (\(18 + 9\sqrt{2}\)) adds only one leg to the hypotenuse instead of both. Choice D (\(36\)) treats the triangle as equilateral with side \(12\).

For the \(30°\)-\(60°\)-\(90°\) triangle: the longer leg equals the short leg times \(\sqrt{3}\), so the short leg \(= \dfrac{6}{\sqrt{3}} = 2\sqrt{3}\), and the hypotenuse \(= 2(2\sqrt{3}) = 4\sqrt{3}\). For the \(45°\)-\(45°\)-\(90°\) triangle: hypotenuse \(= 6\sqrt{2}\). Sum \(= 4\sqrt{3} + 6\sqrt{2}\). Choice B swaps the \(\sqrt{2}\) and \(\sqrt{3}\) multipliers between the two triangles. Choice C (\(6\sqrt{3} + 6\sqrt{2}\)) incorrectly treats the longer leg as the short leg, giving a wrong hypotenuse of \(2 \times 6 = 12\) — not \(6\sqrt{3}\).

Since \(\sin R = \dfrac{ST}{RS} = \dfrac{3}{5}\), leg \(ST = 20 \cdot \dfrac{3}{5} = 12\). Leg \(RT = \sqrt{RS^2 - ST^2} = \sqrt{400 - 144} = \sqrt{256} = 16\). Perimeter \(= 12 + 16 + 20 = 48\). The sides form a \(3\)-\(4\)-\(5\) triple scaled by \(4\). Choice A (\(40\)) omits one leg. Choice D (\(52\)) results from a computation error, perhaps using \(ST = 15\) (taking \(\sin R = \tfrac{3}{4}\) instead of \(\tfrac{3}{5}\)).

By the geometric mean (leg) theorem, each leg is the geometric mean of the full hypotenuse and the adjacent projection segment: \(XZ^2 = XW \cdot XY = 3 \cdot 12 = 36\), so \(XZ = 6\). Choice C (\(3\sqrt{3}\)) results from using the altitude theorem instead: \(ZW^2 = XW \cdot WY = 3 \cdot 9 = 27\), giving \(ZW = 3\sqrt{3}\) — that is the altitude length, not the leg \(XZ\). Choice A (\(4\)) comes from dividing: \(\dfrac{12}{3} = 4\) instead of using the geometric mean.

The altitude of an equilateral triangle with side \(s\) equals \(\dfrac{s\sqrt{3}}{2}\). Setting \(\dfrac{s\sqrt{3}}{2} = 6\sqrt{3}\) gives \(s = 12\). Area \(= \dfrac{\sqrt{3}}{4}s^2 = \dfrac{\sqrt{3}}{4}(144) = 36\sqrt{3}\). Choice A (\(36\)) drops the \(\sqrt{3}\) factor. Choice C (\(72\)) substitutes the side length for the altitude: \(\dfrac{1}{2}(12)(12) = 72\), ignoring that the altitude is \(6\sqrt{3}\), not \(12\). Choice D (\(72\sqrt{3}\)) forgets the \(\dfrac{1}{2}\) in the area formula: \(12 \times 6\sqrt{3} = 72\sqrt{3}\).

Since \(\sin A = \dfrac{BC}{AB} = \dfrac{5}{13}\), leg \(BC = 26 \cdot \dfrac{5}{13} = 10\). The other leg is \(AC = \sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24\). Area \(= \dfrac{1}{2}(10)(24) = 120\). Choice A (\(60\)) is half the correct answer — likely from miscomputing \(AC = 12\) (using \(13^2 - 5^2 = 144\), forgetting to scale by \(2\)). Choice D (\(240\)) omits the \(\dfrac{1}{2}\) factor in the area formula.

Let \(h = BC\) and \(BD = d\). From \(D\): \(\tan 45° = \dfrac{h}{d} = 1\), so \(d = h\). From \(A\): \(\tan 30° = \dfrac{h}{d + 200} = \dfrac{1}{\sqrt{3}}\), giving \(h\sqrt{3} = h + 200\), so \(200 = h(\sqrt{3} - 1)\) and \(h = \dfrac{200}{\sqrt{3} - 1}\). Rationalizing: \(h = \dfrac{200(\sqrt{3}+1)}{(\sqrt{3})^2 - 1^2} = \dfrac{200(\sqrt{3}+1)}{2} = 100(\sqrt{3}+1)\) m. Choice A (\(100(\sqrt{3}-1)\)) is the denominator before rationalizing, not the answer. Choice D (\(100\sqrt{3}\)) ignores the \(+1\) term from rationalizing.

Place \(A\) at the origin: \(A = (0,0)\), \(B = (10,0)\), \(C = (10,8)\), \(D = (0,8)\). The midpoint of \(\overline{CD}\) is \(E = (5, 8)\). By the Pythagorean theorem (distance formula): \(AE = \sqrt{(5-0)^2 + (8-0)^2} = \sqrt{25 + 64} = \sqrt{89}\). Choice B (\(\sqrt{164}\)) is the full diagonal \(AC = \sqrt{10^2 + 8^2} = \sqrt{164}\), not \(AE\). Choice C (\(\sqrt{41}\)) results from treating \(E\) as the center of the rectangle at \((5, 4)\) rather than the midpoint of \(\overline{CD}\) at \((5, 8)\): \(\sqrt{5^2 + 4^2} = \sqrt{41}\).