Precalculus Quiz: Deriving The Triangle Area Formula

What this quiz covers

This quiz focuses on Deriving The Triangle Area Formula, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $3r^2\sin(60°)$ square units (correct answer)
2. $6r^2\sin(60°)$ square units
3. $3r^2\sin(120°)$ square units
4. $6r^2\sin(30°)$ square units

Explanation: A regular hexagon can be divided into 6 congruent triangles, each with two sides of length $r$ (radii) and an included central angle of $\frac{360°}{6} = 60°$. The area of each triangle is $\frac{1}{2}r^2\sin(60°)$. Total hexagon area = $6 \times \frac{1}{2}r^2\sin(60°) = 3r^2\sin(60°)$. Choice B doubles the correct answer, choice C uses $120°$ instead of $60°$, and choice D uses $30°$ and has the wrong coefficient.

Question 2

1. $3000\sin(40°)$ square kilometers
2. $3000\sin(30°)$ square kilometers
3. $1500\sin(70°)$ square kilometers
4. $1500\sin(30°)$ square kilometers (correct answer)

Explanation: This problem tests your ability to apply the triangle area formula in a real-world navigation context. When you see ships traveling at different angles from the same starting point, visualize the triangle formed and identify what information you have about its sides and angles. First, find the distances traveled. After 2 hours, Ship A has traveled $25 \times 2 = 50$ km, and Ship B has traveled $30 \times 2 = 60$ km. These form two sides of our triangle, with the port as the vertex between them. The key insight is finding the angle between the ships' paths. Ship A travels $40°$ north of east, while Ship B travels $70°$ north of east. The angle between their directions is $70° - 40° = 30°$. Using the triangle area formula with two sides and the included angle: Area = $\frac{1}{2}ab\sin(C)$, where $a = 50$, $b = 60$, and $C = 30°$. Therefore: Area = $\frac{1}{2}(50)(60)\sin(30°) = 1500\sin(30°)$. Answer A incorrectly uses $\sin(40°)$ and the wrong coefficient. Answer B uses $\sin(30°)$ correctly but miscalculates the coefficient as 3000 instead of 1500. Answer C uses the wrong angle ($70°$) and wrong coefficient, likely confusing one of the individual ship directions with the angle between them. Remember: when finding the angle between two directions measured from the same reference line, subtract the smaller angle from the larger one. Always double-check your coefficient calculation in the area formula.

Question 3

1. $2am\sin(\beta)$ square units
2. $\frac{1}{2}am\sin(\beta)$ square units
3. $am\sin(\beta)$ square units (correct answer)
4. $am\sin(2\beta)$ square units

Explanation: When you encounter problems involving medians and areas in triangles, remember that a median connects a vertex to the midpoint of the opposite side and always divides the triangle into two equal areas. Let's call the midpoint of side $YZ$ point $M$. The median $XM$ has length $m$ and makes angle $\beta$ with side $XY$. To find the area of triangle $XYZ$, we can calculate the area of triangle $XYM$ and double it (since the median creates two equal areas). In triangle $XYM$, we know two sides: $XY = a$ and $XM = m$, plus the included angle $\beta$. Using the formula for the area of a triangle with two sides and an included angle: Area of $XYM = \frac{1}{2} \cdot XY \cdot XM \cdot \sin(\beta) = \frac{1}{2}am\sin(\beta)$. Since the median divides triangle $XYZ$ into two equal triangles, the total area is: $2 \times \frac{1}{2}am\sin(\beta) = am\sin(\beta)$. Looking at the wrong answers: Choice A gives $2am\sin(\beta)$, which incorrectly doubles the entire calculation instead of just accounting for the two equal parts. Choice B gives $\frac{1}{2}am\sin(\beta)$, which is only the area of one of the two triangles formed by the median. Choice D gives $am\sin(2\beta)$, which incorrectly uses the double-angle formula where it doesn't apply. Study tip: Remember that medians always create two triangles of equal area, so calculate one triangle's area using the two-sides-and-included-angle formula, then double it.

Question 4

1. $150\sqrt{2}\text{ m}^2$
2. $75\sqrt{2}\text{ m}^2$ (correct answer)
3. $150\text{ m}^2$
4. $75\text{ m}^2$

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 15 m, b = 20 m, and included angle C = 45°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(15)(20)·sin(45°) = (1/2)(300)·(√2/2) = 150·(√2/2) = 75√2 square meters. Choice B is correct because it properly applies the formula with the included angle, using sin(45°) = √2/2. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 5

1. $72\text{ cm}^2$
2. $36\text{ cm}^2$ (correct answer)
3. $18\text{ cm}^2$
4. $36\sqrt{3}\text{ cm}^2$

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 9 cm, b = 16 cm, and included angle C = 30°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(9)(16)·sin(30°) = (1/2)(144)·(1/2) = 72·(1/2) = 36 square centimeters. Choice B is correct because it properly applies the formula with the included angle, using sin(30°) = 1/2. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 6

1. $150\sqrt{2}\text{ m}^2$
2. $75\sqrt{2}\text{ m}^2$ (correct answer)
3. $300\text{ m}^2$
4. $150\text{ m}^2$

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 15 m, b = 20 m, and included angle C = 45°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(15)(20)·sin(45°) = (1/2)(300)·(√2/2) = 150·(√2/2) = 75√2 square meters. Choice B is correct because it properly applies the formula with the included angle, using sin(45°) = √2/2. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 7

1. $30^\circ$
2. $45^\circ$
3. $60^\circ$
4. $90^\circ$ (correct answer)

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and when area is maximized. For a triangle with two fixed sides a and b, the area is maximized when the included angle C = 90° because sin(C) reaches its maximum value of 1 at 90°, making the maximum area equal to (1/2)ab. Since sin(C) reaches its maximum value of 1 when C = 90°, the area A = (1/2)ab·sin(C) is maximized when C = 90°, giving maximum area = (1/2)ab = (1/2)(5)(7) = 35/2, which corresponds to a right triangle. Choice D is correct because it correctly identifies the angle that maximizes area. Choice C incorrectly claims the maximum area occurs at 60°, but sin(C) is maximized at C = 90°, not 60°. Maximum area insight: for any two fixed sides, the triangle has maximum area when they meet at a right angle (90°), giving A_max = (1/2)ab, because sin reaches its maximum value of 1 at 90°. The sine formula A = (1/2)ab·sin(C) works for all triangles—acute, right, or obtuse—making it more versatile than the standard base-height formula when you don't know the height directly but do know an angle.

Question 8

1. $\dfrac{143\sqrt{2}}{4}\text{ square units}$ (correct answer)
2. $\dfrac{143\sqrt{2}}{2}\text{ square units}$
3. $\dfrac{143}{4}\text{ square units}$
4. $\dfrac{143}{2}\text{ square units}$

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 11, b = 13, and included angle C = 45°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(11)(13)·sin(45°) = (1/2)(143)·(√2/2) = 143/2·(√2/2) = 143√2/4 square units. Choice A is correct because it properly applies the formula with the included angle, using sin(45°) = √2/2. Choice B forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. The sine formula A = (1/2)ab·sin(C) works for all triangles—acute, right, or obtuse—making it more versatile than the standard base-height formula when you don't know the height directly but do know an angle.

Question 9

1. $\frac{20}{15} = \frac{4}{3}$
2. $\frac{15^2}{20^2} = \frac{9}{16}$
3. $\frac{\sin(60°)}{\sin(60°)} = 1$
4. $\frac{15}{20} = \frac{3}{4}$ (correct answer)

Explanation: When you encounter a triangle with an angle bisector, the key insight is recognizing that the angle bisector theorem applies. This theorem states that an angle bisector divides the opposite side in the same ratio as the adjacent sides. Since $QS$ bisects $\angle PQR$, point $S$ divides side $PR$ such that $\frac{PS}{SR} = \frac{PQ}{QR} = \frac{15}{20} = \frac{3}{4}$. This means that triangles $PQS$ and $QRS$ share the same height from vertex $Q$ to side $PR$, so their areas are proportional to their bases $PS$ and $SR$. Therefore, $\frac{\text{Area of } \triangle PQS}{\text{Area of } \triangle QRS} = \frac{PS}{SR} = \frac{15}{20} = \frac{3}{4}$, making answer choice D correct. Answer choice A reverses the ratio, giving $\frac{QR}{PQ}$ instead of $\frac{PQ}{QR}$. This is a common error when applying the angle bisector theorem. Answer choice B squares the side lengths, creating the ratio $\frac{PQ^2}{QR^2}$. This might tempt students thinking about area formulas, but the angle bisector theorem uses linear ratios, not squared ratios. Answer choice C equals 1, suggesting the triangles have equal areas. This reflects a misconception that the angle bisector creates two congruent triangles, which only happens in isosceles triangles where the two sides forming the bisected angle are equal. Remember: When an angle bisector divides a triangle, the ratio of the resulting triangle areas equals the ratio of the two sides that form the bisected angle.

Question 10

1. $\frac{12}{5}$ units
2. $\frac{24}{5}$ units (correct answer)
3. $\frac{48}{10}$ units
4. $\frac{36}{10}$ units

Explanation: First, check if this is a right triangle: $8^2 + 6^2 = 64 + 36 = 100 = 10^2$, so it's a right triangle with the right angle at $E$. Using the trigonometric formula: Area $= \frac{1}{2} \times 8 \times 6 \times \sin(90°) = 24$. Using base and height: Area $= \frac{1}{2} \times DF \times EG = \frac{1}{2} \times 10 \times EG$. Setting equal: $24 = 5 \times EG$, so $EG = \frac{24}{5}$. Choice A gives half the correct value, choice C simplifies to the same as B, and choice D gives an incorrect calculation.

Question 11

1. $\frac{1}{2}(p + q)h = \frac{1}{2}\sqrt{p^2 + h^2}\sqrt{q^2 + h^2}\sin\theta$ (correct answer)
2. $\frac{1}{2}(p + q)h = \frac{1}{2}pq\sin\theta$
3. $\frac{1}{2}(p + q)h = \frac{1}{2}(p^2 + q^2)\sin\theta$
4. $\frac{1}{2}(p + q)h = \frac{1}{2}\sqrt{(p + q)^2 + h^2}\sin\theta$

Explanation: The area using base and height is $\frac{1}{2}(p + q)h$. Using the trigonometric formula, we need the lengths of the two sides forming angle $\theta$: $AC = \sqrt{p^2 + h^2}$ and $BC = \sqrt{q^2 + h^2}$ by the Pythagorean theorem. Therefore, $A = \frac{1}{2} \cdot AC \cdot BC \cdot \sin\theta = \frac{1}{2}\sqrt{p^2 + h^2}\sqrt{q^2 + h^2}\sin\theta$. Choice B uses incorrect side lengths, choice C uses $p^2 + q^2$ instead of the hypotenuses, and choice D incorrectly combines $p + q$ with $h$.

Question 12

1. $70\sqrt{3}\text{ cm}^2$
2. $35\sqrt{3}\text{ cm}^2$ (correct answer)
3. $70\text{ cm}^2$
4. $140\text{ cm}^2$

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 14 cm, b = 10 cm, and included angle C = 60°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(14)(10)·sin(60°) = (1/2)(140)·(√3/2) = 70·(√3/2) = 35√3 square units. Choice B is correct because it properly applies the formula with the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 13

1. $72\text{ ft}^2$
2. $36\text{ ft}^2$ (correct answer)
3. $144\text{ ft}^2$
4. $18\text{ ft}^2$

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 16, b = 9, and included angle C = 30°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(16)(9)·sin(30°) = (1/2)(144)·(0.5) = 72·0.5 = 36 square feet. Choice B is correct because it properly applies the formula with the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 14

1. $s^2\sin(2\theta)$ square units
2. $2s^2\sin(\theta)$ square units
3. $s^2\sin(\theta)$ square units (correct answer)
4. $\frac{1}{2}s^2\sin(\theta)$ square units

Explanation: When finding the area of a rhombus using triangles, you need to divide the rhombus into triangular pieces and apply the triangle area formula $\text{Area} = \frac{1}{2}ab\sin C$. The most efficient approach is to draw a diagonal from one vertex to the opposite vertex, creating two congruent triangles. Each triangle has two sides of length $s$ (the rhombus's sides) and the included angle $\theta$ between them. Using the triangle area formula, each triangle has area $\frac{1}{2}s \cdot s \cdot \sin(\theta) = \frac{1}{2}s^2\sin(\theta)$. Since the rhombus consists of two such triangles, the total area is $2 \times \frac{1}{2}s^2\sin(\theta) = s^2\sin(\theta)$, confirming answer C. Let's examine why the other choices are incorrect: Answer A gives $s^2\sin(2\theta)$, which results from incorrectly using the double-angle identity. Some students mistakenly think they need to account for "both angles" in the rhombus, but the triangle area formula already uses the single included angle. Answer B gives $2s^2\sin(\theta)$, which happens when you correctly find the area of both triangles but forget to apply the $\frac{1}{2}$ factor in the triangle area formula. Answer D gives $\frac{1}{2}s^2\sin(\theta)$, which is the area of just one triangle—you've forgotten that the rhombus contains two triangles. Strategy tip: When using the triangle area formula for polygons, always count your triangular pieces carefully and remember that $\frac{1}{2}ab\sin C$ applies to each individual triangle, not the entire polygon.

Question 15

1. $A=\tfrac12(9)(14)\sin(45^\circ)$ (correct answer)
2. $A=(9)(14)\sin(45^\circ)$
3. $A=\tfrac12(9)(14)\cos(45^\circ)$
4. $A=\tfrac12(9+14)\sin(45^\circ)$

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. With angle C = 45°, we use sin(45°) = √2/2, so A = (1/2)(9)(14)·sin(45°) = (1/2)(126)·(√2/2) = 63·(√2/2) = (63√2)/2. Choice A is correct because it properly applies the formula with the included angle. Choice C uses cosine instead of sine, computing A = (1/2)ab·cos(C), but the area formula requires the sine of the included angle. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 16

1. It adjusts the product $ab$ to account for how perpendicular the sides are by giving the height ratio: $h=a\sin(C)$ (or $h=b\sin(C)$). (correct answer)
2. It gives the perimeter ratio of the triangle compared to a right triangle with legs $a$ and $b$.
3. It converts the side lengths from linear units to square units.
4. It represents the length of the third side divided by the base.

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how it derives from the base-height formula. The formula A = (1/2)ab·sin(C) derives from the traditional base-height formula A = (1/2)bh by recognizing that if we take side b as the base, the height h equals a·sin(C), giving A = (1/2)b(a·sin(C)) = (1/2)ab·sin(C). Starting with A = (1/2)(base)(height), we let side b be the base and draw an altitude of height h from the opposite vertex. This altitude forms a right triangle where sin(C) = h/a, giving h = a·sin(C). Substituting this into the area formula: A = (1/2)b·h = (1/2)b(a·sin(C)) = (1/2)ab·sin(C). Choice A is correct because it correctly derives using h = a·sin(C). Choice D shows an incorrect derivation step, using tan(C) = h/b so h = b tan(C), but the correct relationship is h = a·sin(C). To remember the formula, think of it as modifying the base-height formula: height h = a·sin(C) when you drop an altitude, so A = (1/2)(base)(height) = (1/2)b(a·sin(C)) = (1/2)ab·sin(C). The sine formula A = (1/2)ab·sin(C) works for all triangles—acute, right, or obtuse—making it more versatile than the standard base-height formula when you don't know the height directly but do know an angle.

Question 17

1. $150\sqrt{2}\text{ m}^2$
2. $75\sqrt{2}\text{ m}^2$ (correct answer)
3. $150\text{ m}^2$
4. $75\text{ m}^2$

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 15, b = 20, and included angle C = 45°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(15)(20)·sin(45°) = (1/2)(300)·(√2/2) = 150·(√2/2) = 75√2 square units. Choice B is correct because it properly applies the formula with the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 18

1. $48\sqrt{3}\text{ in}^2$ (correct answer)
2. $96\sqrt{3}\text{ in}^2$
3. $96\text{ in}^2$
4. $48\text{ in}^2$

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. With angle C = 60°, we use sin(60°) = √3/2, so A = (1/2)(12)(16)·(√3/2) = (96)·(√3/2) = 48√3 square units. Choice A is correct because it properly applies the formula with the included angle. Choice B forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 19

1. $54\text{ square units}$
2. $108\text{ square units}$
3. $54\sqrt{3}\text{ square units}$ (correct answer)
4. $108\sqrt{3}\text{ square units}$

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 18, b = 12, and included angle C = 60°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(18)(12)·sin(60°) = (1/2)(216)·(√3/2) = 108·(√3/2) = 54√3 square units. Choice C is correct because it properly applies the formula with the included angle, using sin(60°) = √3/2. Choice B forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using.

Question 20

1. $A=ab\sin(C)$
2. $A=\tfrac12 ab\cos(C)$
3. $A=\tfrac12 ab\sin(C)$ (correct answer)
4. $A=\tfrac12 (a+b)\sin(C)$

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. The formula A = (1/2)ab·sin(C) derives from the traditional base-height formula A = (1/2)bh by recognizing that if we take side b as the base, the height h equals a·sin(C), giving A = (1/2)b(a·sin(C)) = (1/2)ab·sin(C). Choice C is correct because it properly includes the factor of 1/2 and uses sine of the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. The sine formula A = (1/2)ab·sin(C) works for all triangles—acute, right, or obtuse—making it more versatile than the standard base-height formula when you don't know the height directly but do know an angle.