Math 2 Quiz: Defining Trig Ratios
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Defining Trig RatiosQuestion 1 of 15

A right triangle has one leg of length 232\sqrt{3} and hypotenuse of length 4. If θ\theta is the angle opposite the leg of length 232\sqrt{3}, what is cos⁡θ\cos \theta?

33\frac{\sqrt{3}}{3}
32\frac{\sqrt{3}}{2}
12\frac{1}{2}
233\frac{2\sqrt{3}}{3}
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Math 2 Quiz: Defining Trig Ratios

Practice Defining Trig Ratios in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Defining Trig Ratios, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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Question 1

A right triangle has one leg of length 232\sqrt{3} and hypotenuse of length 4. If θ\theta is the angle opposite the leg of length 232\sqrt{3}, what is cos⁡θ\cos \theta?

  1. 33\frac{\sqrt{3}}{3}
  2. 32\frac{\sqrt{3}}{2}
  3. 12\frac{1}{2} (correct answer)
  4. 233\frac{2\sqrt{3}}{3}
Explanation: When you encounter a right triangle problem with trigonometric functions, start by identifying what information you have and what you need to find. Here, you know one leg (232\sqrt{3}), the hypotenuse (4), and need cos⁡θ\cos \theta where θ\theta is opposite the known leg. First, find the unknown leg using the Pythagorean theorem: a2+b2=c2a^2 + b^2 = c^2. With the known leg as 232\sqrt{3} and hypotenuse as 4: (23)2+b2=42(2\sqrt{3})^2 + b^2 = 4^2 12+b2=1612 + b^2 = 16 b2=4b^2 = 4 b=2b = 2 Now you have all three sides. Since θ\theta is opposite the leg of length 232\sqrt{3}, the adjacent side to θ\theta is the leg of length 2. Using the cosine ratio: cos⁡θ=adjacenthypotenuse=24=12\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2}{4} = \frac{1}{2} Looking at the wrong answers: Choice A (33\frac{\sqrt{3}}{3}) would result from confusing this with sin⁡30°\sin 30° or making calculation errors. Choice B (32\frac{\sqrt{3}}{2}) is cos⁡30°\cos 30°, but this angle is adjacent to the 232\sqrt{3} leg, not opposite it. Choice D (233\frac{2\sqrt{3}}{3}) likely comes from incorrectly using the given leg in the numerator instead of finding the adjacent side. The key strategy here is to always draw and label your triangle completely before applying trigonometric ratios. Make sure you identify which side is adjacent to your angle—it's easy to mix up opposite and adjacent sides.

Question 2

If sin⁡γ=74\sin \gamma = \frac{\sqrt{7}}{4} in a right triangle, what is the exact value of tan⁡γ\tan \gamma?

  1. 74\frac{\sqrt{7}}{4}
  2. 377\frac{3\sqrt{7}}{7}
  3. 73\frac{\sqrt{7}}{3} (correct answer)
  4. 477\frac{4\sqrt{7}}{7}
Explanation: When you encounter a trigonometry problem giving you one trigonometric ratio and asking for another, you need to use the Pythagorean identity and basic trigonometric definitions to find the missing information. Given that sin⁡γ=74\sin \gamma = \frac{\sqrt{7}}{4} in a right triangle, you can use the identity sin⁡2γ+cos⁡2γ=1\sin^2 \gamma + \cos^2 \gamma = 1 to find the cosine. Substituting: (74)2+cos⁡2γ=1\left(\frac{\sqrt{7}}{4}\right)^2 + \cos^2 \gamma = 1, which gives us 716+cos⁡2γ=1\frac{7}{16} + \cos^2 \gamma = 1. Solving for cosine: cos⁡2γ=916\cos^2 \gamma = \frac{9}{16}, so cos⁡γ=34\cos \gamma = \frac{3}{4} (taking the positive value since we're in a right triangle). Now you can find tangent using tan⁡γ=sin⁡γcos⁡γ=7434=74⋅43=73\tan \gamma = \frac{\sin \gamma}{\cos \gamma} = \frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}} = \frac{\sqrt{7}}{4} \cdot \frac{4}{3} = \frac{\sqrt{7}}{3}. Choice A (74\frac{\sqrt{7}}{4}) incorrectly assumes tangent equals sine. Choice B (377\frac{3\sqrt{7}}{7}) appears to come from incorrectly manipulating the fraction 73\frac{\sqrt{7}}{3} by rationalizing when it's not needed. Choice D (477\frac{4\sqrt{7}}{7}) likely results from using the reciprocal relationship incorrectly or computational errors. The correct answer is C: 73\frac{\sqrt{7}}{3}. Study tip: When finding one trig ratio from another, always use the Pythagorean identity first to find the third side, then apply the definition of your target ratio. Don't try shortcuts—work systematically through sine, cosine, then tangent.

Question 3

A ladder leans against a wall forming angle θ\theta with the ground. If the ladder is 15 feet long and reaches 12 feet up the wall, which equation correctly expresses tan⁡θ\tan \theta?

  1. tan⁡θ=1215\tan \theta = \frac{12}{15}
  2. tan⁡θ=129\tan \theta = \frac{12}{9} (correct answer)
  3. tan⁡θ=912\tan \theta = \frac{9}{12}
  4. tan⁡θ=1512\tan \theta = \frac{15}{12}
Explanation: The ladder forms a right triangle with the wall and ground. The ladder (hypotenuse) is 15 feet, the height up the wall (opposite to θ\theta) is 12 feet. Using the Pythagorean theorem, the distance along the ground (adjacent to θ\theta) is 152−122=225−144=81=9\sqrt{15^2 - 12^2} = \sqrt{225 - 144} = \sqrt{81} = 9 feet. Therefore, tan⁡θ=oppositeadjacent=129\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{9}. Choice A gives sin⁡θ\sin \theta, choice C gives cot⁡θ\cot \theta, and choice D is not a standard trigonometric ratio.

Question 4

In a right triangle, the cosine of one acute angle is 725\frac{7}{25}. What is the sine of the other acute angle?

  1. 2425\frac{24}{25}
  2. 725\frac{7}{25} (correct answer)
  3. 1825\frac{18}{25}
  4. 257\frac{25}{7}
Explanation: In a right triangle, the two acute angles are complementary. If cos⁡θ=725\cos \theta = \frac{7}{25}, then sin⁡(90°−θ)=cos⁡θ=725\sin(90° - \theta) = \cos \theta = \frac{7}{25}. Therefore, the sine of the other acute angle is 725\frac{7}{25}.

Question 5

Consider the equation xy=tan⁡θ\frac{x}{y} = \tan \theta where θ\theta is an acute angle in a right triangle. If the hypotenuse of this triangle has length zz, which expression correctly represents cos⁡θ\cos \theta?

  1. yx2+y2\frac{y}{\sqrt{x^2 + y^2}} (correct answer)
  2. xx2+y2\frac{x}{\sqrt{x^2 + y^2}}
  3. x2+y2y\frac{\sqrt{x^2 + y^2}}{y}
  4. yx\frac{y}{x}
Explanation: Since tan⁡θ=xy=oppositeadjacent\tan \theta = \frac{x}{y} = \frac{\text{opposite}}{\text{adjacent}}, we have opposite side = xx and adjacent side = yy. By the Pythagorean theorem, the hypotenuse = x2+y2\sqrt{x^2 + y^2}. Therefore, cos⁡θ=adjacenthypotenuse=yx2+y2\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{y}{\sqrt{x^2 + y^2}}.

Question 6

A student writes: "In any right triangle with acute angle θ\theta, we have tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}." To verify this identity using the definitions of trigonometric ratios, which approach is most direct?

  1. Show that oppositeadjacent×hypotenuseopposite=hypotenuseadjacent\frac{\text{opposite}}{\text{adjacent}} \times \frac{\text{hypotenuse}}{\text{opposite}} = \frac{\text{hypotenuse}}{\text{adjacent}}
  2. Show that adjacenthypotenuse÷oppositehypotenuse=adjacentopposite\frac{\text{adjacent}}{\text{hypotenuse}} \div \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\text{adjacent}}{\text{opposite}}
  3. Show that hypotenuseopposite×hypotenuseadjacent=oppositeadjacent\frac{\text{hypotenuse}}{\text{opposite}} \times \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\text{opposite}}{\text{adjacent}}
  4. Show that oppositehypotenuse÷adjacenthypotenuse=oppositeadjacent\frac{\text{opposite}}{\text{hypotenuse}} \div \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\text{opposite}}{\text{adjacent}} (correct answer)
Explanation: When verifying trigonometric identities, you need to start with the basic definitions and use algebraic manipulation to show equivalence. For a right triangle with acute angle θ\theta, recall that sin⁡θ=oppositehypotenuse\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}, cos⁡θ=adjacenthypotenuse\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}, and tan⁡θ=oppositeadjacent\tan \theta = \frac{\text{opposite}}{\text{adjacent}}. To verify tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}, you need to show that sin⁡θcos⁡θ=oppositeadjacent\frac{\sin \theta}{\cos \theta} = \frac{\text{opposite}}{\text{adjacent}}. This requires dividing one fraction by another: sin⁡θcos⁡θ=opposite/hypotenuseadjacent/hypotenuse\frac{\sin \theta}{\cos \theta} = \frac{\text{opposite}/\text{hypotenuse}}{\text{adjacent}/\text{hypotenuse}}. Choice D correctly represents this division: oppositehypotenuse÷adjacenthypotenuse=oppositeadjacent\frac{\text{opposite}}{\text{hypotenuse}} \div \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\text{opposite}}{\text{adjacent}}. When you divide fractions, you multiply by the reciprocal, giving oppositehypotenuse×hypotenuseadjacent=oppositeadjacent\frac{\text{opposite}}{\text{hypotenuse}} \times \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\text{opposite}}{\text{adjacent}}, which simplifies to tan⁡θ\tan \theta. Choice A multiplies tan⁡θ\tan \theta by csc⁡θ\csc \theta, yielding sec⁡θ\sec \theta, not what we want. Choice B represents cos⁡θsin⁡θ\frac{\cos \theta}{\sin \theta}, which equals cot⁡θ\cot \theta, the reciprocal of tangent. Choice C multiplies two expressions that don't correspond to sine and cosine, producing an incorrect relationship. Key strategy: When verifying trigonometric identities, always substitute the basic ratio definitions first, then use fraction arithmetic carefully. Division of fractions is a common step, so practice converting a/cb/c\frac{a/c}{b/c} to ab\frac{a}{b}.

Question 7

In right triangle ABCABC with right angle at CC, if AB=2ACAB = 2AC, what is the value of sin⁡A\sin A?

  1. 12\frac{1}{2}
  2. 32\frac{\sqrt{3}}{2} (correct answer)
  3. 33\frac{\sqrt{3}}{3}
  4. 233\frac{2\sqrt{3}}{3}
Explanation: Let AC=xAC = x, then AB=2xAB = 2x. Since angle CC is the right angle, ABAB is the hypotenuse. Using the Pythagorean theorem: BC2+AC2=AB2BC^2 + AC^2 = AB^2, so BC2+x2=(2x)2=4x2BC^2 + x^2 = (2x)^2 = 4x^2. Therefore BC2=3x2BC^2 = 3x^2, giving BC=x3BC = x\sqrt{3}. Now sin⁡A=oppositehypotenuse=BCAB=x32x=32\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{x\sqrt{3}}{2x} = \frac{\sqrt{3}}{2}. Choice A would be cos⁡A\cos A, choice C equals 13\frac{1}{\sqrt{3}}, and choice D is not the correct ratio.

Question 8

Which of the following statements about trigonometric ratios in a right triangle is always true?

  1. For any right triangle, sin⁡θ+cos⁡θ=1\sin \theta + \cos \theta = 1 for any acute angle θ\theta
  2. For any acute angle θ\theta, sin⁡θ=cos⁡θ\sin \theta = \cos \theta when θ=45°\theta = 45°
  3. For complementary angles α\alpha and β\beta, tan⁡α=tan⁡β\tan \alpha = \tan \beta
  4. For any acute angle θ\theta, sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 and tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta} (correct answer)
Explanation: When you encounter questions about trigonometric ratios, focus on the fundamental relationships that hold true in all right triangles, not just special cases. The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 is one of the most important relationships in trigonometry. It comes directly from the Pythagorean theorem: if you have a right triangle with hypotenuse of length 1, then (opposite side)² + (adjacent side)² = 1². Since sin⁡θ=oppositehypotenuse\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} and cos⁡θ=adjacenthypotenuse\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}, this gives us the identity. The definition tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta} is also universally true. Answer D states both of these fundamental relationships correctly. Now let's examine why the other options fail. Answer A claims sin⁡θ+cos⁡θ=1\sin \theta + \cos \theta = 1 always, but this is only true for θ=90°\theta = 90° (which isn't acute). For most angles, like 30°30°, we get sin⁡30°+cos⁡30°=12+32≠1\sin 30° + \cos 30° = \frac{1}{2} + \frac{\sqrt{3}}{2} \neq 1. Answer B is actually true—when θ=45°\theta = 45°, both sin⁡45°\sin 45° and cos⁡45°\cos 45° equal 22\frac{\sqrt{2}}{2}—but it's incomplete since it only addresses one specific angle rather than a universal truth. Answer C is false because complementary angles have the property that sin⁡α=cos⁡β\sin \alpha = \cos \beta and cos⁡α=sin⁡β\cos \alpha = \sin \beta, but their tangent values are reciprocals: tan⁡α=1tan⁡β\tan \alpha = \frac{1}{\tan \beta}. Study tip: Memorize the Pythagorean identity and basic trig definitions—they're the foundation for solving most trigonometry problems and appear frequently on standardized tests.

Question 9

In triangle DEFDEF with right angle at EE, if sin⁡D=0.6\sin D = 0.6 and DF=25DF = 25, what is the value of cos⁡F\cos F?

  1. 0.6 (correct answer)
  2. 0.8
  3. 0.75
  4. 1.25
Explanation: Since the right angle is at EE, side DFDF is the hypotenuse. Given sin⁡D=0.6\sin D = 0.6, the side opposite to DD (which is EFEF) has length 25×0.6=1525 \times 0.6 = 15. In a right triangle, complementary angles have the property that sin⁡D=cos⁡F\sin D = \cos F, since the side opposite to DD is the same as the side adjacent to FF. Therefore, cos⁡F=0.6\cos F = 0.6. Choice B would be sin⁡F\sin F, choice C is incorrect, and choice D is greater than 1, which is impossible for a cosine value.

Question 10

If tan⁡β=512\tan \beta = \frac{5}{12} in a right triangle, what is the value of sin⁡βcos⁡β+sin⁡β\frac{\sin \beta}{\cos \beta + \sin \beta}?

  1. 529\frac{5}{29}
  2. 512\frac{5}{12}
  3. 1217\frac{12}{17}
  4. 517\frac{5}{17} (correct answer)
Explanation: When you encounter a trigonometry problem with a given tangent ratio, your first step is to construct or visualize the right triangle to find the other trigonometric ratios. Since tan⁡β=512=oppositeadjacent\tan \beta = \frac{5}{12} = \frac{\text{opposite}}{\text{adjacent}}, you have a right triangle with opposite side 5 and adjacent side 12. To find the hypotenuse, use the Pythagorean theorem: c2=52+122=25+144=169c^2 = 5^2 + 12^2 = 25 + 144 = 169, so c=13c = 13. This gives you sin⁡β=513\sin \beta = \frac{5}{13} and cos⁡β=1213\cos \beta = \frac{12}{13}. Now substitute into the expression: sin⁡βcos⁡β+sin⁡β=5131213+513=5131713=513⋅1317=517\frac{\sin \beta}{\cos \beta + \sin \beta} = \frac{\frac{5}{13}}{\frac{12}{13} + \frac{5}{13}} = \frac{\frac{5}{13}}{\frac{17}{13}} = \frac{5}{13} \cdot \frac{13}{17} = \frac{5}{17} Choice A (529\frac{5}{29}) likely comes from incorrectly adding the numerator and denominator: 5+12+12=295 + 12 + 12 = 29. Choice B (512\frac{5}{12}) is simply the original tangent ratio, suggesting you might have confused the expression with tan⁡β\tan \beta. Choice C (1217\frac{12}{17}) results from mistakenly using cosine in the numerator instead of sine. Remember that when given one trigonometric ratio, you can always find the others by constructing the triangle and using the Pythagorean theorem. Also, be careful with fraction operations—when dividing fractions, multiply by the reciprocal of the denominator.

Question 11

If cos⁡α=45\cos \alpha = \frac{4}{5} in a right triangle, and the side adjacent to angle α\alpha has length 8, what is tan⁡α\tan \alpha?

  1. 34\frac{3}{4} (correct answer)
  2. 43\frac{4}{3}
  3. 35\frac{3}{5}
  4. 53\frac{5}{3}
Explanation: Since cos⁡α=45=adjacenthypotenuse\cos \alpha = \frac{4}{5} = \frac{\text{adjacent}}{\text{hypotenuse}} and the adjacent side is 8, we have 8hypotenuse=45\frac{8}{\text{hypotenuse}} = \frac{4}{5}, so the hypotenuse is 10. Using the Pythagorean theorem, the opposite side is 102−82=36=6\sqrt{10^2 - 8^2} = \sqrt{36} = 6. Therefore, tan⁡α=oppositeadjacent=68=34\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{8} = \frac{3}{4}. Choice B is the reciprocal, choice C is sin⁡α\sin \alpha, and choice D is 1sin⁡α\frac{1}{\sin \alpha}.

Question 12

In right triangle PQRPQR with right angle at QQ, if cos⁡P=35\cos P = \frac{3}{5} and PR=20PR = 20, what is the exact value of sin⁡R+cos⁡R\sin R + \cos R?

  1. 75\frac{7}{5} (correct answer)
  2. 95\frac{9}{5}
  3. 115\frac{11}{5}
  4. 135\frac{13}{5}
Explanation: Since cos⁡P=35=PQPR\cos P = \frac{3}{5} = \frac{PQ}{PR} and PR=20PR = 20, we have PQ=12PQ = 12. Using Pythagorean theorem: QR=16QR = 16. For angle RR: sin⁡R=PQPR=1220=35\sin R = \frac{PQ}{PR} = \frac{12}{20} = \frac{3}{5} and cos⁡R=QRPR=1620=45\cos R = \frac{QR}{PR} = \frac{16}{20} = \frac{4}{5}. Therefore, sin⁡R+cos⁡R=35+45=75\sin R + \cos R = \frac{3}{5} + \frac{4}{5} = \frac{7}{5}.

Question 13

In right triangle ABCABC with right angle at CC, the expression AB⋅sin⁡AAC\frac{AB \cdot \sin A}{AC} simplifies to which of the following?

  1. cos⁡A\cos A
  2. sin⁡A\sin A
  3. tan⁡A\tan A (correct answer)
  4. sec⁡A\sec A
Explanation: When you encounter trigonometric expressions in right triangles, always start by identifying what each trigonometric ratio represents in terms of the triangle's sides. This question tests your ability to simplify expressions using these fundamental relationships. Let's work through AB⋅sin⁡AAC\frac{AB \cdot \sin A}{AC} step by step. In right triangle ABCABC with the right angle at CC, we know that ABAB is the hypotenuse, ACAC is the side adjacent to angle AA, and BCBC is the side opposite to angle AA. The key insight is recognizing that sin⁡A=BCAB\sin A = \frac{BC}{AB} (opposite over hypotenuse). Substituting this into our expression: AB⋅sin⁡AAC=AB⋅BCABAC=BCAC\frac{AB \cdot \sin A}{AC} = \frac{AB \cdot \frac{BC}{AB}}{AC} = \frac{BC}{AC} Since BCBC is opposite to angle AA and ACAC is adjacent to angle AA, we have BCAC=tan⁡A\frac{BC}{AC} = \tan A. Looking at the wrong answers: Choice A (cos⁡A\cos A) equals ACAB\frac{AC}{AB}, which doesn't match our result. Choice B (sin⁡A\sin A) equals BCAB\frac{BC}{AB}, not BCAC\frac{BC}{AC}. Choice D (sec⁡A\sec A) equals ABAC\frac{AB}{AC}, which is the reciprocal of cosine, not our expression. Therefore, the answer is C. Study tip: When simplifying trigonometric expressions, substitute the ratio definitions first, then simplify algebraically. The ABAB terms canceled here, leaving us with a basic tangent ratio. Always double-check by converting back to the fundamental opposite/adjacent/hypotenuse relationships.

Question 14

In right triangle ABCABC with right angle at CC, if sin⁡A=513\sin A = \frac{5}{13} and BC=15BC = 15, what is the value of tan⁡B\tan B?

  1. 512\frac{5}{12}
  2. 125\frac{12}{5} (correct answer)
  3. 1312\frac{13}{12}
  4. 1213\frac{12}{13}
Explanation: Since sin⁡A=513=BCAB\sin A = \frac{5}{13} = \frac{BC}{AB} and BC=15BC = 15, we have 15AB=513\frac{15}{AB} = \frac{5}{13}, so AB=39AB = 39. Using the Pythagorean theorem: AC2+BC2=AB2AC^2 + BC^2 = AB^2, so AC2+152=392AC^2 + 15^2 = 39^2, giving AC=36AC = 36. Therefore, tan⁡B=ACBC=3615=125\tan B = \frac{AC}{BC} = \frac{36}{15} = \frac{12}{5}.

Question 15

In right triangle XYZXYZ with right angle at YY, if tan⁡X=43\tan X = \frac{4}{3} and the perimeter is 84, what is sin⁡Z\sin Z?

  1. 35\frac{3}{5}
  2. 54\frac{5}{4}
  3. 45\frac{4}{5} (correct answer)
  4. 53\frac{5}{3}
Explanation: When you encounter a right triangle problem with trigonometric ratios and perimeter constraints, you need to find the actual side lengths first, then calculate the requested ratio. Given that tan⁡X=43\tan X = \frac{4}{3} in right triangle XYZXYZ with the right angle at YY, this means the ratio of the opposite side to the adjacent side (from angle XX's perspective) is 43\frac{4}{3}. Let's call the sides 4k4k and 3k3k for some scaling factor kk. Using the Pythagorean theorem, the hypotenuse is (4k)2+(3k)2=16k2+9k2=5k\sqrt{(4k)^2 + (3k)^2} = \sqrt{16k^2 + 9k^2} = 5k. Since the perimeter is 84: 3k+4k+5k=843k + 4k + 5k = 84, so 12k=8412k = 84 and k=7k = 7. The actual side lengths are 21, 28, and 35. Now for sin⁡Z\sin Z: this equals the opposite side over the hypotenuse from angle ZZ's perspective. Since ZZ and XX are the two acute angles, the side opposite to ZZ is the same side that's adjacent to XX, which is 21. Therefore, sin⁡Z=2135=35\sin Z = \frac{21}{35} = \frac{3}{5}. Wait - that gives us choice A, but the correct answer is C. Let me reconsider: if the side opposite to XX is 4k=284k = 28 and adjacent is 3k=213k = 21, then the side opposite to ZZ is 28. So sin⁡Z=2835=45\sin Z = \frac{28}{35} = \frac{4}{5}, which is choice C. Choice A (35\frac{3}{5}) would be sin⁡X\sin X, choice B (54\frac{5}{4}) and D (53\frac{5}{3}) are impossible since sine values must be ≤ 1. Study tip: Always sketch the triangle and label sides clearly to avoid mixing up which angle corresponds to which sides.