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$\\theta$."},{"text":"Opposite.","correct":false,"explanation":"The opposite side changes when the chosen acute angle changes. The hypotenuse remains fixed because it is always directly opposite the right angle."},{"text":"Adjacent.","correct":false,"explanation":"The adjacent side depends on which acute angle is labelled $\\theta$. The hypotenuse is fixed because it is always the side opposite the right angle."},{"text":"Perpendicular.","correct":false,"explanation":"Perpendicular describes lines meeting at a right angle rather than a SOHCAHTOA side label. The fixed side is the hypotenuse, opposite the right angle."}]},{"id":"aeede0b2-0b3b-b1a5-2058-edeeb7eaaa37","cardType":"cloze","points":null,"front":"The hypotenuse is always the [ ] side and lies opposite the [ ].","back":"The hypotenuse is always the longest side and lies opposite the right angle.","mcqOptions":[{"text":"The hypotenuse is always the longest side and lies opposite the right angle.","correct":true,"explanation":"The right angle is the triangle’s largest angle, so the side opposite it is the longest. That side is defined as the hypotenuse."},{"text":"The hypotenuse is always the shortest side and lies opposite the right angle.","correct":false,"explanation":"Calling the hypotenuse the shortest side reverses a key property of right-angled triangles. It is always the longest side and lies opposite the right angle."},{"text":"The hypotenuse is always the longest side and lies opposite the chosen angle.","correct":false,"explanation":"The side opposite the chosen angle is labelled opposite, not necessarily hypotenuse. The hypotenuse is the longest side and always lies opposite the right angle."},{"text":"The hypotenuse is always the adjacent side and lies opposite the acute angle.","correct":false,"explanation":"Adjacent is a position relative to a chosen acute angle, not a fixed length property. The hypotenuse is the longest side and lies opposite the right angle."}]},{"id":"2eab4463-9cce-b8d6-4561-220ab847be29","cardType":"basic","points":1,"front":"For a chosen angle $\\theta$, which label goes on the side directly across from $\\theta$?","back":"Opposite.","mcqOptions":[{"text":"Opposite.","correct":true,"explanation":"The side directly across the triangle from the chosen angle $\\theta$ is called the opposite side. Its label changes if a different acute angle is chosen."},{"text":"Adjacent.","correct":false,"explanation":"The adjacent side touches $\\theta$ and the right angle, rather than lying directly across from $\\theta$. The side across from $\\theta$ is labelled opposite."},{"text":"Hypotenuse.","correct":false,"explanation":"The hypotenuse lies opposite the right angle, not necessarily opposite the chosen angle $\\theta$. The side directly across from $\\theta$ is labelled opposite."},{"text":"Perpendicular.","correct":false,"explanation":"Perpendicular describes two lines meeting at $90^\\circ$ and is not the required side label. The side directly across from $\\theta$ is called opposite."}]},{"id":"4cdc5a78-2a17-1df6-1b85-16f9740b6ef0","cardType":"basic","points":1,"front":"For a chosen angle $\\theta$, which label goes on the side connecting $\\theta$ to the right angle?","back":"Adjacent.","mcqOptions":[{"text":"Adjacent.","correct":true,"explanation":"The adjacent side is the non-hypotenuse side next to $\\theta$. It runs from the chosen angle to the right angle of the triangle."},{"text":"Hypotenuse.","correct":false,"explanation":"The hypotenuse touches $\\theta$, but it does not connect $\\theta$ to the right angle as a triangle side. That side is labelled adjacent."},{"text":"Opposite.","correct":false,"explanation":"The opposite side lies directly across from $\\theta$, so it does not touch that angle. The side connecting $\\theta$ to the right angle is adjacent."},{"text":"Perpendicular.","correct":false,"explanation":"Perpendicular describes the relationship between the two shorter sides at the right angle. The side joining $\\theta$ to the right angle is labelled adjacent."}]},{"id":"ccdcd6c1-c439-7149-2b75-4be260e78621","cardType":"cloze","points":null,"front":"The [ ] and [ ] labels depend on the chosen angle $\\theta$.","back":"The opposite and adjacent labels depend on the chosen angle $\\theta$.","mcqOptions":[{"text":"The opposite and adjacent labels depend on the chosen angle $\\theta$.","correct":true,"explanation":"Choosing the other acute angle swaps which shorter side is opposite and which is adjacent. The hypotenuse remains fixed because it stays opposite the right angle."},{"text":"The opposite and hypotenuse labels depend on the chosen angle $\\theta$.","correct":false,"explanation":"The opposite label changes with the chosen acute angle, but the hypotenuse is always opposite the right angle. Opposite and adjacent are the two angle-dependent labels."},{"text":"The hypotenuse and adjacent labels depend on the chosen angle $\\theta$.","correct":false,"explanation":"The adjacent label changes with $\\theta$, but the hypotenuse does not. The two labels that depend on the chosen angle are opposite and adjacent."},{"text":"The hypotenuse and right angle labels depend on the chosen angle $\\theta$.","correct":false,"explanation":"Neither the hypotenuse nor the right angle changes when another acute angle is chosen. The labels that swap according to $\\theta$ are opposite and adjacent."}]},{"id":"d84d65b9-a5f7-66a8-2b4d-f189a0131856","cardType":"cloze","points":null,"front":"SOH: sin uses [ ] over [ ].","back":"SOH: sin uses Opposite over Hypotenuse.","mcqOptions":[{"text":"SOH: sin uses Opposite over Hypotenuse.","correct":true,"explanation":"SOH links sine with opposite and hypotenuse: $\\sin\\theta$ equals the length of the opposite side divided by the length of the hypotenuse."},{"text":"SOH: sin uses Adjacent over Hypotenuse.","correct":false,"explanation":"Adjacent over hypotenuse is the cosine ratio, represented by CAH. SOH states that sine equals the opposite side divided by the hypotenuse."},{"text":"SOH: sin uses Opposite over Adjacent.","correct":false,"explanation":"Opposite over adjacent is the tangent ratio, represented by TOA. For sine, SOH gives opposite divided by hypotenuse."},{"text":"SOH: sin uses Hypotenuse over Opposite.","correct":false,"explanation":"This reverses the sine ratio, producing its reciprocal rather than sine. SOH places opposite in the numerator and hypotenuse in the denominator."}]},{"id":"510a663a-f5a7-5cba-7bd8-dd65ad6ce695","cardType":"cloze","points":null,"front":"CAH: cos uses [ ] over [ ].","back":"CAH: cos uses Adjacent over Hypotenuse.","mcqOptions":[{"text":"CAH: cos uses Adjacent over Hypotenuse.","correct":true,"explanation":"CAH links cosine with adjacent and hypotenuse: $\\cos\\theta$ equals the length of the adjacent side divided by the length of the hypotenuse."},{"text":"CAH: cos uses Opposite over Hypotenuse.","correct":false,"explanation":"Opposite over hypotenuse is the sine ratio, represented by SOH. CAH states that cosine equals adjacent divided by hypotenuse."},{"text":"CAH: cos uses Adjacent over Opposite.","correct":false,"explanation":"Adjacent over opposite is the reciprocal of the tangent ratio, not cosine. CAH gives adjacent divided by hypotenuse for $\\cos\\theta$."},{"text":"CAH: cos uses Hypotenuse over Adjacent.","correct":false,"explanation":"This reverses the cosine ratio, giving its reciprocal rather than cosine. CAH places adjacent in the numerator and hypotenuse in the denominator."}]},{"id":"626109b9-b80d-8167-2e3f-47c5cbf97587","cardType":"cloze","points":null,"front":"TOA: tan uses [ ] over [ ].","back":"TOA: tan uses Opposite over Adjacent.","mcqOptions":[{"text":"TOA: tan uses Opposite over Adjacent.","correct":true,"explanation":"TOA links tangent with opposite and adjacent: $\\tan\\theta$ equals the length of the opposite side divided by the length of the adjacent side."},{"text":"TOA: tan uses Adjacent over Hypotenuse.","correct":false,"explanation":"Adjacent over hypotenuse is the cosine ratio, represented by CAH. TOA states that tangent equals opposite divided by adjacent."},{"text":"TOA: tan uses Opposite over Hypotenuse.","correct":false,"explanation":"Opposite over hypotenuse is the sine ratio, represented by SOH. For tangent, TOA gives opposite divided by adjacent."},{"text":"TOA: tan uses Adjacent over Opposite.","correct":false,"explanation":"This reverses the tangent ratio and produces its reciprocal. TOA places opposite in the numerator and adjacent in the denominator."}]},{"id":"78e0cfc5-42b6-a26a-6477-e1f8c8358f11","cardType":"cloze","points":null,"front":"With a formula triangle, cover [ ] to get the [ ].","back":"With a formula triangle, cover what you want to find to get the rearrangement.","mcqOptions":[{"text":"With a formula triangle, cover what you want to find to get the rearrangement.","correct":true,"explanation":"In a formula triangle, the covered part is the unknown you are trying to make the subject. The uncovered parts then show the correct rearrangement."},{"text":"With a formula triangle, cover what you already know to get the rearrangement.","correct":false,"explanation":"Covering known values is a common mix-up, but the triangle is used by covering the unknown. You cover what you want to find to reveal the rearrangement."},{"text":"With a formula triangle, cover what you want to find to get the substitution.","correct":false,"explanation":"The triangle does not substitute numbers for you; it shows how the formula should be rearranged. After covering the unknown, you then substitute values into that rearrangement."},{"text":"With a formula triangle, cover the largest number to get the calculation.","correct":false,"explanation":"Choosing the largest number is not part of using a formula triangle. You cover the quantity you want to find, and the uncovered parts give the rearrangement."}]},{"id":"de94867d-51fa-2c39-454b-c71792c8852c","cardType":"basic","points":2,"front":"For a 35° angle with hypotenuse 20 cm and opposite side $x$, what equation starts SOH?","back":"Use SOH: $\\sin(35^\\circ)=\\frac{x}{20}$.","mcqOptions":[{"text":"Use SOH: $\\sin(35^\\circ)=\\frac{x}{20}$.","correct":true,"explanation":"SOH means sine equals opposite over hypotenuse. Here the opposite side is $x$ and the hypotenuse is 20, so $\\sin(35^\\circ)=\\frac{x}{20}$."},{"text":"Use CAH: $\\cos(35^\\circ)=\\frac{x}{20}$.","correct":false,"explanation":"Cosine uses adjacent over hypotenuse, not opposite over hypotenuse. Since $x$ is opposite the angle and 20 is the hypotenuse, the correct ratio is $\\sin(35^\\circ)=\\frac{x}{20}$."},{"text":"Use SOH: $\\sin(35^\\circ)=\\frac{20}{x}$.","correct":false,"explanation":"This reverses the fraction in SOH. Sine is opposite over hypotenuse, so the opposite side $x$ goes on top and the hypotenuse 20 goes on the bottom."},{"text":"Use TOA: $\\tan(35^\\circ)=\\frac{x}{20}$.","correct":false,"explanation":"Tangent uses opposite over adjacent, but 20 is the hypotenuse, not the adjacent side. With opposite and hypotenuse, the correct choice is SOH."}]},{"id":"01efe111-d0e6-4671-dd10-ad042f23538d","cardType":"basic","points":2,"front":"From $\\sin(35^\\circ)=\\frac{x}{20}$, how do you isolate $x$?","back":"Multiply by 20: $x=20\\sin(35^\\circ)$.","mcqOptions":[{"text":"Multiply by 20: $x=20\\sin(35^\\circ)$.","correct":true,"explanation":"Since $x$ is divided by 20, multiplying both sides by 20 cancels the denominator. This leaves $x=20\\sin(35^\\circ)$."},{"text":"Divide by 20: $x=\\frac{\\sin(35^\\circ)}{20}$.","correct":false,"explanation":"Dividing by 20 makes the denominator problem worse instead of cancelling it. Because $x$ is already over 20, you must multiply by 20 to isolate $x$."},{"text":"Divide by $\\sin(35^\\circ)$: $x=\\frac{20}{\\sin(35^\\circ)}$.","correct":false,"explanation":"This treats the equation as if 20 were on top and $x$ were connected by multiplication. From $\\sin(35^\\circ)=\\frac{x}{20}$, multiplying by 20 gives $x=20\\sin(35^\\circ)$."},{"text":"Subtract 20: $x=\\sin(35^\\circ)-20$.","correct":false,"explanation":"The 20 is a divisor, not a term being added. To undo division by 20, use multiplication, giving $x=20\\sin(35^\\circ)$."}]},{"id":"2cd54ff5-6f04-1ab9-5975-3d0530e0c235","cardType":"basic","points":2,"front":"What is $20\\sin(35^\\circ)$ to 2 d.p.?","back":"11.47 cm.","mcqOptions":[{"text":"11.47 cm.","correct":true,"explanation":"Calculating $20\\sin(35^\\circ)$ gives approximately 11.4715. Rounded to 2 decimal places, this is 11.47 cm."},{"text":"16.38 cm.","correct":false,"explanation":"This comes from using cosine instead of sine: $20\\cos(35^\\circ)\\approx16.38$. The expression given is $20\\sin(35^\\circ)$, which equals 11.47 cm."},{"text":"14.00 cm.","correct":false,"explanation":"This is the result of using tangent instead of sine: $20\\tan(35^\\circ)\\approx14.00$. The required calculation is $20\\sin(35^\\circ)$, which is 11.47 cm."},{"text":"28.57 cm.","correct":false,"explanation":"This comes from dividing by sine, $20\\div\\sin(35^\\circ)$, instead of multiplying by sine. The expression $20\\sin(35^\\circ)$ gives 11.47 cm."}]},{"id":"5bdeb7c3-27d7-8d82-ecfd-51d0f02fff3a","cardType":"basic","points":2,"front":"For a 42° angle with opposite side 11 cm and adjacent side $x$, what equation starts TOA?","back":"Use TOA: $\\tan(42^\\circ)=\\frac{11}{x}$.","mcqOptions":[{"text":"Use TOA: $\\tan(42^\\circ)=\\frac{11}{x}$.","correct":true,"explanation":"TOA means tangent equals opposite over adjacent. The opposite side is 11 and the adjacent side is $x$, so $\\tan(42^\\circ)=\\frac{11}{x}$."},{"text":"Use TOA: $\\tan(42^\\circ)=\\frac{x}{11}$.","correct":false,"explanation":"This reverses opposite and adjacent. Tangent is opposite over adjacent, so the opposite side 11 must be on top and the adjacent side $x$ on the bottom."},{"text":"Use SOH: $\\sin(42^\\circ)=\\frac{11}{x}$.","correct":false,"explanation":"Sine uses opposite over hypotenuse, but the sides given are opposite and adjacent. With opposite and adjacent, the correct trigonometric ratio is tangent."},{"text":"Use CAH: $\\cos(42^\\circ)=\\frac{11}{x}$.","correct":false,"explanation":"Cosine uses adjacent over hypotenuse, so it does not match the two sides given. Opposite and adjacent belong to TOA, giving $\\tan(42^\\circ)=\\frac{11}{x}$."}]},{"id":"c31cebcd-b52b-c7fa-feb0-1e524a104711","cardType":"basic","points":2,"front":"In $\\tan(42^\\circ)=\\frac{11}{x}$, what first step moves $x$ off the bottom?","back":"Multiply by $x$: $x\\tan(42^\\circ)=11$.","mcqOptions":[{"text":"Multiply by $x$: $x\\tan(42^\\circ)=11$.","correct":true,"explanation":"Because $x$ is in the denominator, multiplying both sides by $x$ cancels the denominator. This moves $x$ to the left as $x\\tan(42^\\circ)=11$."},{"text":"Divide by $x$: $\\frac{\\tan(42^\\circ)}{x}=11$.","correct":false,"explanation":"Dividing by $x$ does not remove $x$ from the denominator; it introduces another division by $x$. To move $x$ off the bottom, multiply both sides by $x$."},{"text":"Multiply by $11$: $11\\tan(42^\\circ)=\\frac{121}{x}$.","correct":false,"explanation":"Multiplying by 11 changes the numerator but leaves $x$ in the denominator. The step that cancels the denominator is multiplying both sides by $x$."},{"text":"Divide by $\\tan(42^\\circ)$: $1=\\frac{11}{x\\tan(42^\\circ)}$.","correct":false,"explanation":"Dividing by tangent isolates neither $x$ nor removes it from the denominator. First multiply by $x$ to get $x\\tan(42^\\circ)=11$."}]},{"id":"db02ead7-cbfa-bad3-5e19-848d4bd47f79","cardType":"basic","points":2,"front":"Given $\\tan(42^\\circ)=\\frac{11}{x}$, what is $x$ to 2 d.p.?","back":"12.22 cm.","mcqOptions":[{"text":"12.22 cm.","correct":true,"explanation":"From $\\tan(42^\\circ)=\\frac{11}{x}$, multiply by $x$ and then divide by $\\tan(42^\\circ)$. So $x=\\frac{11}{\\tan(42^\\circ)}\\approx12.22$ cm."},{"text":"9.90 cm.","correct":false,"explanation":"This comes from multiplying $11\\tan(42^\\circ)$ instead of dividing by tangent. Since $x$ is in the denominator, the correct rearrangement is $x=\\frac{11}{\\tan(42^\\circ)}=12.22$ cm."},{"text":"16.44 cm.","correct":false,"explanation":"This uses sine instead of tangent, giving $11\\div\\sin(42^\\circ)$. The sides are opposite and adjacent, so TOA is needed and $x=\\frac{11}{\\tan(42^\\circ)}=12.22$ cm."},{"text":"14.80 cm.","correct":false,"explanation":"This uses cosine instead of tangent, giving $11\\div\\cos(42^\\circ)$. Opposite and adjacent sides require tangent, so the correct value is 12.22 cm."}]},{"id":"a7ece071-66ef-864f-983e-e6b3d348a394","cardType":"basic","points":2,"front":"With opposite side 5 cm and adjacent side 9 cm, what equation finds angle $\\theta$?","back":"Use TOA: $\\tan(\\theta)=\\frac{5}{9}$.","mcqOptions":[{"text":"Use TOA: $\\tan(\\theta)=\\frac{5}{9}$.","correct":true,"explanation":"TOA means tangent equals opposite over adjacent. With opposite side 5 cm and adjacent side 9 cm, $\\tan(\\theta)=\\frac{5}{9}$."},{"text":"Use TOA: $\\tan(\\theta)=\\frac{9}{5}$.","correct":false,"explanation":"This reverses the tangent ratio. Tangent is opposite over adjacent, so the opposite side 5 goes on top and the adjacent side 9 goes underneath."},{"text":"Use SOH: $\\sin(\\theta)=\\frac{5}{9}$.","correct":false,"explanation":"Sine uses opposite over hypotenuse, but 9 cm is the adjacent side, not the hypotenuse. Opposite and adjacent call for tangent."},{"text":"Use CAH: $\\cos(\\theta)=\\frac{5}{9}$.","correct":false,"explanation":"Cosine uses adjacent over hypotenuse, which does not match these two sides. The given opposite and adjacent sides should be used with tangent: $\\tan(\\theta)=\\frac{5}{9}$."}]},{"id":"bfbb8397-4e54-d32c-1a05-8083adb99de1","cardType":"basic","points":1,"front":"Given $\\tan(\\theta)=\\frac{5}{9}$, what expression gives $\\theta$?","back":"Inverse tan: $\\theta=\\tan^{-1}\\left(\\frac{5}{9}\\right)$.","mcqOptions":[{"text":"Inverse tan: $\\theta=\\tan^{-1}\\left(\\frac{5}{9}\\right)$.","correct":true,"explanation":"Inverse tangent reverses the tangent function. Applying it to both sides isolates the angle, giving $\\theta=\\tan^{-1}\\left(\\frac{5}{9}\\right)$."},{"text":"Inverse sin: $\\theta=\\sin^{-1}\\left(\\frac{5}{9}\\right)$.","correct":false,"explanation":"Inverse sine would apply if $\\sin(\\theta)=\\frac{5}{9}$. Because the given ratio is tangent, the correct expression is $\\theta=\\tan^{-1}\\left(\\frac{5}{9}\\right)$."},{"text":"Inverse cos: $\\theta=\\cos^{-1}\\left(\\frac{5}{9}\\right)$.","correct":false,"explanation":"Inverse cosine reverses cosine, not tangent. Since $\\tan(\\theta)=\\frac{5}{9}$, applying inverse tangent gives $\\theta=\\tan^{-1}\\left(\\frac{5}{9}\\right)$."},{"text":"Reciprocal tan: $\\theta=\\frac{1}{\\tan\\left(\\frac{5}{9}\\right)}$.","correct":false,"explanation":"The superscript $-1$ on $\\tan^{-1}$ means inverse function here, not reciprocal. The angle is found using $\\theta=\\tan^{-1}\\left(\\frac{5}{9}\\right)$."}]},{"id":"45ce4519-4298-222f-90e9-2fa92277423a","cardType":"basic","points":2,"front":"What is $\\tan^{-1}\\left(\\frac{5}{9}\\right)$ to 1 d.p.?","back":"29.1°.","mcqOptions":[{"text":"29.1°.","correct":true,"explanation":"Evaluating the inverse tangent gives approximately $29.0546°$. The hundredths digit is 5, so rounding to one decimal place gives $29.1°$."},{"text":"29.0°.","correct":false,"explanation":"This truncates $29.0546°$ instead of rounding it. Since the hundredths digit is 5, the tenths digit must round up, giving $29.1°$."},{"text":"60.9°.","correct":false,"explanation":"This is the complementary angle, found by subtracting the required angle from $90°$. Directly evaluating $\\tan^{-1}\\left(\\frac{5}{9}\\right)$ gives $29.1°$ to one decimal place."},{"text":"0.6°.","correct":false,"explanation":"This treats the ratio $\\frac{5}{9}\\approx0.6$ as though it were the angle. Applying inverse tangent to the ratio gives $29.1°$ to one decimal place."}]},{"id":"9712fe4f-9c54-095e-e862-85a23023a221","cardType":"basic","points":1,"front":"Given $\\cos(\\theta)=0.5$, which inverse expression finds $\\theta$?","back":"Inverse cos: $\\theta=\\cos^{-1}(0.5)$.","mcqOptions":[{"text":"Inverse cos: $\\theta=\\cos^{-1}(0.5)$.","correct":true,"explanation":"Inverse cosine reverses the cosine function. Applying it to $\\cos(\\theta)=0.5$ isolates the angle, so $\\theta=\\cos^{-1}(0.5)$."},{"text":"Inverse sin: $\\theta=\\sin^{-1}(0.5)$.","correct":false,"explanation":"Inverse sine would reverse a sine equation, not the given cosine equation. The correct operation is inverse cosine, so $\\theta=\\cos^{-1}(0.5)$."},{"text":"Inverse tan: $\\theta=\\tan^{-1}(0.5)$.","correct":false,"explanation":"Inverse tangent applies when a tangent value is given. Since the equation uses cosine, it must be reversed using $\\theta=\\cos^{-1}(0.5)$."},{"text":"Ordinary cos: $\\theta=\\cos(0.5)$.","correct":false,"explanation":"Applying cosine again does not undo the original cosine function. To isolate the angle, use the inverse function: $\\theta=\\cos^{-1}(0.5)$."}]},{"id":"27e57221-ee1c-5592-2b0b-d1c07fb4e71e","cardType":"basic","points":2,"front":"Why can’t $\\theta$ be found by dividing $\\tan(\\theta)=\\frac{5}{9}$ by “tan”?","back":"

tan is a function, not a number.
Use inverse tan to extract $\\theta$.

","mcqOptions":[{"text":"

tan is a function, not a number.
Use inverse tan to extract $\\theta$.

","correct":true,"explanation":"Tangent acts on an angle to produce a ratio, so it is not an algebraic factor that can be divided away. Inverse tangent reverses the function and isolates $\\theta$."},{"text":"

tan is a number multiplying $\\theta$.
Divide both sides by tan to find $\\theta$.

","correct":false,"explanation":"This mistakes function notation for multiplication. Tangent acts on $\\theta$ rather than multiplying it, so applying inverse tangent is required to isolate the angle."},{"text":"

tan is the reciprocal of an angle.
Use ordinary tan to extract $\\theta$.

","correct":false,"explanation":"Tangent is a trigonometric function, not the reciprocal of an angle. Applying tangent again does not undo it; inverse tangent must be used to find $\\theta$."},{"text":"

tan can be cancelled from the left side.
Multiply by tan to extract $\\theta$.

","correct":false,"explanation":"This treats $\\tan(\\theta)$ like a product with a cancellable factor. Because tangent is a function, the correct inverse operation is inverse tangent, not multiplication."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Component 1"},"heading":"SOHCAHTOA (Trigonometry)","title":"SOHCAHTOA (Trigonometry) Flashcards","topbar":{"backHref":"/gcse/maths/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"efe71b2a-4bbe-4e22-bc7a-0b66ce6a7fb0","walled":"$L39"}],"$L3a","$L3b"]