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MathProblem-Solving and Data AnalysisHard

A store raises the price of a jacket by 20%. Later, it lowers the new price by 20%. The final price is what percent of the original price?

A80%
B96%Correct
C100%The trap
D104%

Check with $100: $100then$120then$96

  1. Turn each change into a multiplier.

    Up 20% means × 1.20. Down 20% means × 0.80.

  2. Apply them in order.

    1.20 × 0.80 = 0.96

  3. Read the answer.

    The final price is 96% of the original. That’s B.

  4. Why not 100%?

    The 20% cut is taken from the higher price, so it removes more than the 20% increase added.

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  1. Reading & Writing · Module 1

    27 questions32 min

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98 questions · 2 hr 14 min of testing, plus the break

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The Daily, plus 36 practice questions every day: 12 easy, 12 medium and 12 hard, each with a full explanation. Full-length Test A, Challenge Test 1 and all 1,023 vocabulary words are free too. No card needed.

Pro is $20 a month or $200 a year: unlimited practice with no daily limit, plus every full-length test and Challenge test. Peak is $29 a month or $280 a year: everything in Pro, plus the full Learn course and Peak Plan. Cancel anytime from your account.

98 questions across four modules, in the digital SAT’s order: Reading & Writing Module 1 and Module 2 (27 questions each), a 10-minute break, then Math Module 1 and Module 2 (22 questions each). On official timing that’s 2 hr 14 min of testing. You get estimated section scores and a total at the end.

Yes. In the standard full-length tests, Module 2 of each section is the harder version if you get 60% or more of Module 1 right, and the easier version if you don’t. Challenge tests use the hardest material throughout.

Official timing, extended 1.5× time, or untimed, on every test, including the free ones. Extended time allows 48 minutes for each Reading & Writing module and about 52 for each Math module. Untimed removes the clock so you can focus on accuracy.

A quick run with three hearts. Each right answer builds your streak: the questions get harder as it grows, and your points multiply (×2 from three in a row, ×3 from six). A miss costs a heart and drops you back to easy. Best is your longest streak.

Every day at midnight. You get a fresh 12 easy, 12 medium and 12 hard.

No. Peakscor is an independent SAT practice tool, not affiliated with the College Board. Scores shown are estimates.

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Right triangles and trigonometry · Lesson 3

Sine, cosine and tangent

Pick one of the acute angles in a right triangle, say angle AA. The hypotenuse is still the side across from the right angle. The opposite side is the leg across from angle AA. The adjacent side is the leg that touches angle AA.

The three trigonometric ratios compare these sides: sin⁡A=oppositehypotenuse\sin A = \frac{\text{opposite}}{\text{hypotenuse}}, cos⁡A=adjacenthypotenuse\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} and tan⁡A=oppositeadjacent\tan A = \frac{\text{opposite}}{\text{adjacent}}. Remember them as SOH CAH TOA. They are not on the reference sheet.

In the 3-4-5 triangle from Lesson 1, take the angle across from the side of length 3. Its sine is 35\frac{3}{5}, its cosine is 45\frac{4}{5} and its tangent is 34\frac{3}{4}. For the other acute angle, the legs trade roles: its sine is 45\frac{4}{5} and its tangent is 43\frac{4}{3}.

A ratio depends only on the angle, not on the size of the triangle. A 6-8-10 triangle has the same angles as the 3-4-5 triangle and the same ratios. So sin⁡A=35\sin A = \frac{3}{5} tells you the sides are in the ratio 3 to 5, not that they are 3 and 5. To find real lengths, multiply by the scale factor.

When you're given one ratio and asked for another, sketch a right triangle with the given sides, find the third side with the Pythagorean theorem, then read off the ratio you need. Because the hypotenuse is the longest side, the sine and cosine of an acute angle are always between 0 and 1. The tangent can be any positive number.

Naming the sides from angle θθadjacent to θopposite θhypotenuse
Naming the sides from angle θ A right triangle with the right angle at the bottom left, marked with a square. Angle θ is at the right end of the horizontal leg. The horizontal leg, which touches θ, is labelled adjacent to θ. The vertical leg, across from θ, is labelled opposite θ. The slanted side is labelled hypotenuse.

Worked example Easy

A right triangle with angle x°x°452853
A right triangle with angle x° A right triangle with the right angle at the bottom left, marked with a square. The angle at the right end of the horizontal leg is marked x°. The horizontal leg is labelled 45, the vertical leg is labelled 28, and the hypotenuse is labelled 53.

In the right triangle shown, what is the value of cos⁡(x∘)\cos(x^\circ)?

  1. 2853\frac{28}{53}
  2. 4553\frac{45}{53}Answer
  3. 2845\frac{28}{45}
  4. 5345\frac{53}{45}

How to solve it

  1. Name the sides from the angle marked x∘x^\circ. The hypotenuse is 53. The leg that touches the angle is 45, so 45 is adjacent. The leg across from it is 28, so 28 is opposite.
  2. Cosine is adjacent over hypotenuse (CAH): cos⁡(x∘)=4553\cos(x^\circ) = \frac{45}{53}.

Why each choice is right or wrong

  • A. Incorrect. 2853\frac{28}{53} is opposite over hypotenuse, which is sin⁡(x∘)\sin(x^\circ). Cosine uses the adjacent side, 45.
  • B. Correct. The side of length 45 touches the angle, so it is adjacent, and the hypotenuse is 53: cos⁡(x∘)=4553\cos(x^\circ) = \frac{45}{53}.
  • C. Incorrect. 2845\frac{28}{45} is opposite over adjacent, which is tan⁡(x∘)\tan(x^\circ).
  • D. Incorrect. 5345\frac{53}{45} puts the hypotenuse on top. Cosine is adjacent over hypotenuse, so the cosine of an acute angle is less than 1.

Worked example Medium

In right triangle PQRPQR, angle QQ is a right angle and tan⁡P=1160\tan P = \frac{11}{60}. What is the value of cos⁡P\cos P?

  1. 1161\frac{11}{61}
  2. 6061\frac{60}{61}Answer
  3. 6160\frac{61}{60}
  4. 6011\frac{60}{11}

How to solve it

  1. tan⁡P=oppositeadjacent\tan P = \frac{\text{opposite}}{\text{adjacent}}, so sketch a right triangle whose leg across from PP is 11 and whose leg touching PP is 60. The real triangle may be a multiple of this one, but the ratios are the same.
  2. Find the hypotenuse: 112+602=121+3,600=3,721=61\sqrt{11^2 + 60^2} = \sqrt{121 + 3{,}600} = \sqrt{3{,}721} = 61.
  3. cos⁡P=adjacenthypotenuse=6061\cos P = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{60}{61}.

Why each choice is right or wrong

  • A. Incorrect. 1161\frac{11}{61} is opposite over hypotenuse, which is sin⁡P\sin P.
  • B. Correct. With legs 11 and 60, the hypotenuse is 61, and cos⁡P=6061\cos P = \frac{60}{61}.
  • C. Incorrect. 6160\frac{61}{60} puts the hypotenuse over the adjacent side. Cosine is adjacent over hypotenuse.
  • D. Incorrect. This flips the given tangent. 6011\frac{60}{11} is adjacent over opposite, and cosine needs the hypotenuse on the bottom.

Trap Opposite and adjacent belong to one angle

Opposite and adjacent change when you switch angles; the hypotenuse never does. Before you write a ratio, find the angle the question names, then find the leg that touches it and the leg across from it.

Trap Treating a ratio as the side lengths

A ratio such as sin⁡A=35\sin A = \frac{3}{5} fixes the shape, not the size. If the hypotenuse is 20, the side opposite AA is 35⋅20=12\frac{3}{5} \cdot 20 = 12, not 3.

Trap Flipping a ratio

The sine and cosine of an acute angle are less than 1, because the hypotenuse is on the bottom. A choice with the hypotenuse on top can't be a sine or a cosine.

Desmos When Desmos isn’t faster

Desmos can't tell which side is opposite or adjacent; that comes from the figure, and once the sides are named the ratio is one fraction. Doing it by hand is faster.

If the choices are decimals, type the fraction, such as 45/53, and Desmos shows about 0.849. To see whether two fractions are equal, type each one on its own line and compare the decimals.

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Same 98-question format, but every question is drawn from the toughest SAT material: Challenging-tier math throughout and the hardest reading.

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