SAT right triangles and trigonometry: review and practice

• Use the Pythagorean theorem to find a missing side, a diagonal or a perimeter.
• Use the side ratios of 30°–60°–90° and 45°–45°–90° triangles.
• Use sine, cosine and tangent to connect an angle to side lengths.
• Use sin x = cos(90° − x), with angles in degrees or radians.
• Read sine and cosine as the coordinates of a point on the unit circle.

• Giving a side length when the question asks for perimeter. If a rectangle's diagonal is 13 and one side is 5, the other side is 12 and the perimeter is 34.
• Adding sides instead of squares. A right triangle with legs 8 and 15 has hypotenuse √(64 + 225) = 17, not 23.
• Labeling from the wrong angle. The side opposite angle A is adjacent to angle B. If the question names angle B, relabel.
• Mixing up the 30°–60°–90° sides. The hypotenuse is twice the short leg, and the long leg is the short leg times √3. If the short leg is 5, the long leg is 5√3, not 10.
• Leaving the calculator in degree mode on a radian question. π/6 radians is 30°, but cos(π/6) typed in degree mode gives the wrong value.
• Forgetting the acute angles are complementary. In a right triangle, sin A = cos B. If sin A = 5/13, then cos B = 5/13 with no more work.

1. Question 1 · Easy

   An altitude of an equilateral triangle measures 38√3 centimeters. What is the length, in centimeters, of a side of the triangle?

   • A38
   • B38√3
   • C76
   • D76√3
   Show the answer

   C. An altitude bisects a side and forms two 30°–60°–90° triangles; since the longer leg is 38√3, half of an original side is 38, and the full side length is 2(38) = 76. Choice A reports half a side, choice B reports the height, and choice D doubles the height instead of the shorter leg.

2. Question 2 · Medium

   The length of a diagonal of a rectangle is sqrt{205} inches, and the length of one side is 6 inches. What is the perimeter, in inches, of the rectangle?

   • A19
   • B38
   • C78
   • D205
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   B. The other side has length sqrt{205 − 6²} = sqrt{169} = 13 inches, so the perimeter is 2(6 + 13) = 38 inches. Choice A is half the perimeter, choice C is the area, and choice D is the square of the diagonal.

3. Question 3 · Medium

   Angle a measures π/78 radians. The measure of angle b is 65 times the measure of angle a. What is the value of cos b?

   • A−√3/2
   • B−1/2
   • C1/2
   • D√3/2
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   A. Angle b measures 65π/78 = 5π/6 radians, which has reference angle π/6 and lies in the second quadrant, so cos b = −√3/2. The other choices use the complementary reference angle with a negative sign, report sin b, or ignore the negative sign of cosine in the second quadrant.

4. Question 4 · Medium

   In triangle DEF, angle E is a right angle. The length of side DE is 7sqrt{3}, and the length of side EF is 24sqrt{3}. What is the length of side DF?

   • A25
   • B17sqrt{3}
   • C25sqrt{3}
   • D31sqrt{3}
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   C. The hypotenuse is DF = sqrt{(7sqrt{3})² + (24sqrt{3})²} = sqrt{3(49 + 576)} = 25sqrt{3}. Choice A omits the common factor sqrt{3}, choice B subtracts the leg coefficients, and choice D adds the coefficients instead of their squares.

5. Question 5 · Hard

   The area of circle C is 1,156π square units. Rectangle JKLM has all four vertices on C. A diagonal of JKLM is twice as long as its shorter side. What is the perimeter, in units, of JKLM?

   • A68 + 68√3
   • B204
   • C68 + 68√5
   • D136 + 136√3
   Show the answer

   A. From πr² = 1,156π, the radius is 34; the rectangle’s diagonal is a diameter of length 68, so its shorter side is 34. The other side is √(68² − 34²) = 34√3, giving perimeter 2(34 + 34√3) = 68 + 68√3. Choice B uses the diagonal as the longer side, choice C adds the squares to get √(68² + 34²) = 34√5 for the other side, and choice D doubles both correctly found side lengths.

6. Question 6 · Hard

   In rectangle ABCD, diagonal BD is drawn. The sine of angle ABD is 56/65. What is the value of cos(∠BDC)?

   • A9/65
   • B33/65
   • C33/56
   • D56/65
   Show the answer

   B. Because AB is parallel to DC, angles ABD and BDC are alternate interior angles, so they are equal. For angle ABD, the opposite-to-hypotenuse ratio 56:65 gives an adjacent leg ratio of √(65² − 56²) = 33, so cos(∠BDC) = cos(∠ABD) = 33/65. The other choices subtract 56 from 65 instead of using the Pythagorean theorem, give the tangent instead of the cosine, or treat the two angles as complementary.

For a missing leg, type √(17² − 15²) with your own numbers: hypotenuse squared minus known leg squared. Before typing sine, cosine or tangent, open the wrench menu and set degrees or radians to match the question. If the choices are exact values like 5√3, type each one to compare its decimal with your answer.

Ashore's SAT course starts with a 15-minute check that finds the question types costing you the most points, then gives you lessons and practice for each.

• Linear Equations and Inequalities
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• Ratios, Rates, Percentages, and Units
• One- and Two-Variable Data
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