Right Triangle Trigonometry

The answer is $\frac{7}{25}$. As always, draw the triangle. If you remember that the hypotenuse (longest side) of the triangle must be across from the right angle, that will help you draw the figure more quickly. Alternatively, you can note that side $\overline{DE}$ must be across from angle $F$, and “across” is precisely what we are seeking to find sine (opposite/hypotenuse). Whether you reason this way or use the diagram, the opposite is $14$ and the hypotenuse must be $50$. $14/50$ reduces to $7/25$ when you divide by two.

One final note: it’s always worth noting Pythagorean triples when they appear, and we have one here: $7:24:25$ is one of the less common triples, but a good one to know nevertheless!

The answer is $\frac{\sqrt{3}}{3}$. The key to this question is recognizing the $30-60-90$ special right triangle implied by the angles given. Angle $C$ must be the right angle, so we know that angle $B$ is the smallest angle. This means (as shown in our lesson on triangles) that it must be across from the smallest side. As shown in the SAT reference list (though ideally also solidly memorized by you!), the smallest side is represented by the $1$ in the $1:\sqrt{3}:2$ ratio. This means that the ratio of the opposite in this case (smallest side) to the adjacent (middle side) is $1:\sqrt{3}$. Since opposite/adjacent is what we need for tangent, we are close to our answer.

What remains to be done? The rules of algebra tell us that a radical (in this case, a square root) in the denominator is not considered simplified, so we need to rationalize the denominator by multiplying both top and bottom of the fraction by that denominator. That gives us the following:

$\frac{1}{\sqrt{3}}$ x $\frac{\sqrt{3}}{\sqrt{3}}$ = $\frac{\sqrt{3}}{3}$

The answer is $\frac{120}{169}$. This problem is simpler than it might first appear. Remember that the cosine (as well as the sine and the tangent) of an angle is always the same if the angle is the same measure. For example, the sine of $60°$ is always $1/2$, even if we’re talking about two different $60°$ angles in two different triangles. So if two angles correspond, they must have all the same trigonometric ratios (sine, cosine, and tangent), because they have, by definition, the same measure. This means that the only question we need to answer here is, Do angles $C$ and $F$ correspond? According to the definition given at the beginning of the problem and the SAT’s consistent use of alphabetical order to denote corresponding angles, we know that they do. So the cosine of one angle must be the same as the others.

It’s worth briefly addressing the wrong answers here. The answer of $\frac{119}{169}$ could be a trap if you do the hard work of the Pythagorean theorem on your calculator. There is actually a very large Pythagorean triple at play here: the values $119$, $120$, and $169$ work in the Pythagorean theorem. So $\frac{119}{169}$ is a valid ratio for this triangle, but it is the sine, not the cosine.

The two answers in which the numerator is larger than the denominator can be ruled out based on the general rule that sine and cosine can never be greater than $1$. This is because the leg of a triangle can never be larger than the hypotenuse. Of the three major trig ratios, only tangent can be greater than one, so these two answer choices are only possible if we are looking for tangent.

The answer is $\frac{20}{29}$. The key to understanding this problem is realizing that the two values in the sine ratio give you, effectively, two of the three parts needed for the Pythagorean theorem, and you can solve for them. Remember: trig ratios are ratios only; we do NOT know that two sides of the triangle are exactly $21$ and $29$. But because we know they exist in a ratio of $21/29$, we can “pretend” those are the actual values as we seek to calculate a different trig ratio in the same triangle.

If one of the legs, therefore, is $21$ for our purposes, and the hypotenuse is $29$ (remember that $\sin =\frac{opp}{hyp}$, so we know $29$ is the hypotenuse), we can use the Pythagorean theorem ($a^2+b^2=c^2$, given in the SAT reference list) to solve for the third side. Doing so reveals that we actually have an unusual Pythagorean triple here: one in the ratio of $20:21:29$.

How do we know the value of $\sin S$? If the answer doesn’t present itself to you, take a moment to sketch a right triangle in which the right angle is angle $R$. The sine of $21/29$ already given means that side $\overline{RS}$ must be $21$, while we know that hypotenuse $\overline{PS}$ is $29$. That leaves side $\overline{PR}$ which, by process of elimination, must be $20$. So, from the point of view of angle $S$, $\overline{PR}$ is the opposite. This means that $\sin S=\frac{20}{29}$.

The answer is $\frac{\sqrt{39}}{8}$. Even the hardest right triangle trig questions on the SAT can sometimes be solved simply if you know the important trig identity. If you weren’t sure what identity to apply in this case, go back and review the complementary relationship of sine and cosine described under Flashcard Fodder. It makes things surprisingly simple! Whenever the two acute angles in a right triangle, we know that the sine of one equals the cosine of the other, and vice versa. The right answer here must be identical to the sine already given!

The wrong answer choices largely come from other trigonometric ratios in this particular triangle. If you were to use the Pythagorean theorem to discover that the third side of this triangle has a length of $5$, you would see that $\frac{5}{\sqrt{39}}$ is the tangent of angle H. $\frac{5}{8}$, meanwhile, it the sine of H. $\frac{\sqrt{39}}{5}$ is the cotangent, in case you’re curious, but cotangent rarely to never pops up on the SAT. In any case, you don’t need to worry about any of these wrong answers if you simply apply the trig identity concerning the complementary relationship of sine and cosine.