Using Sine, Cosine, & Tangent - SAT Math

Because these triangles are similar, they will have the exact same ratio between their sides, meaning that the sine of Z will be the same as the sine of C. And since sine = opposite/hypotenuse, you can calculate that as , which reduces to .

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles .

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question, since , and have to be the two smaller angles of the right triangle. Therefore, based on the rules explained above, and .

Since , this means that also! Thus the correct answer choice is .

This is a classic trick question because the test question writers know that most students will not be expecting the correct answer to be exactly the same as the other value given in the question! So students who don’t know the rules or who aren’t 100% confident in their knowledge will be afraid to choose that answer and will be likely to guess a different answer. This is the challenge that the SAT question writers present to students. You must know these rules very well in order to successfully pass this challenge and choose the correct answer!

First, you have to sort out the corresponding angles in the similar triangles. Clearly since , and are the right angles. Thus we know that within each triangle, and . The key extra piece of information in the question about the angles is that . This means that and . Now we know exactly which angles in the similar triangles are congruent (equal) to each other.

Now the key step of the question is to use the value to find the value of . Since we now know that , we know that . Now we can focus only on the triangle with angles , , and .

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles .

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question, since , and have to be the two smaller angles of the right triangle. Therefore, based on the rules explained above, and .

Since , this means that also! Thus the correct answer choice is 0.8.

This is a classic trick question because the test question writers know that most students will not be expecting the correct answer to be exactly the same as the other value given in the question! So students who don’t know the rules or who aren’t 100% confident in their knowledge will be afraid to choose that answer and will be likely to guess a different answer. This is the challenge that the SAT question writers present to students. You must know these rules very well in order to successfully pass this challenge and choose the correct answer!

First of all, watch out for the trap answer choice “The answer cannot be determined from the information given in the question”!! NEVER GUESS THIS ANSWER CHOICE if you are not completely confident that it MUST be true! Most often the SAT uses this answer choice as a trap for students who don’t know the advanced method that does exist and is necessary to find the correct answer, which can be determined from the information given in the question if you know the correct method. If you don’t understand how to answer the question and you have to guess, guess one of the other three answer choices!

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles .

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question, and have to be the two smaller angles of the right triangle. Therefore, based on the rules explained above, and .

Therefore, . Thus the correct answer choice is 0.

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles .

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question, we have to think about the rules explained above in another way: Since the rules tell us that , and the question tells us that , we can draw the logical conclusion that . \[The question also adds the information that both angles are acute, which means , so we know we are dealing with “normal” angle measures between and , and not some more complicated angle measures greater than for example.\] Now we can rearrange in the simpler form .

Now we can simply substitute in the values and into the equation . Continuing by adding 25 to both sides, we get . Dividing both sides by 10, we get . Therefore, the correct answer choice is 11.5.

If you mistakenly think you would get the wrong answer choice 20.5. If you mistakenly think you would get the wrong answer choice 38.5. This large value 38.5 is included among the answer choices mainly in order to make the more tempting wrong answer choice 20.5 appear to look more “reasonable” by comparison, to trap more students into guessing 20.5 incorrectly.

The SAT likes to pack many different concepts into a single question, to force you to use all of your knowledge of math to answer the question. Here you have to begin with the information that DE is perpendicular to AB, to establish that is a right angle and so triangle ADE is a right triangle. Further, since DE and BC are parallel, that means must also be a right angle, and triangle ABC is a right triangle.

Since triangles ADE and ABC also share the same angle A, their third angles and must also be equal, because the sum of the three angles of each triangle must add up to . Now that you know all three angles of triangles ADE and ABC are the same, this means they are similar triangles.

The next key piece of information in this question is . Knowing SOH-CAH-TOA, you know that . In these triangles, the adjacent sides to A are AD in the small triangle and AB in the large triangle. The hypotenuses are AE in the small triangle and AC in the large triangle. Therefore you know that .

Now you should recognize 12 and 13 as values in the “Pythagorean triple” of the 5-12-13 right triangle! Recall that these are the lengths of the sides of a right triangle because . However, be careful! The cosine value only tells you the ratio of the side lengths, not the actual side length measures themselves. They could be 12 and 13, 24 and 26, 120 and 130, 1.2 and 1.3, 0.12 and 0.13, or any other pair of lengths with a ratio of .

The information in the question that helps us determine the actual side lengths. Since BC must be the shortest side of a 5-12-13 ratio right triangle, you can see that the actual side lengths of triangle ABC must be two times 5, 12, and 13: Thus, and . Further, the question tells you that . Therefore, subtracting the length of EC from the length of AC, you know that . This means that and . Since and , this means that .

Now you finally have all the side lengths of the trapezoid DBCE, and you are ready to calculate its area. You need to know the formula for the area of a trapezoid: . This looks complicated, but the logical way to understand it and remember it is to realize that is just the average of the two bases of the trapezoid. Always keep in mind that the bases are the two sides that are parallel to each other, regardless of which direction they are oriented in the diagram. Here the two bases are DE and BC. Their lengths are 5 and 10, so their average is 7.5. The height of a trapezoid is the altitude from one base to the other that is perpendicular to both bases. In this case, since and are right angles DB is perpendicular to DE and BC, and so DB itself is the height of the trapezoid. Thus the height is 12, and the area of the trapezoid is , so the correct answer choice is 90.

Although the SAT can sometimes present geometry questions like this one without giving you a diagram, often you should draw a diagram yourself to visualize what is going on in the question!

Here the question tells you that LM is perpendicular to MN. Perpendicular lines meet at a right angle, so this is a fancy hidden way to tell you that is a right angle. Now you know that LMN is a right triangle, so you should draw your own diagram of it:

Next, the question tells you that . Knowing SOH-CAH-TOA, you know that . The opposite side from L is MN, and the adjacent side to L is LM, so now you know that . Technically we do not know the actual side lengths, only the ratio of one side length to the other: They could be 4 and 3, or 8 and 6, or 400 and 300, or 0.8 and 0.6, or any two lengths with a ratio of . But in this particular question that does not matter, since the question is asking you for the value of , which is also a ratio of side lengths just like tan L is. So in this question, you can just keep it simple and assume the simplest case of and .

You know the 3-4-5 right triangle, so now you can assume .

Again from SOH-CAH-TOA, you know that . Make sure that you switch your perspective now from angle L to angle N! The question asks you for the value of . From the perspective of angle N, the adjacent side is MN and the hypotenuse is LN. So . Therefore the correct answer choice is .

First of all, watch out for the trap answer choice “The answer cannot be determined from the information given in the question”!! NEVER GUESS THIS ANSWER CHOICE if you are not completely confident that it MUST be true! Most often the SAT uses this answer choice as a trap for students who don’t know the advanced method that does exist and is necessary to find the correct answer, which can be determined from the information given in the question if you know the correct method. If you don’t understand how to answer the question and you have to guess, guess one of the other three answer choices!

In this question, the SAT is trying to confuse you with the extra numbers and algebra and parentheses in the first equation they give you. Don’t be fooled or confused! There is no trick to the parentheses: You can just add all the terms together and get . Now just subtract 10 from both sides and you get .

Now you can see the connection to a question about tangent values: Since , and can be the two smaller angles of a right triangle!

Pro Tip: At this point the most experienced math students will already see the answer instantly: Since you know from SOH-CAH-TOA that , and switching from one smaller angle of a right triangle to the other smaller angle simply swaps the angles’ opposite sides and adjacent sides, the tangent of one angle is always the reciprocal of the tangent of the other angle! So, since , the most experienced math students know right away that , which is the correct answer choice.

If you don’t see this right away, the other way to finish solving the question is to draw your own diagram:

Now you can see that , so you can assume the simplest case that side and side . (Keep in mind that they can actually be any side lengths that have a ratio of 2:3.) Now switching to look at the triangle from the perspective of angle , you see that now the opposite side is and the adjacent side is . So , which is the correct answer choice.

The question tells you that the tangent of angle a is 1. We’ll want to keep in mind that a tangent is a ratio: it’s the length of the opposite side divided by the length of the adjacent side. When does a ratio equal 1? When both figures are the same. So, side XY = side YZ, meaning that you’re dealing with an isosceles right triangle (and that’s where the “not drawn to scale” notation really comes in handy since you wouldn’t have thought that to be true from the picture). Since it’s an isosceles right triangle, you know the side ratio is: . At this point, even though you don’t now know the actual lengths, all you need to do is apply that ratio. The sine of b is the opposite over the hypotenuse, which is a ratio of . From here you have to “rationalize the denominator” by multiplying by 1 (in this case, ). Once simplified, this leaves you with .

In this example, we’re being tasked to apply our understanding of the ratio-driven relationships we express with SOH-CAH-TOA, our sin, cosine, and tangent. Keep in mind that these ratios are as follows:

From this, if we’re told that the tangent (the ) of is , then the tangent of must be the reciprocal since the opposite of a° is the adjacent of b°, and vice versa. In many cases, recognizing the way SOH-CAH-TOA manipulates the same relationships and sides of a triangle in SAT questions can simplify the extent to which we really end up needing to “do the math.” Here, by recognizing that the tangent of is simply the reciprocal of the tangent of , this question becomes extremely efficient, and allows us to save time for more step-by-step questions elsewhere on the exam.

Because these triangles are similar, they will have the exact same ratio between their sides, meaning that the sine of Z will be the same as the sine of C. And since sine = opposite/hypotenuse, you can calculate that as , which reduces to .

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles .

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question, since , and have to be the two smaller angles of the right triangle. Therefore, based on the rules explained above, and .

Since , this means that also! Thus the correct answer choice is .

This is a classic trick question because the test question writers know that most students will not be expecting the correct answer to be exactly the same as the other value given in the question! So students who don’t know the rules or who aren’t 100% confident in their knowledge will be afraid to choose that answer and will be likely to guess a different answer. This is the challenge that the SAT question writers present to students. You must know these rules very well in order to successfully pass this challenge and choose the correct answer!

First, you have to sort out the corresponding angles in the similar triangles. Clearly since , and are the right angles. Thus we know that within each triangle, and . The key extra piece of information in the question about the angles is that . This means that and . Now we know exactly which angles in the similar triangles are congruent (equal) to each other.

Now the key step of the question is to use the value to find the value of . Since we now know that , we know that . Now we can focus only on the triangle with angles , , and .

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles .

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question, since , and have to be the two smaller angles of the right triangle. Therefore, based on the rules explained above, and .

Since , this means that also! Thus the correct answer choice is 0.8.

This is a classic trick question because the test question writers know that most students will not be expecting the correct answer to be exactly the same as the other value given in the question! So students who don’t know the rules or who aren’t 100% confident in their knowledge will be afraid to choose that answer and will be likely to guess a different answer. This is the challenge that the SAT question writers present to students. You must know these rules very well in order to successfully pass this challenge and choose the correct answer!

First of all, watch out for the trap answer choice “The answer cannot be determined from the information given in the question”!! NEVER GUESS THIS ANSWER CHOICE if you are not completely confident that it MUST be true! Most often the SAT uses this answer choice as a trap for students who don’t know the advanced method that does exist and is necessary to find the correct answer, which can be determined from the information given in the question if you know the correct method. If you don’t understand how to answer the question and you have to guess, guess one of the other three answer choices!

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles .

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question, and have to be the two smaller angles of the right triangle. Therefore, based on the rules explained above, and .

Therefore, . Thus the correct answer choice is 0.

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles .

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question, we have to think about the rules explained above in another way: Since the rules tell us that , and the question tells us that , we can draw the logical conclusion that . \[The question also adds the information that both angles are acute, which means , so we know we are dealing with “normal” angle measures between and , and not some more complicated angle measures greater than for example.\] Now we can rearrange in the simpler form .

Now we can simply substitute in the values and into the equation . Continuing by adding 25 to both sides, we get . Dividing both sides by 10, we get . Therefore, the correct answer choice is 11.5.

If you mistakenly think you would get the wrong answer choice 20.5. If you mistakenly think you would get the wrong answer choice 38.5. This large value 38.5 is included among the answer choices mainly in order to make the more tempting wrong answer choice 20.5 appear to look more “reasonable” by comparison, to trap more students into guessing 20.5 incorrectly.

The SAT likes to pack many different concepts into a single question, to force you to use all of your knowledge of math to answer the question. Here you have to begin with the information that DE is perpendicular to AB, to establish that is a right angle and so triangle ADE is a right triangle. Further, since DE and BC are parallel, that means must also be a right angle, and triangle ABC is a right triangle.

Since triangles ADE and ABC also share the same angle A, their third angles and must also be equal, because the sum of the three angles of each triangle must add up to . Now that you know all three angles of triangles ADE and ABC are the same, this means they are similar triangles.

The next key piece of information in this question is . Knowing SOH-CAH-TOA, you know that . In these triangles, the adjacent sides to A are AD in the small triangle and AB in the large triangle. The hypotenuses are AE in the small triangle and AC in the large triangle. Therefore you know that .

Now you should recognize 12 and 13 as values in the “Pythagorean triple” of the 5-12-13 right triangle! Recall that these are the lengths of the sides of a right triangle because . However, be careful! The cosine value only tells you the ratio of the side lengths, not the actual side length measures themselves. They could be 12 and 13, 24 and 26, 120 and 130, 1.2 and 1.3, 0.12 and 0.13, or any other pair of lengths with a ratio of .

The information in the question that helps us determine the actual side lengths. Since BC must be the shortest side of a 5-12-13 ratio right triangle, you can see that the actual side lengths of triangle ABC must be two times 5, 12, and 13: Thus, and . Further, the question tells you that . Therefore, subtracting the length of EC from the length of AC, you know that . This means that and . Since and , this means that .

Now you finally have all the side lengths of the trapezoid DBCE, and you are ready to calculate its area. You need to know the formula for the area of a trapezoid: . This looks complicated, but the logical way to understand it and remember it is to realize that is just the average of the two bases of the trapezoid. Always keep in mind that the bases are the two sides that are parallel to each other, regardless of which direction they are oriented in the diagram. Here the two bases are DE and BC. Their lengths are 5 and 10, so their average is 7.5. The height of a trapezoid is the altitude from one base to the other that is perpendicular to both bases. In this case, since and are right angles DB is perpendicular to DE and BC, and so DB itself is the height of the trapezoid. Thus the height is 12, and the area of the trapezoid is , so the correct answer choice is 90.

Although the SAT can sometimes present geometry questions like this one without giving you a diagram, often you should draw a diagram yourself to visualize what is going on in the question!

Here the question tells you that LM is perpendicular to MN. Perpendicular lines meet at a right angle, so this is a fancy hidden way to tell you that is a right angle. Now you know that LMN is a right triangle, so you should draw your own diagram of it:

Next, the question tells you that . Knowing SOH-CAH-TOA, you know that . The opposite side from L is MN, and the adjacent side to L is LM, so now you know that . Technically we do not know the actual side lengths, only the ratio of one side length to the other: They could be 4 and 3, or 8 and 6, or 400 and 300, or 0.8 and 0.6, or any two lengths with a ratio of . But in this particular question that does not matter, since the question is asking you for the value of , which is also a ratio of side lengths just like tan L is. So in this question, you can just keep it simple and assume the simplest case of and .

You know the 3-4-5 right triangle, so now you can assume .

Again from SOH-CAH-TOA, you know that . Make sure that you switch your perspective now from angle L to angle N! The question asks you for the value of . From the perspective of angle N, the adjacent side is MN and the hypotenuse is LN. So . Therefore the correct answer choice is .

First of all, watch out for the trap answer choice “The answer cannot be determined from the information given in the question”!! NEVER GUESS THIS ANSWER CHOICE if you are not completely confident that it MUST be true! Most often the SAT uses this answer choice as a trap for students who don’t know the advanced method that does exist and is necessary to find the correct answer, which can be determined from the information given in the question if you know the correct method. If you don’t understand how to answer the question and you have to guess, guess one of the other three answer choices!

In this question, the SAT is trying to confuse you with the extra numbers and algebra and parentheses in the first equation they give you. Don’t be fooled or confused! There is no trick to the parentheses: You can just add all the terms together and get . Now just subtract 10 from both sides and you get .

Now you can see the connection to a question about tangent values: Since , and can be the two smaller angles of a right triangle!

Pro Tip: At this point the most experienced math students will already see the answer instantly: Since you know from SOH-CAH-TOA that , and switching from one smaller angle of a right triangle to the other smaller angle simply swaps the angles’ opposite sides and adjacent sides, the tangent of one angle is always the reciprocal of the tangent of the other angle! So, since , the most experienced math students know right away that , which is the correct answer choice.

If you don’t see this right away, the other way to finish solving the question is to draw your own diagram:

Now you can see that , so you can assume the simplest case that side and side . (Keep in mind that they can actually be any side lengths that have a ratio of 2:3.) Now switching to look at the triangle from the perspective of angle , you see that now the opposite side is and the adjacent side is . So , which is the correct answer choice.

The question tells you that the tangent of angle a is 1. We’ll want to keep in mind that a tangent is a ratio: it’s the length of the opposite side divided by the length of the adjacent side. When does a ratio equal 1? When both figures are the same. So, side XY = side YZ, meaning that you’re dealing with an isosceles right triangle (and that’s where the “not drawn to scale” notation really comes in handy since you wouldn’t have thought that to be true from the picture). Since it’s an isosceles right triangle, you know the side ratio is: . At this point, even though you don’t now know the actual lengths, all you need to do is apply that ratio. The sine of b is the opposite over the hypotenuse, which is a ratio of . From here you have to “rationalize the denominator” by multiplying by 1 (in this case, ). Once simplified, this leaves you with .

In this example, we’re being tasked to apply our understanding of the ratio-driven relationships we express with SOH-CAH-TOA, our sin, cosine, and tangent. Keep in mind that these ratios are as follows:

From this, if we’re told that the tangent (the ) of is , then the tangent of must be the reciprocal since the opposite of a° is the adjacent of b°, and vice versa. In many cases, recognizing the way SOH-CAH-TOA manipulates the same relationships and sides of a triangle in SAT questions can simplify the extent to which we really end up needing to “do the math.” Here, by recognizing that the tangent of is simply the reciprocal of the tangent of , this question becomes extremely efficient, and allows us to save time for more step-by-step questions elsewhere on the exam.