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Flashcards: Using Sine, Cosine, & Tangent

The three angles of a right triangle have measures $a^{\circ}$ , $b^{\circ}$ , and $c^{\circ}$ . If c=90 and $\cos a^{\circ}=\frac{3}{5}$ , what is $\sin b^{\circ}$ ?

$\frac{4}{5}$

$\frac{5}{3}$

$\frac{3}{4}$

$\frac{3}{5}$

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Flashcards: Using Sine, Cosine, & Tangent

The three angles of a right triangle have measures $a^{\circ}$ , $b^{\circ}$ , and $c^{\circ}$ . If c=90 and $\cos a^{\circ}=\frac{3}{5}$ , what is $\sin b^{\circ}$ ?

Your answer:

$\frac{4}{5}$

Your answer:

$\frac{5}{3}$

Your answer:

$\frac{3}{4}$

Correct answer:

$\frac{3}{5}$

Explanation:

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:

$\sin x^{\circ}=\cos (90-x)^{\circ}$

$\cos x^{\circ}=\sin (90-x)^{\circ}$

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 $< 90^{\circ}$ .

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 c=90 , $a^{\circ}$ and $b^{\circ}$ have to be the two smaller angles of the right triangle. Therefore, based on the rules explained above, $\sin a^{\circ}=\cos b^{\circ}$ and $\cos a^{\circ}=\sin b^{\circ}$ .

Since $\cos a^{\circ}=\frac{3}{5}$ , this means that $\sin b^{\circ}=\frac{3}{5}$ also! Thus the correct answer choice is $\frac{3}{5}$ .

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!

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All SAT Mathematics Resources

Flashcards: Using Sine, Cosine, & Tangent

Flashcards: Using Sine, Cosine, & Tangent

Flashcards: Using Sine, Cosine, & Tangent

Flashcards: Using Sine, Cosine, & Tangent

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All SAT Mathematics Resources

Flashcards: Using Sine, Cosine, & Tangent

Flashcards: Using Sine, Cosine, & Tangent

Flashcards: Using Sine, Cosine, & Tangent

Flashcards: Using Sine, Cosine, & Tangent

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DMCA Complaint

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