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Flashcards: Sine and Cosine Relationship of Complementary Angles: CCSS.Math.Content.HSG-SRT.C.7

True or False: The sine of any acute angle is equal to the cosine of its complement.

True

False

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Flashcards: Sine and Cosine Relationship of Complementary Angles: CCSS.Math.Content.HSG-SRT.C.7

True or False: The sine of any acute angle is equal to the cosine of its complement.

True

False

Report an Error Copyright Notice

Flashcards: Sine and Cosine Relationship of Complementary Angles: CCSS.Math.Content.HSG-SRT.C.7

True or False: The sine of any acute angle is equal to the cosine of its complement.

Your answer:

False

Correct answer:

True

Explanation:

Consider the triangle below

First we must understand the definition of complementary angles. Complementary angles are a pair of angles that add up to 90 degrees. Looking at the triangle above, we see that $\angle BAC$ $\displaystyle \angle BAC$and $\angle ACB$ $\displaystyle \angle ACB$ add up to 90, so they are complementary.

Recall that the definition of sine is $sin = \frac{opposite}{hypotenuse}$ $\displaystyle sin = \frac{opposite}{hypotenuse}$ and cosine is $cos = \frac{adjacent}{hypotenuse}$ $\displaystyle cos = \frac{adjacent}{hypotenuse}$.

Next, list out the sine and cosine of $\angle BAC$ $\displaystyle \angle BAC$ and $\angle ACB$ $\displaystyle \angle ACB$.

$sin(\angle BAC) = \frac{10.83}{11.39}$ $\displaystyle sin(\angle BAC) = \frac{10.83}{11.39}$

$sin(\angle ACB) = \frac{3.52}{11.39}$ $\displaystyle sin(\angle ACB) = \frac{3.52}{11.39}$

$cos(\angle BAC) = \frac{3.52}{11.39}$ $\displaystyle cos(\angle BAC) = \frac{3.52}{11.39}$

$cos(\angle ACB) = \frac{10.83}{11.39}$ $\displaystyle cos(\angle ACB) = \frac{10.83}{11.39}$

Notice that $sin(\angle BAC) = \frac{10.83}{11.39} = cos(\angle ACB)$ $\displaystyle sin(\angle BAC) = \frac{10.83}{11.39} = cos(\angle ACB)$

And $sin(\angle ACB) = \frac{3.52}{11.39} = cos(\angle BAC)$ $\displaystyle sin(\angle ACB) = \frac{3.52}{11.39} = cos(\angle BAC)$

We have just shown that the sine of the two angles is equal to the cosines of their complements. This is true for any acute angles.

Flashcards: Sine and Cosine Relationship of Complementary Angles: CCSS.Math.Content.HSG-SRT.C.7

You have two complementary angles, $\angle A$ $\displaystyle \angle A$ and $\angle B$ $\displaystyle \angle B$. You know that $\angle A = 30$ $\displaystyle \angle A = 30$. Using your sine and cosine relationships, what is $\angle B$ $\displaystyle \angle B$ in degrees?

$\angle B = 30$ $\displaystyle \angle B = 30$

$\angle B = 90$ $\displaystyle \angle B = 90$

$\angle B = 60$ $\displaystyle \angle B = 60$

$\angle B = 45$ $\displaystyle \angle B = 45$

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Flashcards: Sine and Cosine Relationship of Complementary Angles: CCSS.Math.Content.HSG-SRT.C.7

Your answer:

$\angle B = 30$ $\displaystyle \angle B = 30$

Your answer:

$\angle B = 90$ $\displaystyle \angle B = 90$

Your answer:

$\angle B = 45$ $\displaystyle \angle B = 45$

Correct answer:

$\angle B = 60$ $\displaystyle \angle B = 60$

Explanation:

You know that $\angle A = 30$ $\displaystyle \angle A = 30$. We use this information to set up our work in the following way:

$sin(\angle A) = cos(\angle B)$ $\displaystyle sin(\angle A) = cos(\angle B)$ $sin(\angle A) = cos(\angle B)$ $\displaystyle sin(\angle A) = cos(\angle B)$

$\implies sin(30) = cos(\angle B)$ $\displaystyle \implies sin(30) = cos(\angle B)$

$\implies \frac{1}{2} = cos(\angle B)$ $\displaystyle \implies \frac{1}{2} = cos(\angle B)$

$\implies cos^{-1}(\frac{1}{2}) = \angle B$ $\displaystyle \implies cos^{-1}(\frac{1}{2}) = \angle B$

$\implies 60 = \angle B$ $\displaystyle \implies 60 = \angle B$

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Flashcards: Sine and Cosine Relationship of Complementary Angles: CCSS.Math.Content.HSG-SRT.C.7

All Common Core: High School - Geometry Resources

Flashcards: Sine and Cosine Relationship of Complementary Angles: CCSS.Math.Content.HSG-SRT.C.7

Flashcards: Sine and Cosine Relationship of Complementary Angles: CCSS.Math.Content.HSG-SRT.C.7

Flashcards: Sine and Cosine Relationship of Complementary Angles: CCSS.Math.Content.HSG-SRT.C.7

Flashcards: Sine and Cosine Relationship of Complementary Angles: CCSS.Math.Content.HSG-SRT.C.7

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