Geometry - Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

Which of the following expressions properly states the Law of Sines? Use the following triangle for reference.

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$\frac{A}{cos(\gamma)} = \frac{B}{cos(\alpha)} = \frac{C}{cos(\beta)}$

$\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$

$A^2 + B^2 -2ABcos(\gamma) = C^2$

$A^2 + B^2 - 2ABsin(\gamma) = C^2$

Explanation

The Law of Sines uses ratios of a triangle’s sides and angles to allow us to be able to solve for unknown sides and/or angles. This relationship is true for any triangle.

Using the Law of Sines, solve for and . Round to the second decimal place.

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Explanation

We begin by labeling our sides and angles according the variables in the equation for the Law of Sines: $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$

We will let

$\alpha = 75$

$\beta = 65$

$\gamma = 40$

We will begin by solving for . From our equation $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$ , we will leave out the term to solve for since it is not needed at this time.

$\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$

$\implies \frac{6}{sin(40)} = \frac{x}{sin(75)}$

$\implies (6)sin(75) = (x)sin(40)$ (by cross-multiplication)

$\implies (6)(0.9659) = (x)(0.6428)$

$\implies 5.7954 = (x)(0.6428)$

$\implies 9.02 = x$

Now we will solve for . We could use $\frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$ for our relationship, but we can also use $\frac{A}{sin(\gamma)} = \frac{C}{sin(\beta)}$ .

$\frac{A}{sin(\gamma)} = \frac{C}{sin(\beta)}$

$\implies \frac{6}{sin(40)} = \frac{y}{sin(65)}$

$\implies (6)sin(65) = (y)sin(40)$ (by cross-multiplication)

$\implies 5.4378 = (y)(0.6428)$

$\implies 8.46 = y$

In a triangle where the side opposite a $45 ^{\circ}$ has length 13 find the side opposite a $65 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Explanation

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 45 for , 13 for and 65 for .
Now our equation becomes

$\left(\frac{\sin( 45 )}{ 13 }\right)= \left(\frac{\sin( 65 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 13 \sin( 65 )}{\sin( 45 )}\right) \\b = -15.9863244465377$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Using the Law of Cosines, solve for . Round to the second decimal place.

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Explanation

We will begin by defining the variables in our equation $A^2 + B^2 -2ABcos(\gamma) = C^2$ .

We will let

$\gamma = x$

Next we plug our values into our formulas and solve for .

$A^2 + B^2 -2ABcos(\gamma) = C^2$

$\implies 16^2 + 12^2 - 2(16)(12)cos(x) = 8^2$

$\implies 256 + 144 - 384cos(x) = 64$

$\implies -384cos(x) = -336$

$\implies cos(x) = 0.875$

$\implies x = cos^{-1}(0.875)$

$\implies x = 28.96$

Use the Law of Sines to find the missing side lengths and angles. Round to the second decimal place.

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$C = 9.17, \alpha = 70, B = 15.63$

$C = 12.13, \alpha = 45, B = 23.45$

$C = 10.15, \alpha = 70, B = 19.08$

$C = 19.08, \alpha = 90, B = 10.15$

Explanation

We know for the Law of Sines our equation is $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac {C}{sin(\beta)}$ . We begin by assigning our values on the figure to variables in the equation.

We will let:

$\gamma = 80$

$\beta = 30$

We will begin by solving for side .

$\frac{A}{sin(\gamma)} = \frac{C}{sin(\beta)}$

$\implies \frac{20}{sin(80)} = \frac{C}{sin(30)}$

$\implies (20)sin(30) = Csin(80)$

$\implies (20)(0.5) = C(0.9848)$

$\implies 10.15 = C$

Using the Law of Sines we need to solve for $\alpha$ and using the relation $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$ . We do not know either or $\alpha$ . We are able to solve for $\alpha$ because we know that the angles of a triangle must add up to 180.

$30 + 80 + \alpha = 180$

Now we have enough information to solve for using the Law of Sines

$\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$

$\implies \frac{20}{sin(80)} = \frac{B}{sin(70)}$

$\implies 20sin(70) = Bsin(80)$

$\implies (20)(0.9397) = B(0.9848)$

$\implies 18.7939 = B(0.9848)$

$\implies 19.08 = B$

In a triangle where the side opposite a $11 ^{\circ}$ has length 5 find the side opposite a $64 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Explanation

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 11 for , 5 for and 64 for .
Now our equation becomes

$\left(\frac{\sin( 11 )}{ 5 }\right)= \left(\frac{\sin( 64 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 5 \sin( 64 )}{\sin( 11 )}\right) \\b = -3.41136278039312$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

True or False: You are looking at an obtuse triangle. You are given two side lengths and their adjoining angle. You are able to figure out the third side length using the Law of Cosines.

True

False

Explanation

Just looking at the formula we can see that this is similar to the Pythagorean Theorem. We commonly use the Pythagorean Theorem to solve for side , which is what this problem is asking us to do. Consider the picture below. We are able to solve for the third side using the information from the picture. This is the same information given in the question.

True or False: We are given the length of one side and one angle of a triangle. This is enough information to use the Law of Sines to solve for the other sides and angles.

True

False

Explanation

Consider the following relationship: $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$ . Let’s say we know side and $\alpha = 60$ . Plugging this into our relationship we have $\frac{9}{sin(\gamma)} = \frac{B}{sin(60)}$ . We still have two unknowns and no other information to solve for these. This is not enough information to use the Law of Sines to solve for the unknown sides and angles.

Use the Law of Cosine to find the missing side. Round to the second decimal place.

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Explanation

We are able to use our equation from the Law of Cosine . We will start by assigning the values in our figure to the variables in the formula.

We will let:

Now we can plug these values into our formula.

$\implies 3^2 + 4^2 - (2)(3)(4)cos(100) = C^2$

$\implies 9 + 16 - 24cos(100) = C^2$

$\implies 25 - (24)(-0.1736) = C^2$ $\implies 25 - (24)(-0.1736) = C^2$

$\implies 25 - (-4.1676) = C^2$

$\implies 29.1676 = C^2$

$\implies 5.40 = C$

In a triangle where the side opposite a $41 ^{\circ}$ has length 6 find the side opposite a $39 ^{\circ}$ angle. Round you answer to the nearest hundredth.